MCA NIMCET Mathematics Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Mathematics PYQ

Qus : 1 MCA NIMCET PYQ

$$ \frac{d^{2}x}{dy^{2}} \; = \; ? $$

$ (a) -\left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1}\left(\dfrac{dy}{dx}\right)^{-3} $

$ (b) \left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-2} $

$ (c) -\left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-3} $

$ (d) \left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1} $

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MCA NIMCET Mathematics PYQ MCA NIMCET Differentiation (Disha JEE) PYQ

Solution

Qus : 2 MCA NIMCET PYQ

If $\lim_{x\to\infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2x}=e^2$ then the values of a and b are:

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MCA NIMCET Mathematics PYQ MCA NIMCET Limit Continuity Differentiability (Disha JEE) PYQ

Solution

Qus : 3 MCA NIMCET PYQ

The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 4 MCA NIMCET PYQ

If $\vec{a}=4\hat{j}$ and $\vec{b}=3\hat{j}+4\hat{k}$ , then the vector form of the component of $\vec{a}$ alond $\vec{b}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 5 MCA NIMCET PYQ

If $\vec{a}=\hat{i}-\hat{k}$, $\vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$, then $\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}}\end{bmatrix}$ depends on

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Given:

$ \vec{a} = \hat{i} - \hat{k}, \quad $ $\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k},$ $ \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} $

Form the matrix:

$ M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} $

Find the determinant:

$ \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 $

Since the determinant is constant and non-zero, the vectors are linearly independent.

$ \boxed{\text{The matrix does not depend on } x \text{ or } y} $

Qus : 6 MCA NIMCET PYQ

Angle of elevation of the top of the tower from 3 points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60°, respectively. The ratio of AB and BC is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

According to the given information, the figure should be as follows.

Let the height of tower = h

Qus : 7 MCA NIMCET PYQ

If $\vec{a}$ and $\vec{b}$ in space, given by $\vec{a}=\frac{\hat{i}-2\hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt{14}}$ , then the value of $(2\vec{a}+\vec{b}).[(\vec{a} \times \vec{b}) \times (\vec{a}-2\vec{b})]$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

If $\vec{a}, \vec{b}$ are unit vectors such that $2\vec{a}+\vec{b} =3$ then which of the following statement is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Quick Solution

Given: $ \vec{a}, \vec{b} $ are unit vectors and

$ 2\vec{a} + \vec{b} = 3 $

Take magnitude on both sides:

$ |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 $

Use identity:

\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

Set equal to 9:

\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]

Qus : 9 MCA NIMCET PYQ

The area enclosed between the curves y² = x and y = |x| is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Solving y² = x and y = x,

we get, y = 0, x = 0, y = 1, x = 1 Therefore,

Qus : 10 MCA NIMCET PYQ

$\int f(x)\mathrm{d}x=g(x)$, then $\int {x}^5f({x}^3)\mathrm{d}x$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Quick Solution

Given:

$ \int f(x)\, dx = g(x) $

Required: $ \int x^5 f(x^3)\, dx $

Use substitution:

Let $ u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} $

Now rewrite the integral:

\[ \int x^5 f(x^3)\, dx = \int x^5 f(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int x^3 f(u)\, du \]

But $ x^3 = u $, so:

\[ \frac{1}{3} \int u f(u)\, du \]

Now integrate by parts or use the identity:

\[ \int u f(u)\, du = u g(u) - \int g(u)\, du \]

Final answer:

\[ \int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right] = x^3 g(x^3) - \int x^2 g(x^3)\, dx \]

\[ \boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx } \]

Qus : 11 MCA NIMCET PYQ

Test the continuity of the function at x = 2

$f(x)= \begin{cases} \frac{5}{2}-x & \text{ if } x<2 \\ 1 & \text{ if } x=2 \\ x-\frac{3}{2}& \text{ if } x>2 \end{cases}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

LHL ≠ f(2)

Qus : 12 MCA NIMCET PYQ

$\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$, then find k

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

Qus : 13 MCA NIMCET PYQ

The value of

2tan^-1[cosec(tan^-1x) - tan(cot^-1x)]

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

The graph of function $f(x)=\log _e({x}^3+\sqrt[]{{x}^6+1})$ is symmetric about:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 15 MCA NIMCET PYQ

If $3 sin x + 4 cos x = 5$, then $6tan\frac{x}{2}-9tan^2\frac{x}{2}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

If the equation $|x^2 – 6x + 8| = a$ has four real solution then find the value of $a$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 17 MCA NIMCET PYQ

If A is a subset of B and B is a subset of C, then cardinality of A ∪ B ∪ C is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

If A ⊆ B and B ⊆ C, then find the cardinality of A ∪ B ∪ C.

Explanation

Since A is a subset of B, everything in A is already in B.
Since B is a subset of C, everything in B (and hence A) is already in C.
Therefore, A ∪ B ∪ C = C.

Solution

Probability — No Black Ball is Chosen

Given:

Yellow balls = 5
Black balls = 4
Green balls = 3
Total balls = 5 + 4 + 3 = 12

We are to find:

Probability that no black ball is selected when 3 balls are drawn at random.

Step 1: Total number of ways to choose any 3 balls from 12:

\[ \text{Total ways} = \binom{12}{3} = 220 \]

Step 2: Ways to choose 3 balls such that no black ball is chosen:

Only yellow and green balls are allowed ⇒ Total = 5 (yellow) + 3 (green) = 8
\[ \text{Favorable ways} = \binom{8}{3} = 56 \]

Step 3: Probability

\[ P(\text{no black ball}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{56}{220} = \frac{14}{55} \]

\[ \boxed{\text{Probability} = \frac{14}{55}} \]

Qus : 22 MCA NIMCET PYQ

Between any two real roots of the equation $e^x sin x = 1$, the equation $e^x cos x = –1$ has

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Number of Roots

Given:

$ e^x \sin x = 1 $ has two real roots → say $ x_1 $ and $ x_2 $

Apply Rolle’s Theorem:

Since $ f(x) = e^x \sin x $ is continuous and differentiable, and $ f(x_1) = f(x_2) $, ⇒ There exists $ c \in (x_1, x_2) $ such that $ f'(c) = 0 $

Compute:

\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]

Qus : 23 MCA NIMCET PYQ

If f(x) is a polynomial of degree 4, f(n) = n + 1 & f(0) = 25, then find f(5) = ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Correct Shortcut Method — Find $ f(5) $

Step 1: Define a helper polynomial:

\[ g(x) = f(x) - (x + 1) \]

Given: $ f(1) = 2, f(2) = 3, f(3) = 4, f(4) = 5 \Rightarrow g(1) = g(2) = g(3) = g(4) = 0 $

So, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]

Step 2: Use $ f(0) = 25 $ to find A:

\[ f(0) = A(-1)(-2)(-3)(-4) + (0 + 1) = 24A + 1 = 25 \Rightarrow A = 1 \]

Step 3: Compute $ f(5) $:

\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]

✅ Final Answer: $ \boxed{f(5) = 30} $

Qus : 24 MCA NIMCET PYQ

The maximum value of $f(x) = (x – 1)^2 (x + 1)^3$ is equal to $\frac{2^p3^q}{3125}$ then the ordered pair of (p, q) will be

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Maximum Value of $ f(x) = (x - 1)^2(x + 1)^3 $

Step 1: Let’s define the function:

\[ f(x) = (x - 1)^2 (x + 1)^3 \]

Step 2: Take derivative to find critical points

Use product rule:
Let $ u = (x - 1)^2 $, $ v = (x + 1)^3 $
\[ f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2 \] \[ f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)] \] \[ f'(x) = (x - 1)(x + 1)^2 (5x - 1) \]

Step 3: Find critical points

Set $ f'(x) = 0 $: \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]

Step 4: Evaluate $ f(x) $ at these points

$ f(1) = 0 $
$ f(-1) = 0 $
$ f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 $

\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]

Step 5: Compare with given form:

It is given that maximum value is $ \frac{3456}{3125} = 2^p \cdot 3^q / 3125 $

Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]

✅ Final Answer: $ \boxed{(p, q) = (7,\ 3)} $

Qus : 25 MCA NIMCET PYQ

The coefficient of $x^{50}$ in the expression of ${(1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + ...... + 1001x^{1000}}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Easiest Method — Coefficient of $ x^{50} $

Given Expression:

\[ (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + \cdots + 1001x^{1000} \]

This follows a known identity that simplifies the full expression to:

\[ f(x) = (1 + x)^{1002} \]

Now: The coefficient of $ x^{50} $ in $ f(x) $ is:

\[ \boxed{\binom{1002}{50}} \]

✅ Final Answer: $ \boxed{\binom{1002}{50}} $

Qus : 26 MCA NIMCET PYQ

If ${{x}}_k=\cos \Bigg{(}\frac{2\pi k}{n}\Bigg{)}+i\sin \Bigg{(}\frac{2\pi k}{n}\Bigg{)}$ , then $\sum ^n_{k=1}({{x}}_k)=?$

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Solution

Sum of Complex Roots of Unity

Given:

\[ x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n} \]

Required: Find: \[ \sum_{k=1}^{n} x_k \]

This is the sum of all $ n^\text{th} $ roots of unity (from $ k = 1 $ to $ n $).

We know: \[ \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 \] So shifting index from $ k = 1 $ to $ n $ just cycles the same roots: \[ \sum_{k=1}^{n} e^{2\pi i k/n} = 0 \]

✅ Final Answer: $ \boxed{0} $

Qus : 27 MCA NIMCET PYQ

Number of point of which f(x) is not differentiable $f(x)=|cosx|+3$ in $[-\pi, \pi]$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Points of Non-Differentiability of $ f(x) = |\cos x| + 3 $

Step 1: $ \cos x $ is differentiable everywhere, but $ |\cos x| $ is not differentiable where $ \cos x = 0 $.

Step 2: In the interval $ [-\pi, \pi] $, we have:

\[ \cos x = 0 \Rightarrow x = -\frac{\pi}{2},\ \frac{\pi}{2} \]

So $ f(x) = |\cos x| + 3 $ is not differentiable at these two points due to sharp turns.

✅ Final Answer: $ \boxed{2 \text{ points}} $

Qus : 28 MCA NIMCET PYQ

If n₁ and n₂ are the number of real valued solutions x = | sin^–1 x | & x = sin (x) respectively, then the value of n₂– n₁ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 29 MCA NIMCET PYQ

The negation of $\sim S\vee(\sim R\wedge S)$ is equivalent to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 30 MCA NIMCET PYQ

A point P in the first quadrant, lies on $y^2 = 4ax$, a > 0, and keeps a distance of 5a units from its focus. Which of the following points lies on the locus of P?

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Solution

Locus of Point on Parabola

Given: Point on parabola $ y^2 = 4a x $ is at distance $ 5a $ from focus $ (a, 0) $.

Distance Equation:

\[ (x - a)^2 + y^2 = 25a^2 \] \[ \Rightarrow (x - a)^2 + 4a x = 25a^2 \] \[ \Rightarrow x^2 + 2a x - 24a^2 = 0 \]

Solving gives: $ x = 4a $, $ y = 4a $

✅ Final Answer: $ \boxed{(4a,\ 4a)} $

Qus : 31 MCA NIMCET PYQ

If $\int x\, \sin x\, sec^3x\, dx=\frac{1}{2}\Bigg{[}f(x){se}c^2x+g(x)\Bigg{(}\frac{\tan x}{x}\Bigg{)}\Bigg{]}+C$, then which of the following is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 32 MCA NIMCET PYQ

Let a, b, c, d be no zero numbers. If the point of intersection of the line 4ax + 2ay + c = 0 & 5bx + 2by + d=0 lies in the fourth quadrant and is equidistance from the two are then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Given: Lines $4ax+2ay+c=0$ and $5bx+2by+d=0$.

Condition: Intersection lies in the 4th quadrant and is equidistant from the axes $\Rightarrow$ point is of the form $(t,-t)$ with $t>0$.

Substitute $y=-x$ in the two lines:
$4ax+2a(-x)+c=0 \;\Rightarrow\; 2x+\dfrac{c}{2a}=0 \;\Rightarrow\; x=-\dfrac{c}{2a}.$
$5bx+2b(-x)+d=0 \;\Rightarrow\; 3x+\dfrac{d}{b}=0 \;\Rightarrow\; x=-\dfrac{d}{3b}.$

Equate $x$ from both: $-\dfrac{c}{2a}=-\dfrac{d}{3b} \;\Rightarrow\; \boxed{3bc=2ad}.$

(QIV check: $x>0,\ y<0 \;\Rightarrow\; \dfrac{c}{a}<0$ and $\dfrac{d}{b}<0$ which is consistent.)

Qus : 33 MCA NIMCET PYQ

$\theta={\cos }^{-1}\Bigg{(}\frac{3}{\sqrt[]{10}}\Bigg{)}$ is the angle between $\vec{a}=\hat{i}-2x\hat{j}+2y\hat{k}$ & $\vec{b}=x\hat{i}+\hat{j}+y\hat{k}$ then possible values of (x,y) that lie on the locus

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Solution

Qus : 34 MCA NIMCET PYQ

Let R be reflexive relation on the finite set a having 10 elements and if m is the number of ordered pair in R, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Explanation

Total possible ordered pairs on a set of 10 elements = 10² = 100.
A reflexive relation must include all pairs of the form (a,a) for all a ∈ A.
Thus, at least 10 pairs must be in R ⇒ m ≥ 10.
The maximum relation is the universal relation with all 100 pairs ⇒ m ≤ 100.

✅ Final Answer

The number of ordered pairs m can take any value in the range:
10 ≤ m ≤ 100

Qus : 35 MCA NIMCET PYQ

If $| x - 6|= | x - 4x | -| x^2- 5x +6 |$ , where x is a real variable

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 36 MCA NIMCET PYQ

The range of values of θ in the interval (0, π) such that the points (3,5) and (sinθ, cosθ) lie on the same side of the line x + y − 1 = 0, is

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Solution

Same Side of a Line — Geometric Condition

Line: $ x + y - 1 = 0 $
Point 1: (3, 2) → Lies on the side where value is positive:
\[ f(3, 2) = 3 + 2 - 1 = 4 > 0 \]

Point 2: $ (\cos\theta, \sin\theta) $ lies on same side if: \[ \cos\theta + \sin\theta > 1 \] Using identity: \[ \cos\theta + \sin\theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \Rightarrow \sin\left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}} \] So: \[ \theta \in \left(0,\ \frac{\pi}{2}\right) \]

✅ Final Answer: $ \boxed{\theta \in \left(0,\ \frac{\pi}{2}\right)} $

Qus : 37 MCA NIMCET PYQ

Which of the following number is the coefficient of $x^{100}$ in the expansion of $\log _e\Bigg{(}\frac{1+x}{1+{x}^2}\Bigg{)},\, |x|{\lt}1$ ?

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Solution

Qus : 38 MCA NIMCET PYQ

A real valued function f is defined as $f(x)=\begin{cases}{-1} & {-2\leq x\leq0} \\ {x-1} & {0\leq x\leq2}\end{cases}$. Which of the following statement is FALSE?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 39 MCA NIMCET PYQ

If $ \theta = \tan^{-1}\dfrac{1}{1+2} + \tan^{-1}\dfrac{1}{1+2\cdot3} + \tan^{-1}\dfrac{1}{1+3\cdot4} + \ldots + \tan^{-1}\dfrac{1}{1+n(n+1)} $, then $\tan\theta$ is equal to:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

We use identity $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}$

Here each term satisfies

$\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1) - \tan^{-1}(k)$

So the series telescopes:

$\theta = \tan^{-1}(n+1) - \tan^{-1}(1)$

Thus

$\tan\theta = \dfrac{(n+1)-1}{1+(n+1)(1)} = \dfrac{n}{n+2}$

Qus : 40 MCA NIMCET PYQ

A line segment AB of length 10 meters is passing through the foot of the perpendicular of a pillar, which is standing at right angle to the ground. Top of the pillar subtends angles $tan^{–1}$ 3 and $tan^{–1} 2$ at A and B respectively. Which of the following choice represents the height of the pillar?

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Solution

Qus : 41 MCA NIMCET PYQ

If $(1 + x - 2x^2)^6 = 1 + a_1 x + a_2 x^2 + \ldots + a_{12} x^{12}$, then the value of $a_2 + a_4 + a_6 + \ldots + a_{12}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Even–power coefficient sum = $\dfrac{f(1) + f(-1)}{2}$

$f(1) = (1 + 1 - 2)^6 = 0^6 = 0$

$f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64$

Required sum $= \dfrac{0 + 64}{2} = 32$

Qus : 42 MCA NIMCET PYQ

If a vector having magnitude of 5 units, makes equal angle with each of the three mutually perpendicular axes, then the sum of the magnitude of the projections on each of the axis is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Vector Projection Problem

Given: A vector of magnitude 5 makes equal angles with x, y, and z axes.
To Find: Sum of magnitudes of projections on each axis.

Let angle with each axis be $ \alpha $. Then, from direction cosine identity: \[ \cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1 \Rightarrow 3\cos^2\alpha = 1 \Rightarrow \cos\alpha = \frac{1}{\sqrt{3}} \]

Projection on each axis: $ 5 \cdot \frac{1}{\sqrt{3}} $
Sum = $ 3 \cdot \frac{5}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \boxed{5\sqrt{3}} $

✅ Final Answer: $ \boxed{5\sqrt{3}} $

Qus : 43 MCA NIMCET PYQ

A square with side $a$ is revolved about its centre through $45^\circ$. What is the area common to both the squares?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

The overlapping area of a square rotated by $45^\circ$ is: Common area $= 2(\sqrt{2}-1)a^2$

Qus : 44 MCA NIMCET PYQ

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ballsis transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred is red, is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Conditional Probability – Bayes' Theorem

Given: One ball is transferred from Bag I to Bag II, and then a ball is drawn from Bag II and is black.

Goal: Find the probability that the transferred ball was red, given that a black ball was drawn.

Using Bayes' theorem: \[ P(R|A) = \frac{P(R \cap A)}{P(A)} = \frac{\frac{3}{10} \cdot \frac{5}{10}}{\frac{3}{10} \cdot \frac{5}{10} + \frac{4}{10} \cdot \frac{6}{10} + \frac{3}{10} \cdot \frac{5}{10}} = \frac{15}{54} = \boxed{\frac{5}{18}} \]

✅ Final Answer: $ \boxed{\frac{5}{18}} $

Qus : 45 MCA NIMCET PYQ

A and B throw a die in succession to win a bet with A starting first. Whoever throws ‘1’ first wins Rs. 110. What are the respective expectations of A and B?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Let $P_A$ = probability A wins.

$P_A = \dfrac{1}{6} + \dfrac{5}{6}\cdot\dfrac{5}{6} P_A$

Solving: $P_A = \dfrac{6}{11}$

$P_B = \dfrac{5}{11}$

Expected money:

$E_A = 110 \cdot \dfrac{6}{11} = 60$

$E_B = 110 \cdot \dfrac{5}{11} = 50$

Qus : 46 MCA NIMCET PYQ

Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Given $f(x)=\dfrac{x^2-1}{|x|-1}$

$x=-2$

So $f(x)=\dfrac{(x-1)(x+1)}{-(x+1)}$ Cancel $(x+1)$: $f(x)=-(x-1)=1-x$

Now take the limit:

$\lim_{x\to -1}(1-x)$

$=1-(-1)=2$

Qus : 47 MCA NIMCET PYQ

How many different paths in the $xy$-plane are there from $(1,3)$ to $(5,6)$, if a path proceeds one step at a time either right (R) or upward (U)?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Right steps = $5-1 = 4$

Up steps = $6-3 = 3$

Total steps = $7$

Number of paths = $\binom{7}{3} = 35$

Qus : 48 MCA NIMCET PYQ

If the distance of $(x,y)$ from the origin is defined as $d(x,y) = \max(|x|,|y|)$, then the locus of points where $d(x,y)=1$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

$\max(|x|,|y|)=1$ describes a square with vertices $(\pm1,\pm1)$.

Side length = $2$

Area = $4$

Qus : 49 MCA NIMCET PYQ

If $\sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(1-x)$ then $x$ satisfies the equation:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Simplifying the given inverse–trigonometric equation gives: $2x^2 - 3x = 0$

Qus : 50 MCA NIMCET PYQ

A and B are independent witnesses. Probability A speaks the truth = $x$, Probability B speaks the truth = $y$. If both agree on a statement, the probability that the statement is true is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Relations are subsets of $A \times (A \times A)$

Size = $m \cdot m^2 = m^3$

Total relations $= 2^{m^3}$ Not present in options.

Qus : 54 MCA NIMCET PYQ

Water runs into a conical tank of radius $5$ ft and height $10$ ft at a constant rate of $2\text{ ft}^3/\text{min}$. How fast is the water level rising when the water is $6$ ft deep?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Cone volume: $V = \dfrac{1}{3}\pi r^2 h$

Similarity: $r = \dfrac{h}{2}$

So, $V = \dfrac{\pi}{12}h^3$

Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi}{4}h^2 \dfrac{dh}{dt}$

Put values: $2 = \dfrac{\pi}{4}(36)\dfrac{dh}{dt}$ $2 = 9\pi\dfrac{dh}{dt}$ $\displaystyle \dfrac{dh}{dt} = \dfrac{2}{9\pi}$

Qus : 55 MCA NIMCET PYQ

The probability that a man who is 85 yrs old will die before attaining the age of 90 is $1/3$. $A_1, A_2, A_3, A_4$ are four persons aged 85 yrs. The probability that $A_1$ will die before attaining 90 and will be the first to die is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Given:

P(dies before 90) $= \dfrac{1}{3}$ P(survives till 90) $= \dfrac{2}{3}$

Key Idea — Symmetry:

If $k$ persons die before 90, each is equally likely to be first.

\[P(A_1 \text{ is first to die}) = \sum_{k=1}^{4} P(\text{exactly } k \text{ persons die}) \times \frac{1}{k}\]

Case k = 1:

\[P = \binom{3}{0}\left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 \times \frac{1}{1} = \frac{8}{81}\]

Case k = 2:

\[P = \binom{3}{1}\left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 \times \frac{1}{2} = \frac{2}{27}\]

Case k = 3:

\[P = \binom{3}{2}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^1 \times \frac{1}{3} = \frac{2}{81}\]

Case k = 4:

\[P = \binom{3}{3}\left(\frac{1}{3}\right)^4 \left(\frac{2}{3}\right)^0 \times \frac{1}{4} = \frac{1}{324}\]

Final Answer:

\[P = \frac{32}{324} + \frac{24}{324} + \frac{8}{324} + \frac{1}{324}\]

\[\boxed{P = \frac{65}{324}}\]

Qus : 56 MCA NIMCET PYQ

A box open at the top is made by cutting squares from the four corners of a $6 \times 6$ m sheet. The height of the box for maximum volume is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Volume: $V = x(6-2x)^2$

Differentiate: $\dfrac{dV}{dx} = 0$ gives $x = 1$

So height = $1$ m → not in options.

Closest correct is None of these.

Qus : 57 MCA NIMCET PYQ

If the mean of the squares of first $n$ natural numbers be $11$, then $n$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Mean of squares $=\dfrac{n(n+1)(2n+1)}{6n}=\dfrac{(n+1)(2n+1)}{6}=11$ Solve: $(n+1)(2n+1)=66$ $2n^2+3n+1-66=0$ $2n^2+3n-65=0$ $n=5$

Qus : 58 MCA NIMCET PYQ

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, then $|\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2$ does not exceed:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Expand

\[|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}\]

\[|\vec{b}-\vec{c}|^2 = 2 - 2\vec{b}\cdot\vec{c}\]

\[|\vec{c}-\vec{a}|^2 = 2 - 2\vec{c}\cdot\vec{a}\]

Add:

\[= 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\]

Find Maximum :

\[|\vec{a} + \vec{b} + \vec{c}|^2 \geq 0\]

\[3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \geq 0\]

\[\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \geq -\frac{3}{2}\]

Final Answer:

\[6 - 2 \times \left(-\frac{3}{2}\right) = 6 + 3 = \boxed{9}\]

✅ Expression does not exceed 9

Qus : 59 MCA NIMCET PYQ

If $H$ is the harmonic mean between $P$ and $Q$, then $\dfrac{H}{P} + \dfrac{H}{Q}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Harmonic mean of $P$ and $Q$ is $H = \dfrac{2PQ}{P+Q}$

Now,

$\dfrac{H}{P} + \dfrac{H}{Q} = H\left(\dfrac{1}{P} + \dfrac{1}{Q}\right) = H \cdot \dfrac{P+Q}{PQ}$

Substitute

$H = \dfrac{2PQ}{P+Q}$:

$\dfrac{H}{P} + \dfrac{H}{Q} = \dfrac{2PQ}{P+Q} \cdot \dfrac{P+Q}{PQ} = 2$

So the correct Answer is $2$.

Qus : 60 MCA NIMCET PYQ

Probability a blade is defective $=0.002$, packet of $10$ blades. Find packets with no defective blades in $10000$ packets.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$P(\text{good})=0.998$ Probability all 10 good $=(0.998)^{10}\approx 0.9802$ So packets $=10000\times0.9802\approx9802$

Qus : 61 MCA NIMCET PYQ

Let $f(x) = \lfloor x^2 - 3 \rfloor$ where $\lfloor \cdot \rfloor$ is the greatest integer function. Number of points in $(1,2)$ where $f$ is discontinuous:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

This is a symmetric linear system.

Solving gives the unique solution:

$x = y = z = 0$

Since $a+b+c \neq 0$, determinant $\neq 0$ → unique solution.

Qus : 65 MCA NIMCET PYQ

The sum $^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use Pascal identity: $^{n}C_r + ^{n}C_{r+1} = ^{n+1}C_{r+1}$

So, $^{20}C_8 + ^{20}C_9 = ^{21}C_9$

Then, $^{21}C_9 + ^{21}C_{10} = ^{22}C_{10}$

Next, $^{22}C_{10} + ^{22}C_{11} = ^{23}C_{11}$

Thus the expression becomes:

$^{23}C_{11} - ^{23}C_{11} = 0$

$\therefore$ the answer is 0.

Qus : 66 MCA NIMCET PYQ

Car travels half distance with $v_1$, half with $v_2$. Average speed is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

For equal distances: Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$

Qus : 67 MCA NIMCET PYQ

If $y = f(x)$ is odd and differentiable on $(-\infty,\infty)$ such that $f'(3) = -2$, then $f'(-3)$ equals:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

If $f(x)$ is odd, then:

$f(-x) = -f(x)$

Differentiate both sides:

$f'(-x)(-1) = -f'(x)$

$f'(-x) = f'(x)$

So derivative is even.

Therefore:

$f'(-3) = f'(3) = -2$

Qus : 68 MCA NIMCET PYQ

The value of $\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use identity for inverse cotangent:

$\cot^{-1}a + \cot^{-1}b = \cot^{-1}\left(\dfrac{ab - 1}{a + b}\right)$

Apply to the first two terms:

$\cot^{-1}(21) + \cot^{-1}(13)$

$= \cot^{-1}\left(\dfrac{21 \cdot 13 - 1}{21 + 13}\right)$

$= \cot^{-1}\left(\dfrac{273 - 1}{34}\right)$

$= \cot^{-1}\left(\dfrac{272}{34}\right)$

$= \cot^{-1}(8)$

Now the expression becomes:

$\cot^{-1}(8) + \cot^{-1}(-8)$

But we know: $\cot^{-1}(x) + \cot^{-1}(-x) = \pi$

So, $\cot^{-1}(8) + \cot^{-1}(-8) = \pi$

$\therefore$ The value is $\pi$.

Qus : 69 MCA NIMCET PYQ

Mean of first $n$ natural numbers $=\dfrac{n+7}{3}$. Find $n$:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Mean $=\dfrac{n+1}{2}=\dfrac{n+7}{3}$ Cross-multiply: $3(n+1)=2(n+7)$ $3n+3=2n+14$ $n=11$

Qus : 70 MCA NIMCET PYQ

The value of $\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x},dx$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Let $I = \displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x}dx$

Use property: $I = \int_{0}^{\pi} f(x)dx = \int_{0}^{\pi} f(\pi - x)dx$

Compute

$f(\pi - x) = \dfrac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} = \dfrac{(\pi - x)\sin x}{1 + \cos^2 x}$

Add them: $f(x) + f(\pi - x) = \dfrac{\pi \sin x}{1 + \cos^2 x}$

So, $2I = \displaystyle \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x}dx$

Let $u = \cos x$, $du = -\sin x,dx$.

When $x=0$, $u=1$, and when $x=\pi$, $u=-1$:

$2I = \pi \displaystyle \int_{1}^{-1} \frac{-du}{1+u^2}$

$2I = \pi \displaystyle \int_{-1}^{1} \frac{du}{1+u^2}$

This equals: $2I = \pi\left[\tan^{-1}u\right]_{-1}^{1} = \pi\left(\dfrac{\pi}{4} - \left(-\dfrac{\pi}{4}\right)\right)$

$2I = \pi \cdot \dfrac{\pi}{2} = \dfrac{\pi^2}{2}$

So, $I = \dfrac{\pi^2}{4}$

Qus : 71 MCA NIMCET PYQ

Normal to the curve $y = x^3 - 3x + 2$ at the point $(2,4)$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Differentiate:

$\dfrac{dy}{dx} = 3x^2 - 3$

At $x = 2$:

$\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = 9$

So slope of tangent $m_t = 9$

Slope of normal: $m_n = -\dfrac{1}{9}$

Equation of normal at $(2,4)$:

$y - 4 = -\dfrac{1}{9}(x - 2)$

Multiply by 9:

$9y - 36 = -(x - 2)$

$9y - 36 = -x + 2$

$ x + 9y - 38 = 0$

Answer: (c) $x + 9y - 38 = 0$

Qus : 72 MCA NIMCET PYQ

Find least integer $k$ such that $(k-2)x^2 + k + 8x + 4 > 0$ for all $x\in\mathbb{R}$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

For quadratic $ax^2+bx+c>0$ for all $x$: $a>0$ → $k-2>0$ → $k>2$ Discriminant $<0$ $D=b^2-4ac=8^2-4(k-2)(k+4)$ Compute: $D=64-4(k^2+2k-8)=64-4k^2-8k+32$ $D=96-4k^2-8k<0$ Divide by $-4$: $k^2+2k-24>0$ $(k+?)(k+?)$ → roots $4$ and $-6$ So $k>4$ or $k<-6$ Combine with $k>2$ ⇒ $k>4$ Least integer = $5$

Qus : 73 MCA NIMCET PYQ

If $\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}$, then $x$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Use formula: $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$

So: $\tan^{-1}\left(\dfrac{2x+3x}{1 - 6x^2}\right) = \dfrac{\pi}{4}$

Thus: $\dfrac{5x}{1 - 6x^2} = 1$

Solve: $5x = 1 - 6x^2$

$6x^2 + 5x - 1 = 0$

Quadratic gives roots: $x = \dfrac{1}{3}$ or $x = -\dfrac{1}{2}$

Qus : 74 MCA NIMCET PYQ

The value of $\displaystyle \lim_{n\to\infty} \frac{\pi}{n}\left[\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin\frac{(n-1)\pi}{n}\right]$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

This is a Riemann sum.

Rewrite the expression:

$\displaystyle \frac{\pi}{n}\sum_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$

Let $x_k = \frac{k\pi}{n}$

Then spacing $\Delta x = \frac{\pi}{n}$

As $n\to\infty$, this becomes:

$\displaystyle \int_{0}^{\pi} \sin x , dx$

Now evaluate: $\displaystyle \int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}$

$= (-\cos\pi) - (-\cos 0)$

$= -(-1) - (-1)$

$= 1 + 1 = 2$

$\therefore$ the value of the limit is 2.

Qus : 75 MCA NIMCET PYQ

If $\displaystyle \sum_{K=0}^{2n}(-1)^K\binom{2n}{K}^2 = A$, find $\displaystyle \sum_{K=0}^{2n}(-1)^K(K-2n)\binom{2n}{K}^2$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$\displaystyle \sum (-1)^K (K-n)\binom{2n}{K}^2 = 0$ Shifting to $(K-2n)$ keeps symmetry ⇒ still $0$.

Qus : 76 MCA NIMCET PYQ

The equation $\sin^4 x + \cos^4 x + \sin 2x + \alpha = 0$ is solvable for:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Simplify: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$

$= 1 - \dfrac{1}{2}\sin^2(2x)$

So equation becomes:

$1 - \dfrac{1}{2}\sin^2(2x) + \sin(2x) + \alpha = 0$

Let $t = \sin(2x)$, where $t \in [-1,1]$

Expression becomes: $1 + \alpha + t - \dfrac{t^2}{2} = 0$

Multiply by 2:

$t^2 - 2t - 2(1+\alpha) = 0$

For real $t$ in $[-1,1]$, discriminant must allow roots inside interval. Solving gives the range:

${-\dfrac{3}{2} \le \alpha \le \dfrac{1}{2}}$

Qus : 77 MCA NIMCET PYQ

The point on the curve $y = 6x - x^2$ where the tangent is parallel to the x-axis is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Tangent ∥ x-axis ⇒ slope = 0

Differentiate: $\dfrac{dy}{dx} = 6 - 2x$

Set derivative equal to zero:

$6 - 2x = 0$ $2x = 6$

$x = 3$

Now find $y$:

$y = 6(3) - (3)^2 = 18 - 9 = 9$

$a,b,c$ in H.P. ⇒

$\dfrac{1}{b} = \dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)$

⇒ $2bc = ac + ab$

Simplify gives identity:

$(a+c)(a-2b+c) = (c-b)^2$

Take log:

$\log(a+c) + \log(a-2b+c) $

$= \log\big((c-b)^2\big)$

$=2\log(c-b)$

Qus : 82 MCA NIMCET PYQ

The vector $\vec{B} = 3\vec{i} + 4\vec{k}$ is to be written as the sum of a vector $\vec{B_1}$ parallel to $\vec{A} = \vec{i} + \vec{j}$ and a vector $\vec{B_2}$ perpendicular to $\vec{A}$. Then $\vec{B_1}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Parallel component formula: $\displaystyle \vec{B_1} = \frac{\vec{B}\cdot \vec{A}}{\vec{A}\cdot \vec{A}} \vec{A}$

Compute:

$\vec{B} = (3,0,4)$

$\vec{A} = (1,1,0)$

$\vec{B}\cdot\vec{A} = 3\cdot 1 + 0\cdot 1 + 4\cdot 0 = 3$

$\vec{A}\cdot\vec{A} = 1^2 + 1^2 = 2$

Thus: $\vec{B_1} = \dfrac{3}{2}(\vec{i} + \vec{j})$

Qus : 83 MCA NIMCET PYQ

The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Let $I = \displaystyle \int_{0}^{\pi/2} \log \tan x dx$

Using property $\displaystyle \int_{0}^{\pi/2} f(x)dx = \int_{0}^{\pi/2} f\left(\dfrac{\pi}{2}-x\right)dx$

$I = \displaystyle \int_{0}^{\pi/2} \log \tan\left(\dfrac{\pi}{2}-x\right) dx$

$= \displaystyle \int_{0}^{\pi/2} \log \cot x dx$

$= \displaystyle \int_{0}^{\pi/2} \log\left(\dfrac{1}{\tan x}\right) dx$

$= \displaystyle \int_{0}^{\pi/2} [-\log \tan x] dx$

$I= -I$

$\Rightarrow I + I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0$

Answer: $0$

Qus : 84 MCA NIMCET PYQ

Area enclosed by $|x|+|y|=1$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

This is a diamond (square rotated 45°) with diagonals length $2$ and $2$. Area $=\dfrac12 d_1 d_2=\dfrac12\cdot2\cdot2=2$

Qus : 85 MCA NIMCET PYQ

Find $k$ in the equation $x^3 - 6x^2 + kx + 64 = 0$ if roots are in geometric progression.

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Solution

Let roots be $a, ar, ar^2$.

Sum of roots:

$a + ar + ar^2 = 6$ $a(1+r+r^2) = 6$ ...(1)

Product of roots:

$ar \cdot ar^2 \cdot a = a^3 r^3 = -64$

$\Rightarrow (ar)^3 = -64$ $\Rightarrow ar = -4$ ...(2)

Middle coefficient relation:

Sum of pairwise products:

$k = a(ar) + ar(ar^2) + ar^2(a)$

$k = a^2 r + a^2 r^3 + a^2 r^2$

Factor: $k = a^2 (r + r^2 + r^3)$

$k = ar \cdot a(r + r^2 + r^3)$

Using (2): $ar = -4$

Also: $r + r^2 + r^3 = r(1 + r + r^2)$

Thus: $k = -4a \cdot r(1 + r + r^2)$

But from (1): $a(1+r+r^2) = 6$

So: $k = -4r \cdot 6 = -24r$

Now solve $ar = -4$ and equation (1).

Standard GP root problem yields $r = 1$ or $r = -1$.

Check sign consistency → $r = 1$.

So: $k = -24(1)$

$k = -24$

Qus : 86 MCA NIMCET PYQ

A determinant is chosen at random from the set of all determinants of matrices of order 2 with elements 0 and 1 only. The probability that the determinant chosen is non-zero is:

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Solution

Total $2\times2$ matrices with entries $0$ or $1$: $2^4 = 16$

Determinant: $ad - bc$

Non-zero determinant occurs in two cases.

Case 1: $ad = 1$ and $bc = 0$

$a = 1, d = 1$ $(b,c)$ can be $(0,0), (0,1), (1,0)$

→ 3 matrices

Case 2: $bc = 1$ and $ad = 0$

$b = 1,c = 1$ $(a,d)$ can be $(0,0), (0,1), (1,0)$

→ 3 matrices

Total non-zero determinants = $3 + 3 = 6$

Since $I + A + A^2 + \cdots = (I - A)^{-1}$,

$I - A = \begin{bmatrix} 0 & -2 \\ -3 & -3 \end{bmatrix}$

$(I - A)^{-1} = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{1}{3} \\ -\tfrac{1}{2} & 0 \end{bmatrix}$.

Qus : 92 MCA NIMCET PYQ

The equation of the plane passing through the point $(1,2,3)$ and having the normal vector $N = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ is:

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Solution

Equation of plane with normal $(a,b,c)$ passing through $(x_1, y_1, z_1)$ is:

$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$

Here: $a = 3; b = -1; c = 2$

$(x_1, y_1, z_1) = (1,2,3)$

So: $3(x - 1) - 1(y - 2) + 2(z - 3) = 0$

Expand: $3x - 3 - y + 2 + 2z - 6 = 0$

Combine constants: $3x - y + 2z - 7 = 0$

$3x - y + 2z = 7$

Thus the correct answer is: $3x - y + 2z = 7$

Qus : 93 MCA NIMCET PYQ

In a college of $300$ students, every student reads $5$ newspapers and every newspaper is read by $60$ students. The number of newspapers is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Total readings $=300\times5=1500$ If $N$ is number of newspapers, each is read by $60$ students: $60N=1500 \Rightarrow N=25$

Qus : 94 MCA NIMCET PYQ

$A_1, A_2, A_3, A_4$ are subsets of $U$ (75 elements). Each $A_i$ has 28 elements. Any two intersect in 12 elements. Any three intersect in 5 elements. All four intersect in 1 element. Find the number of elements belonging to none of the four subsets.

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Solution

Use Inclusion–Exclusion:

$|A_1 \cup A_2 \cup A_3 \cup A_4|$ $= \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - |A_1 \cap A_2 \cap A_3 \cap A_4|$

Substitute values:

$\sum |A_i| = 4 \cdot 28 = 112$

$\sum |A_i \cap A_j| = \binom{4}{2} \cdot 12 = 6 \cdot 12 = 72$

$\sum |A_i \cap A_j \cap A_k| = \binom{4}{3} \cdot 5 = 4 \cdot 5 = 20$

Intersection of four = 1

So: $|A_1 \cup A_2 \cup A_3 \cup A_4| = 112 - 72 + 20 - 1 = 59$

Elements belonging to none: $75 - 59 = 16$

Qus : 95 MCA NIMCET PYQ

The value of $\displaystyle \int_{0}^{\sin^2 x} \sin^{-1}\sqrt{t} dt + \int_{0}^{\cos^2 x} \cos^{-1}\sqrt{t} dt$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use the identity:

$\sin^{-1}y+\cos^{-1}y=\dfrac{\pi}{2}$

So: $\cos^{-1}\sqrt{t}=\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}$

Now substitute: $I=\displaystyle\int_0^{\sin^2 x}\sin^{-1}\sqrt{t} dt + \int_0^{\cos^2 x} \left(\dfrac{\pi}{2}-\sin^{-1}\sqrt{t}\right) dt$

$I=\dfrac{\pi}{2}\cos^2 x + \int_0^{\sin^2 x}\sin^{-1}\sqrt{t}dt - \int_0^{\cos^2 x}\sin^{-1}\sqrt{t} dt$

Combine integrals: $I=\dfrac{\pi}{2}\cos^2 x + \int_{\cos^2 x}^{\sin^2 x}\sin^{-1}\sqrt{t} dt$

But: $\sin^2 x + \cos^2 x = 1$

Limits become from $1$ to $0$:

$I=\dfrac{\pi}{2}(1 - \sin^2 x) + \int_{1}^{0}\sin^{-1}\sqrt{t} dt$

$I=\dfrac{\pi}{2} - \int_0^{1}\sin^{-1}\sqrt{t} dt$

Let $u=\sqrt{t}$,

$dt=2udu$:

$\displaystyle \int_0^{1}\sin^{-1}\sqrt{t} dt = 2\int_0^{1} u\sin^{-1}udu$

Standard result: $\displaystyle 2\int_0^{1} u\sin^{-1}u du = \dfrac{\pi}{4}$

Thus: $I = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$

Qus : 96 MCA NIMCET PYQ

The number of ways of forming different $9$-digit numbers from $223355588$ by rearranging digits so that odd digits occupy even positions is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Odd digits in the number: $3,3,5,5,5$ (total $5$ odd digits) Even positions in a $9$-digit number = $4$ positions. Choose $4$ odd digits out of $5$: $\binom{5}{4}=5$ Arrange those $4$ chosen digits in $4!$ ways but with repetition: If digits chosen are $3,3,5,5$: arrangements $=\dfrac{4!}{2!,2!}=6$ If chosen are $3,5,5,5$: arrangements $=\dfrac{4!}{3!}=4$ Total arrangements for odd positions: $1$ way with $(3,3,5,5)$ giving $6$ $4$ ways with $(3,5,5,5)$ each giving $4$ Total $=6 + 4\cdot4 = 22$ Even digits $2,2,8,8$ fill $5$ positions → contradiction unless a specific interpretation (official key gives $60$). (We keep official expected answer.)

Qus : 97 MCA NIMCET PYQ

$ABC$ is isosceles with $AB = AC$. $BC$ is parallel to x-axis. $m_1, m_2$ are slopes of the medians from $B$ and $C$. Then:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Let coordinates: $B(-a, 0)$ $C(a, 0)$ $A(0, h)$

Median from $B$ goes to midpoint of $AC$: $(0, h/2)$

Slope: $m_1 = \dfrac{h/2 - 0}{0 - (-a)} = \dfrac{h}{2a}$

Median from $C$ goes to midpoint of $AB$: $(0, h/2)$

Slope: $m_2 = \dfrac{h/2 - 0}{0 - a} = -\dfrac{h}{2a}$

Thus: $m_1 + m_2 = 0$

Qus : 98 MCA NIMCET PYQ

Coefficients a, b, c of $ax^2 + bx + c = 0$ are chosen by tossing 3 fair coins. Head means 1, Tail means 2. Find the probability that the roots are imaginary

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Solution

Since each of $a,b,c$ can be $1$ or $2$:

Total possible triples $(a,b,c)$: $2^3 = 8$.

Roots are imaginary if: $b^2 - 4ac < 0$

So we check all $8$ cases.

List all possibilities

$(1,1,1)$: $1^2 - 4(1)(1) = 1 - 4 = -3 < 0$ ✔

$(1,1,2)$: $1 - 8 = -7 < 0$ ✔

$(1,2,1)$: $4 - 4 = 0$ ✘ (not imaginary)

$(1,2,2)$: $4 - 8 = -4 < 0$ ✔

$(2,1,1)$: $1 - 8 = -7 < 0$ ✔

$(2,1,2)$: $1 - 16 = -15 < 0$ ✔

$(2,2,1)$: $4 - 8 = -4 < 0$ ✔

$(2,2,2)$: $4 - 16 = -12 < 0$ ✔

Count imaginary root cases

Total imaginary cases = $7$ out of $8$.

So probability: $\displaystyle P = \frac{7}{8}$

Qus : 99 MCA NIMCET PYQ

An anti-aircraft gun fires at a plane. Probabilities of hitting at slots 1,2,3,4 are $0.4,;0.3,;0.2,;0.1$. Probability that the gun hits the plane is

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Solution

Probability of miss in all four shots: $(1-0.4)(1-0.3)(1-0.2)(1-0.1)$ $=0.6 \times 0.7 \times 0.8 \times 0.9$ $=0.3024$ Probability of at least one hit: $1-0.3024 = 0.6976$

Qus : 100 MCA NIMCET PYQ

The smaller area bounded by $y = 2 - x$ and $x^2 + y^2 = 4$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Line intersects circle at two symmetric points.

Required smaller area = circular segment area.

Solve intersection:

$y = 2 - x$

$x^2 + (2 - x)^2 = 4$

$x^2 + x^2 - 4x + 4 = 4$

$2x^2 - 4x = 0$

$2x(x - 2) = 0$

So intersection points at $(0, 2)$ and $(2, 0)$.

Using known segment area formula → the smaller area = $\pi - 2$.

Qus : 101 MCA NIMCET PYQ

In a class of 100 students, 55 passed in Mathematics and 67 passed in Physics.The number of students who passed in Physics only is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Use the formula:

$\text{Physics only} = n(P) - n(M \cap P)$

First find $n(M \cap P$) using:

$n(M) + n(P) - n(M \cap P) = 100$

So: $55 + 67 - n(M \cap P) = 100$

$122 - n(M \cap P) = 100$

$n(M \cap P) = 22$

Now: $\text{Physics only} = 67 - 22 = 45$

Qus : 102 MCA NIMCET PYQ

The mean of 5 observation is 5 and their variance is 12.4. If three of the observations are 1,2 and 6; then the mean deviation from the mean of the data is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Given: 5 observations, mean = 5 ⇒ total sum = $5\times5=25$. Three values are 1, 2, 6. Let the other two be $a,b$.

From mean: $a+b=25-(1+2+6)=16$.

Variance (about mean): $12.4$ ⇒ $\sum (x_i-5)^2 = 5\times12.4 = 62$.
Known part: $(1-5)^2+(2-5)^2+(6-5)^2=16+9+1=26$.
Hence $(a-5)^2+(b-5)^2 = 62-26 = 36$.

Let $u=a-5,\ v=b-5$. Then $u+v=(a+b)-10=6$ and $u^2+v^2=36$.
$(u+v)^2 = u^2+v^2+2uv \Rightarrow 36 = 36 + 2uv \Rightarrow uv=0$.
So one of $u,v$ is 0 ⇒ one of $a,b$ is 5, the other is $16-5=11$.

Mean deviation about mean:
$\displaystyle \text{MD}=\frac{1}{5}\big(|1-5|+|2-5|+|6-5|+|5-5|+|11-5|\big) $ $ =\frac{1}{5}(4+3+1+0+6)=\frac{14}{5}=2.8.$

Answer: 2.8

Qus : 103 MCA NIMCET PYQ

The minimum value of $px + qy$ when $xy=r^2$ and $p,q,x,y$ are positive numbers is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Use AM-GM or substitution $y=\dfrac{r^2}{x}$: $px + q\dfrac{r^2}{x}$ Min occurs when derivatives equal: $px = q\dfrac{r^2}{x}$ $\Rightarrow x^2 = \dfrac{qr^2}{p}$ Compute minimum: $2r\sqrt{pq}$

Qus : 104 MCA NIMCET PYQ

There are 10 points, out of which 6 are collinear. Number of triangles formed:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Total triangles from 10 points: $\binom{10}{3} = 120$

Invalid triangles from 6 collinear points: $\binom{6}{3} = 20$

So valid triangles = $120 - 20 = 100$

Qus : 105 MCA NIMCET PYQ

If $(4,-3)$ and $(-9,7)$ are two vertices of a triangle and $(1,4)$ is its centroid, find the area of the triangle.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Let the third vertex be $(x,y)$. Centroid formula: $\displaystyle \left(\frac{4 + (-9) + x}{3}, \frac{-3 + 7 + y}{3}\right) = (1,4)$

From x–coordinate: $\dfrac{4 - 9 + x}{3} = 1$

$x - 5 = 3$

$x = 8$

From y–coordinate: $\dfrac{-3 + 7 + y}{3} = 4$

$y + 4 = 12$

$y = 8$

So third vertex is: $(8,8)$

Vertices: $A(4,-3), B(-9,7), C(8,8)$

Area: $\displaystyle \text{Area} = \frac{1}{2}\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|$ Substitute:

$\displaystyle = \frac{1}{2}\left|4(7-8) + (-9)(8+3) + 8(-3-7) \right|$

$\displaystyle = \frac{1}{2}\left| 4(-1) - 9(11) + 8(-10) \right|$

$\displaystyle = \frac{1}{2}\left| -4 - 99 - 80 \right|$

$\displaystyle = \frac{1}{2} \times 183$ $\displaystyle = \frac{183}{2}$

Qus : 106 MCA NIMCET PYQ

In a beauty contest, half the number of experts voted Mr. A and two thirds voted for Mr. B 10 voted for both and 6 did not for either. How may experts were there in all.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Let the total number of experts be

N

E

is the set of experts who voted for miss

A

F

is the set of experts who voted for miss

B

.
Since

6

did not vote for either, n

(E \cup F) = N - 6

n (E) = \frac{N}{2}, n (F) = \frac{2}{3} N

and

n (E \cap F) = 10

.
So,

N - 6 = \frac{N}{2} + \frac{2}{3} N - 10

Solving the above equation gives

\frac{N}{6} = 4 \Rightarrow N = 24

Qus : 107 MCA NIMCET PYQ

A circle touches the x–axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 108 MCA NIMCET PYQ

If $a$ is a positive integer, then the number of values satisfying $ \displaystyle \int_{0}^{\pi/2} \left[ a^{2}\left(\frac{\cos 3x}{4}+\frac{3}{4}\cos x\right)+a\sin x - 20\cos x \right] dx \le -\frac{a^{2}}{3} $ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ \displaystyle \int_{0}^{\pi/2} \cos 3x, dx = \frac{1}{3},;; \int_{0}^{\pi/2} \cos x, dx = 1,;; \int_{0}^{\pi/2} \sin x, dx = 1 $ So integral becomes $ \displaystyle a^{2}\left(\frac{1}{12}+\frac{3}{4}\right)+a - 20 = \frac{5a^{2}}{6} + a - 20 $ Given $ \displaystyle \frac{5a^{2}}{6}+a-20 \le -\frac{a^{2}}{3} $ $ \displaystyle \Rightarrow \frac{7a^{2}}{6} + a - 20 \le 0 $ Multiply by 6: $ 7a^{2} + 6a - 120 \le 0 $ Roots: $ a = \frac{26}{7} \approx 3.714 $ So valid positive integers: $ a = 1,,2,,3 $

Qus : 109 MCA NIMCET PYQ

Number of distinct integer values of $a$ satisfying $2^{2a} - 3(2^{a+2}) + 25 = 0$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Rewrite: $2^{2a} - 12\cdot 2^a + 25 = 0$

Let $t = 2^a$ (positive).

Equation becomes: $t^2 - 12t + 25 = 0$

$t = \dfrac{12 \pm \sqrt{44}}{2} = 6 \pm \sqrt{11}$

We need $t = 2^a$, a power of 2.

Check if $6 + \sqrt{11}$ or $6 - \sqrt{11}$ is a power of 2.

Values: $6 + \sqrt{11} \approx 9.316$ → not power of 2

$6 - \sqrt{11} \approx 2.684$ → not power of 2 No integer $a$ exists.

Qus : 110 MCA NIMCET PYQ

The equation of ellipse with major axis along the x–axis and passes through the point $(4,3)$ and $(-1,4)$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Ellipse form:

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\quad a^2 > b^2$

Using points: $\displaystyle \frac{16}{a^2} + \frac{9}{b^2} = 1$

$\displaystyle \frac{1}{a^2} + \frac{16}{b^2} = 1$

Solve gives:

$\displaystyle 7x^2 + 15y^2 = 247$

Qus : 111 MCA NIMCET PYQ

The value of non-zero scalars α and β such that for all vectors $\vec{a}$ and $\vec{b}$ such that $\alpha (2\vec{a}-\vec{b})+\beta (\vec{a}+2\vec{b})=8\vec{b}-\vec{a}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Given: $\alpha (2\vec a-\vec b)+\beta (\vec a+2\vec b)=8\vec b-\vec a$ for all vectors $\vec a,\vec b$.

Combine like terms: $(2\alpha+\beta)\vec a+(-\alpha+2\beta)\vec b=-\vec a+8\vec b$.

Equate coefficients:
$2\alpha+\beta=-1$,
$-\alpha+2\beta=8$.

Solve:
From $2\alpha+\beta=-1 \Rightarrow \beta=-1-2\alpha$.
Substitute in $-\alpha+2\beta=8$:
$-\alpha+2(-1-2\alpha)=8 \Rightarrow -\alpha-2-4\alpha=8 \Rightarrow -5\alpha=10 \Rightarrow \alpha=-2$.
Then $\beta=-1-2(-2)=3$.

Answer: $\alpha=-2,\ \beta=3$.

Qus : 112 MCA NIMCET PYQ

Find $ \displaystyle \frac{d}{dx}\left( \sqrt{x} - \frac{5}{\sqrt{x}} \right) $

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ \displaystyle \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}},;; \frac{d}{dx}\left(\frac{5}{\sqrt{x}}\right)= 5\cdot\left(-\frac{1}{2}x^{-3/2}\right) $ So $ \displaystyle \frac{d}{dx}\left(\sqrt{x}-\frac{5}{\sqrt{x}}\right) = \frac{1}{2\sqrt{x}} + \frac{5}{2}x^{-3/2} $

Qus : 113 MCA NIMCET PYQ

A man has 5 coins: 2 double-headed 1 double-tailed 2 normal He randomly picks a coin and tosses it. Probability that the lower face is a head is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

Double-headed coin (2 coins):

lower face = Head always → probability = 1

Double-tailed coin (1 coin):

lower face = Head → probability = 0

Normal coin (2 coins):

lower face = Head → probability = $\tfrac{1}{2}$

Total probability: $\displaystyle P = \frac{2}{5}(1) + \frac{1}{5}(0) + \frac{2}{5}\left(\frac{1}{2}\right)$

$P = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$

Qus : 114 MCA NIMCET PYQ

If the circles $ x^2 + y^2 + 2x + 2ky + 6 = 0$ and $x^2 + y^2 + 2ky + k = 0$ intersect orthogonally, then $k$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Two circles: $C_1:\; x^2 + y^2 + 2x + 2ky + 6 = 0$

$C_2:\; x^2 + y^2 + 2ky + k = 0$

Orthogonality condition for circles

$x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$

and

$x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ is:

$2(g_1g_2 + f_1f_2) = c_1 + c_2$

From $C_1$: $g_1 = 1,\; f_1 = k,\; c_1 = 6$

From $C_2$: $g_2 = 0,\; f_2 = k,\; c_2 = k$

Apply formula:

$2(1\cdot 0 + k\cdot k) = 6 + k$

$2k^2 = 6 + k$

$2k^2 - k - 6 = 0$

$(2k + 3)(k - 2) = 0$

So: $k = 2 \;\text{or}\; -\dfrac{3}{2}$

Qus : 115 MCA NIMCET PYQ

A force of 78 grams acts at the point (2,3,5). The direction ratios of the line of action being 2,2,1 . The magnitude of its moment about the line joining the origin to the point (12,3,4) is

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Solution

Given: Force magnitude $=78$ (grams), point of application $\mathbf{r}=(2,3,5)$, force direction ratios $(2,2,1)$. Line (about which moment is required) is along $(12,3,4)$ through the origin.

Force vector: Unit along $(2,2,1)$ is $\dfrac{1}{\sqrt{2^2+2^2+1^2}}(2,2,1)=\dfrac{1}{3}(2,2,1)$. Hence \[ \mathbf{F}=78\cdot \frac{1}{3}(2,2,1)=(52,52,26). \]

Torque about origin: \[ \boldsymbol{\tau}=\mathbf{r}\times \mathbf{F}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&3&5\\ 52&52&26 \end{vmatrix} =(-182,\,208,\,-52). \]

Moment about the line with direction $\mathbf{L}=(12,3,4)$: Let $\hat{\ell}=\dfrac{\mathbf{L}}{\lVert \mathbf{L}\rVert}$. The required magnitude is the component of $\boldsymbol{\tau}$ along $\hat{\ell}$: \[ M=\big|\boldsymbol{\tau}\cdot \hat{\ell}\big|=\frac{\big|\boldsymbol{\tau}\cdot \mathbf{L}\big|}{\lVert \mathbf{L}\rVert} =\frac{|(-182,208,-52)\cdot(12,3,4)|}{\sqrt{12^2+3^2+4^2}} =\frac{|{-1768}|}{13}=136. \]

Answer: $ \boxed{136} $ (in gram–unit of length).

Qus : 116 MCA NIMCET PYQ

$ \displaystyle \lim_{x\to 0} \frac{x+\sin x}{\sqrt{x}-\cos x} $

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Use expansions: $\sin x = x - \frac{x^{3}}{6} + \cdots$ $\cos x = 1 - \frac{x^{2}}{2} + \cdots$ $\sqrt{x}$ near $0$ goes to $0$ Numerator: $ x + (x - \frac{x^{3}}{6}) = 2x + O(x^{3}) $ Denominator: $ \sqrt{x} - \cos x = \sqrt{x} - 1 + \frac{x^{2}}{2} + \cdots $ As $x\to 0$, denominator → $-1$ So limit = $0$.

Qus : 117 MCA NIMCET PYQ

If $A = \cos^2\theta + \sin^4\theta$, then for all values of $\theta$:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

$A = \cos^2\theta + \sin^4\theta$

Let $\sin^2\theta = t$, $0 \le t \le 1$

Then $A = (1 - t) + t^2 = t^2 - t + 1$

This is a quadratic in $t$.

Minimum value at $t = \dfrac{1}{2}$:

$A_{\min} = \left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4}$

Maximum value at endpoints $t=0$ or $t=1$:

$A_{\max} = 1$

Thus: ${\dfrac{3}{4} \le A \le 1}$

Qus : 118 MCA NIMCET PYQ

Focus of the parabola $x^2 + y^2 - 2xy - 4(x + y - 1) = 0$ is:

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Solution

Given equation:

$x^2 + y^2 - 2xy - 4x - 4y + 4 = 0$

Group terms:

$(x - y)^2 - 4(x + y) + 4 = 0$

Let: $u = x - y,\;\; v = x + y$

Then equation becomes:

$u^2 - 4v + 4 = 0$

$u^2 = 4(v - 1)$

This is the standard parabola:

$u^2 = 4p(v - 1)$

Comparing gives:

$4p = 4 \Rightarrow p = 1$

Vertex in $(u,v)$: $(0,1)$

Focus in $(u,v)$: $(0, 1 + p) = (0,2)$

Convert to $(x,y)$:

$x - y = 0$

$x + y = 2$

Solving: $x = 1,\; y = 1$

Therefore, the focus is: $\boxed{(1,1)}$

Qus : 119 MCA NIMCET PYQ

Number of real solutions of the equation $\sin\!\left(e^x\right) = 5^x + 5^{-x}$ is

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Solution

Number of real solutions

Equation: $$\sin\!\left(e^x\right) \;=\; 5^x + 5^{-x}$$

Reasoning

For all real $x$, $\sin(e^x)\in[-1,1]$.
By AM–GM, $5^x + 5^{-x} \ge 2$ (with equality only when $5^x=5^{-x}\Rightarrow x=0$).
At $x=0$: LHS $=\sin(1)\approx 0.84$, RHS $=2$ ⇒ not equal.
Hence RHS is always $\ge 2$ while LHS is always $\le 1$ → they can never match.

Thus: $3 + 2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = 0$

$2(\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a}) = -3$

$\mathbf{a}\cdot\mathbf{b} + \mathbf{b}\cdot\mathbf{c} + \mathbf{c}\cdot\mathbf{a} = -\dfrac{3}{2}$

Therefore: ${-\dfrac{3}{2}}$

Qus : 123 MCA NIMCET PYQ

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function $f(x)=x^3+3x-9$ in the interval [-2,3] and the difference between the first two terms is $f'(0)$. Then the common ratio of GP is

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Solution

Given:

$f(x)=x^3+3x-9$ on $[-2,3]$.

Since $f'(x)=3x^2+3>0$, $f$ is increasing, so the greatest value is at $x=3$:

$f(3)=27$.

Sum to infinity of decreasing GP:

$S=\dfrac{a}{1-r}=27$ with

Also $f'(0)=3\Rightarrow$ difference of first two terms:

$a-ar=a(1-r)=3$.

From $a=\;27(1-r)$,

plug into $a(1-r)=3$: $27(1-r)^2=3\Rightarrow(1-r)^2=\dfrac{1}{9}\Rightarrow 1-r=\dfrac{1}{3}$ (take positive)

Common ratio: $r=1-\dfrac{1}{3}=\boxed{\dfrac{2}{3}}$.

Qus : 124 MCA NIMCET PYQ

The value of $ \sin 30^\circ \cos 45^\circ + \cos 30^\circ \sin 45^\circ $

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Expression is $\sin(30^\circ + 45^\circ)$ $ = \sin 75^\circ = \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}} $

Qus : 125 MCA NIMCET PYQ

Number of solutions for $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \dfrac{\pi}{2}$ is:

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Solution

Both square-root expressions must be real.

Domain 1: $x(x+1) \ge 0 \Rightarrow x \le -1 \text{ or } x \ge 0$

Domain 2: $x^2 + x + 1 > 0$ for all $x$ (always positive)

Equation transforms into identity impossible to satisfy over real domain.

Therefore no real solution.

Qus : 126 MCA NIMCET PYQ

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Solution

Qus : 127 MCA NIMCET PYQ

Number of onto (surjective) functions from A to B if n(A)=6 and n(B)=3, is

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Solution

Given: $ n(A)=6 $ and $ n(B)=3 $.

Formula: The number of onto (surjective) functions from a set of size $m$ to a set of size $n$ is \[ n! \, S(m,n) \] where $S(m,n)$ is the Stirling number of the second kind (number of ways to partition $m$ elements into $n$ non-empty subsets).

We can also use the Inclusion–Exclusion Principle: \[ n! \, S(m,n) = \sum_{k=0}^{n} (-1)^k \binom{n}{k}(n-k)^m \] For $m=6,\ n=3$: \[ N = 3^6 - 3\times 2^6 + 3\times 1^6 \]

Calculation: \[ 3^6 = 729,\quad 2^6 = 64 \] \[ N = 729 - 3(64) + 3(1) = 729 - 192 + 3 = 540. \]

Answer: The number of onto functions is \[ \boxed{540}. \]

Qus : 128 MCA NIMCET PYQ

In $\triangle ABC$, $B = 45^\circ, C = 105^\circ, c=\sqrt{2}$. Find side $a$ and $b$.

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Solution

$A = 180 - (45 + 105) = 30^\circ$ Using Law of Sines: $ \displaystyle \frac{a}{\sin A} = \frac{c}{\sin C} $ $ \displaystyle a = \frac{\sin 30^\circ}{\sin 105^\circ}\cdot \sqrt{2} $ Compute: $\sin 30^\circ = \frac12$ $\sin 105^\circ = \sin(60+45)=\frac{\sqrt{6}+\sqrt{2}}{4}$ $ \displaystyle a = \sqrt{2}\cdot \frac{1/2}{(\sqrt{6}+\sqrt{2})/4} = \frac{\sqrt{2}}{(\sqrt{6}+\sqrt{2})/2} = \frac{2\sqrt{2}}{\sqrt{6}+\sqrt{2}} $ After rationalizing: $ \displaystyle a=\sqrt{3}-1 $ Similarly, $ \displaystyle b=\sqrt{2}(\sqrt{3}-1) $

Qus : 129 MCA NIMCET PYQ

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar unit vectors and $\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{\vec{b} + \vec{c}}{\sqrt{2}}$, then the angle between $\vec{a}$ and $\vec{b}$ is:

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Solution

Vector triple product identity:

$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a}\cdot \vec{c})\vec{b} - (\vec{a}\cdot \vec{b})\vec{c}$

Given equals $\dfrac{\vec{b} + \vec{c}}{\sqrt{2}}$

So compare coefficients:

$\vec{b}$ coefficient:

$a\cdot c = \dfrac{1}{\sqrt{2}}$ $\vec{c}$ coefficient: $-(a\cdot b) = \dfrac{1}{\sqrt{2}}$ $\Rightarrow a\cdot b = -\dfrac{1}{\sqrt{2}}$ Thus angle $\theta$ between $a$ and $b$ satisfies: $\cos\theta = -\dfrac{1}{\sqrt{2}}$ $\Rightarrow \theta = \dfrac{3\pi}{4}$

Qus : 130 MCA NIMCET PYQ

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Intersection point $(x,y)$ satisfies:

$\dfrac{x}{a} + \dfrac{y}{b} = k$

$\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{k}$

Subtract → inconsistent unless the meeting point satisfies: Multiply equations: $\left(\dfrac{x}{a} + \dfrac{y}{b}\right)\left(\dfrac{x}{a} + \dfrac{y}{b}\right) = 1$

Thus locus: $\left(\dfrac{x}{a} + \dfrac{y}{b}\right)^2 = 1$

This is a pair of parallel lines (degenerate hyperbola) → classified as hyperbola.

Qus : 134 MCA NIMCET PYQ

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Solution

Qus : 135 MCA NIMCET PYQ

A computer producing factory has only two plants T₁ and T₂ produces 20% and plant T₂ produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant T₁ 10P(computer turns out to be defective given that it is produced in plant T₂ ). A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T₂ is

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Solution

Given: $P(T_1)=0.2,\ P(T_2)=0.8,\ P(D)=0.07,$ and $P(D\mid T_1)=10\,P(D\mid T_2)$.

Let $p_2=P(D\mid T_2)$. Then $P(D\mid T_1)=10p_2$. Using total probability: \[ 0.07=P(D)=0.2(10p_2)+0.8(p_2)=(2+0.8)p_2=2.8p_2 \Rightarrow p_2=\frac{0.07}{2.8}=0.025. \] Hence $P(D\mid T_1)=0.25$.

We need: $P(T_2\mid \overline D)=\dfrac{P(T_2)\,P(\overline D\mid T_2)}{P(\overline D)}$, where $P(\overline D)=1-0.07=0.93$ and $P(\overline D\mid T_2)=1-0.025=0.975$.

\[ P(T_2\mid \overline D)=\frac{0.8\times 0.975}{0.93} =\frac{0.78}{0.93} =\frac{26}{31}\approx 0.8387. \]

Answer: $\boxed{\dfrac{26}{31}}$.

Qus : 136 MCA NIMCET PYQ

The general solution of $ \sqrt{3}\cos x + \sin x = 3 $ is:

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Solution

Maximum of $ \sqrt{3}\cos x + \sin x $ is $ \sqrt{(\sqrt{3})^{2} + 1^{2}} = 2 $ Since $2 < 3$, the equation cannot be satisfied.

Qus : 137 MCA NIMCET PYQ

Events $A$ and $B$ satisfy: $P(A \cup B) = \dfrac{1}{6}$, $P(A \cap B) = \dfrac{1}{4}$, $P(A) = \dfrac{1}{4}$ Then events $A$ and $B$ are:

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Solution

Qus : 138 MCA NIMCET PYQ

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Solution

Qus : 139 MCA NIMCET PYQ

If A > 0, B > 0 and A + B = $\frac{\pi}{6}$ , then the minimum value of $ \tan A + \tan B$

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Solution

Given $A,B>0$ and $A+B=\dfrac{\pi}{6}$. Using \[ \tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}, \] with $\sin(A+B)=\sin\frac{\pi}{6}=\dfrac12$. To minimize $\tan A+\tan B$, maximize $\cos A\cos B$ subject to $A+B=\dfrac{\pi}{6}$.

The product $\cos A\cos B$ (with fixed sum) is maximized at $A=B=\dfrac{\pi}{12}$. Thus \[ \cos A\cos B\le \cos^2\!\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{2} =\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}. \] Hence \[ \min(\tan A+\tan B)=\frac{\frac12}{\frac{2+\sqrt3}{4}} =\frac{2}{2+\sqrt3} =\boxed{\,4-2\sqrt3\,}. \]

Qus : 140 MCA NIMCET PYQ

Inverse of the function $f(x)=\frac{10^x-10^{-x}}{10^{x}+10^{-x}}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Hit at least once = 1 − (miss all shots)

Miss probabilities:

1st: $(1 - 0.4) = 0.6$

2nd: $(1 - 0.3) = 0.7$

3rd: $(1 - 0.2) = 0.8$

4th: $(1 - 0.1) = 0.9$

Letters and frequencies:

$M=2,\ A=2,\ T=2,\ H=1,\ E=1,\ I=1,\ C=1,\ S=1$ (total $11$ letters)

Condition: Starts with T and ends with T

Fix $T$ at first and last position.

Remaining letters = $9$ letters: $M=2,\ A=2,\ H=1,\ E=1,\ I=1,\ C=1,\ S=1$

Number of distinct arrangements of these $9$ letters:

Using division method $\displaystyle \frac{9!}{2!,2!}$

Final answer: ${\frac{9!}{2!,2!} = 90720}$

Qus : 148 MCA NIMCET PYQ

The integral $\int \sqrt{1+2 cot x(cosec x+cotx)} dx$ , $(0<x<\frac{\pi}{2})$ (where C is a constant of integration) is equal to

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Solution

Qus : 149 MCA NIMCET PYQ

If area between $ y=x^{2} $ and $ y=x $ is $ A $, then area between $ y=x^{2} $ and $ y=1 $ is:

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Solution

$ A = \int_{0}^{1} (x - x^{2}), dx = \frac{1}{6} $ Area between $1$ and $x^{2}$: $ \int_{0}^{1} (1 - x^{2}) dx = \frac{2}{3} = 4A $

Qus : 150 MCA NIMCET PYQ

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Solution

Qus : 151 MCA NIMCET PYQ

If all the words, with or without meaning, are written using the letters of the word QUEEN add are arranged as in English Dictionary, then the position of the word QUEEN is

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Solution

Word: QUEEN (letters: E, E, N, Q, U). Arrange in dictionary order: E < N < Q < U.

1st position < Q:
- Start with E: permutations of {E,N,Q,U} = $ \frac{4!}{1!}=24 $
- Start with N: permutations of {E,E,Q,U} = $ \frac{4!}{2!}=12 $
Total before Q… = $24+12=36$
Fix Q, 2nd position < U:
- Q E _ _ _: perms of {E,N,U} = $ \frac{3!}{1!}=6 $
- Q N _ _ _: perms of {E,E,U} = $ \frac{3!}{2!}=3 $
More before QU… = $6+3=9$
Fix QU: Remaining {E,E,N}. Next letter is E (smallest), so add 0.
Fix QUE: Remaining {E,N}. Next letter is E (smallest), so add 0.

Total before “QUEEN” = $36+9=45$. Hence rank = $45+1=\boxed{46}$.

Qus : 152 MCA NIMCET PYQ

If $ a,b,c $ are coplanar, evaluate $ [,2a - b, 2b - c,2c - a,] $

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Solution

Scalar triple product of coplanar vectors $=0$.

Qus : 153 MCA NIMCET PYQ

Let $P(E)$ denote the probability of event $E$. Given $P(A) = 1$, $P(B) = \frac{1}{2}$, the values of $P(A \mid B)$ and $P(B \mid A)$ respectively are

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Solution

Since $P(A)=1$, event $A$ is a sure event.

Therefore for any event $B$,

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$

Also, $P(B \mid A)=\frac{P(A \cap B)}{P(A)}=\frac{P(B)}{1}=\frac{1}{2}$

Thus the correct pair is:

$P(A \mid B)=1$ and $P(B \mid A)=\frac{1}{2}$

Qus : 154 MCA NIMCET PYQ

The curve satisfying the differential equation ydx-(x+3y²)dy=0 and passing through the point (1,1) also passes through the point __________

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Solution

Given: $y\,dx-(x+3y^{2})\,dy=0 \;\Rightarrow\; y\,\dfrac{dx}{dy}=x+3y^{2}$.

So $ \dfrac{dx}{dy}-\dfrac{1}{y}x=3y$ (linear). Integrating factor $=\exp\!\int\!-\dfrac{1}{y}dy=\dfrac{1}{y}$.

$\displaystyle \frac{d}{dy}\!\left(\frac{x}{y}\right)=3 \;\Rightarrow\; \frac{x}{y}=3y+C \;\Rightarrow\; x=3y^{2}+Cy.$

Through $(1,1)$: $1=3(1)+C(1)\Rightarrow C=-2$. Hence curve: $x=3y^{2}-2y$.

Qus : 155 MCA NIMCET PYQ

$ \vec{a} = x\hat{i} - 3\hat{j} - \hat{k},\quad \vec{b} = 2x\hat{i} + x\hat{j} - \hat{k} $ Angle between $ \vec{a} $ and $ \vec{b} $ is acute and angle between $ \vec{b} $ and $ +y $ axis lies in $ \left(\dfrac{\pi}{2}, \pi\right) $ Find $x$.

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Solution

$ \vec{a}\cdot\vec{b} = 2x^{2} - 3x + 1 > 0 \Rightarrow x < \frac12 \text{ or } x > 1 $ Angle with $+y$ axis obtuse: $ \cos\theta = \dfrac{x}{\sqrt{5x^{2}+1}} < 0 \Rightarrow x < 0 $ Combined: $ x < 0 $

Qus : 156 MCA NIMCET PYQ

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Solution

Qus : 157 MCA NIMCET PYQ

$\lim_{x\to3} \dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ is equal to

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Solution

Rationalize numerator and denominator:

\[ \frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} =\frac{\dfrac{3(x-3)}{\sqrt{3x}+3}}{\dfrac{2(x-3)}{\sqrt{2x-4}+\sqrt{2}}} =\frac{3}{\sqrt{3x}+3}\cdot\frac{\sqrt{2x-4}+\sqrt{2}}{2}. \]

Now let $x\to 3$: $\sqrt{3x}\to 3$ and $\sqrt{2x-4}\to \sqrt{2}$. Hence

\[ \lim_{x\to 3}=\frac{3}{3+3}\cdot\frac{\sqrt{2}+\sqrt{2}}{2} =\frac{1}{2}\cdot\sqrt{2}=\boxed{\frac{1}{2\sqrt{2}}}. \]

Qus : 158 MCA NIMCET PYQ

Lines $2x + 3y - 6 = 0$ and $9x + 6y - 18 = 0$ cut coordinate axes in concyclic points. Center of circle is:

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Solution

Axis intercepts of both lines lie on same circle. Correct center is $ \left(\dfrac52,\dfrac52\right) $.

Qus : 159 MCA NIMCET PYQ

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Solution

Qus : 160 MCA NIMCET PYQ

If S and S' are foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, B is the end of the minor axis and BSS' is an equilateral triangle, then the eccentricity of the ellipse is

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Solution

The foci are at: $$S(ae,0), \quad S'(-ae,0)$$ and the end of the minor axis is: $$B(0,b), \quad \text{where } b^2 = a^2(1-e^2).$$

In an equilateral triangle ∆BSS′: $$BS = SS'.$$

Now, $$BS = \sqrt{(ae)^2 + b^2}, \quad SS' = 2ae.$$ Hence, $$\sqrt{a^2e^2 + b^2} = 2ae.$$

But, $$b^2 = a^2(1-e^2).$$

So, $$\sqrt{a^2e^2 + a^2(1-e^2)} = 2ae,$$ $$\sqrt{a^2} = 2ae,$$ $$a = 2ae \;\;\Rightarrow\;\; e = \tfrac{1}{2}.$$

Final Answer: The eccentricity of the ellipse is $$\boxed{\tfrac{1}{2}}.$$

Qus : 175 MCA NIMCET PYQ

Find the number of elements in the union of 4 sets A, B, C and D having 150, 180, 210 and 240 elements respectively, given that each pair of sets has 15 elements in common. Each triple of sets has 3 elements in common and $A \cap B \cap C \cap D = \phi$

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Solution

Given: $|A|=150,\ |B|=180,\ |C|=210,\ |D|=240$; every pair has 15 common elements; every triple has 3 common elements; and $A\cap B\cap C\cap D=\varnothing$.

Use Inclusion–Exclusion for 4 sets:

\[ \begin{aligned} |A\cup B\cup C\cup D| &= \sum |A_i| \;-\; \sum |A_i\cap A_j| \;+\; \sum |A_i\cap A_j\cap A_k| \;-\; |A\cap B\cap C\cap D|\\ &= (150+180+210+240)\;-\; \binom{4}{2}\cdot 15 \;+\; \binom{4}{3}\cdot 3 \;-\; 0\\ &= 780 \;-\; 6\cdot 15 \;+\; 4\cdot 3\\ &= 780 - 90 + 12\\ &= \boxed{702}. \end{aligned} \]

Answer: 702

Let $P(\text{odd}) = 3k$ and $P(\text{even}) = k$.

Since $P(\text{odd}) + P(\text{even}) = 1$:

$3k + k = 1$

$4k = 1$

$k = \frac{1}{4}$

So, $P(\text{odd}) = \frac{3}{4}$ $P(\text{even}) = \frac{1}{4}$

The sum of two numbers is even if:

1. both are odd, or

2. both are even.

Therefore, $P(\text{even sum}) = P(\text{odd})^2 + P(\text{even})^2$ $= \left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)^2$

$= \frac{9}{16} + \frac{1}{16}$

$= \frac{10}{16}$

$= \frac{5}{8}$

So the probability that the sum is even is $\frac{5}{8}$.

Person A can hit the target with probability $\frac{4}{5}$

Person B with probability $\frac{3}{4}$

Person C with probability $\frac{2}{3}$

They fire together.

We need:

Probability(target is hit at least two times)

This means: exactly 2 hits OR exactly 3 hits.

Let $p_A=\frac{4}{5}$, $p_B=\frac{3}{4}$, $p_C=\frac{2}{3}$ and miss probabilities:

$q_A=\frac{1}{5}$, $q_B=\frac{1}{4}$, $q_C=\frac{1}{3}$.

1.Exactly 2 hits:

-A and B hit, C misses: $p_A p_B q_C$

- A and C hit, B misses: $p_A q_B p_C$

- B and C hit, A misses: $q_A p_B p_C$

So, $P(\text{exactly 2}) = p_A p_B q_C + p_A q_B p_C + q_A p_B p_C$

$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \;+\; \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{2}{3} \;+\; \frac{1}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$

$= \frac{1}{5} + \frac{2}{15} + \frac{2}{20}$

$= \frac{4}{20} + \frac{8}{60} + \frac{6}{60}$

$= \frac{12}{60} + \frac{14}{60}$

$= \frac{26}{60} = \frac{13}{30}$

2. Exactly 3 hits: $P(\text{all hit}) = p_A p_B p_C$

$= \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3}$

$= \frac{2}{5}$

Therefore,

$P(\text{at least 2 hits}) = P(\text{exactly 2}) + P(\text{exactly 3})$

$= \frac{13}{30} + \frac{2}{5}$

$= \frac{13}{30} + \frac{12}{30}$

$= \frac{25}{30}$

$= \frac{5}{6}$

Final Answer: $\frac{5}{6}$

Qus : 196 MCA NIMCET PYQ

If $ |\vec{a}\times \vec{b}| = |\vec{a}\cdot \vec{b}| $, then angle $\theta$ between $\vec{a},\vec{b}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ |\vec{a}\times \vec{b}| = ab\sin\theta $ $ |\vec{a}\cdot \vec{b}| = ab|\cos\theta| $ Equation: $ \sin\theta = |\cos\theta| $ Thus: $ \tan\theta = 1 $ In first quadrant: $ \theta = \pi/4 $

Qus : 197 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 198 MCA NIMCET PYQ

The $sin^2 x tanx + cos^2 x cot x-sin2x=1+tanx+cotx $, $x \in (0 , \pi)$, then x

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 207 MCA NIMCET PYQ

If the vector $2\hat{i}-3\hat{j}$ , $\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}-\hat{k}$ form three conterminous edges of a parallelepiped, then thevolume of parallelepiped is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 208 MCA NIMCET PYQ

Triangle sides: $x^{2}+x+1,\ 2x+1,\ x^{2}-1$. Largest angle?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Largest side ≈ $x^{2}+x+1$. Use cosine rule, solve quadratic relationships → angle opposite largest side = $150^\circ$

Qus : 209 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 210 MCA NIMCET PYQ

Suppose A₁, A₂, ... ₃₀ are thirty sets, each with five elements and B₁, B₂, ...., B_n are n sets each with three elements. Let $\bigcup_{i=1}^{30} A_i= \bigcup_{j=1}^{n} Bj= S$. If each element of S belongs to exactly ten of the Ai' s and exactly nine of the Bj' s then n=

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Let $|S|=m$.

Count incidences via the $A_i$: There are 30 sets each of size 5, so total memberships $=30\times 5=150$. Each element of $S$ lies in exactly 10 of the $A_i$, so also $= m\times 10$. Hence $m=\dfrac{150}{10}=15$.

Count incidences via the $B_j$: There are $n$ sets each of size 3, so total memberships $=n\times 3$. Each element of $S$ lies in exactly 9 of the $B_j$, so also $= m\times 9 = 15\times 9=135$.

Thus $n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.$

Qus : 211 MCA NIMCET PYQ

In a G.P. consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of the G.P. is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 212 MCA NIMCET PYQ

Solve: $ 2\sin^{2}\theta - 3\sin\theta - 2 = 0$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Solve quadratic in $\sin\theta$: $ 2s^{2} - 3s - 2 = 0 $ $ s = \frac{3 \pm 5}{4} $ Thus $ \sin\theta = 2 $ (reject) or $ \sin\theta = -\frac12 $ So general solution: $ \theta = n\pi + (-1)^{n}\frac{7\pi}{6} $

Qus : 213 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 214 MCA NIMCET PYQ

Let

,where [x]denotes the greatest integer

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 215 MCA NIMCET PYQ

If $f(x)=tan^{-1}\left [ \frac{sinx}{1+cosx} \right ]$ , then what is the first derivative of $f(x)$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 216 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 217 MCA NIMCET PYQ

Let U and V be two events of a sample space S and P(A) denote the probability of an event A. Which of the following statements is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Statement 1 is false:

Two events can have equal probability without being identical.

Statement 2 is false:

If $P(U)=0$, it only means $U$ is a null-probability event, not that $U^{c}=S$.

Statement 3 is incomplete, so it cannot be true.

Statement 4 is true:

If $U$ and $V$ are independent,
then $P(U \cap V) = P(U)P(V)$
We check independence of complements:

$P(U^{c} \cap V^{c}) = 1 - P(U) - P(V) + P(U \cap V)$

$= 1 - P(U) - P(V) + P(U)P(V)$

$= (1 - P(U))(1 - P(V))$

$= P(U^{c}) P(V^{c})$

Hence, $U^{c}$ and $V^{c}$ are independent.

Qus : 218 MCA NIMCET PYQ

The solution of $\sin x +1 = \cos x $ such that $0\leq x\leq 2\pi$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 219 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 220 MCA NIMCET PYQ

If a man purchases a raffle ticket, he can win a first prize of Rs.5,000 or a second prize of Rs.2,000 with probabilities 0.001 and 0.003 respectively. What should be a fair price to pay for the ticket?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 221 MCA NIMCET PYQ

Let $T_n$ denote the number of triangles which can be formed by using the vertices of a regular polygon of $n$ sides. If $T_{n+1} - T_{n} = 21$ then $n$ equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

$T_{n+1} - T_n = 21$

$\binom{n+1}{3} - \binom{n}{3} = 21$

$\frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 21$

$\frac{n(n-1)}{6}\left[(n+1)-(n-2)\right] = 21$

$\frac{n(n-1)}{6}\times 3 = 21$

$\frac{n(n-1)}{2} = 21$

$n(n-1) = 42$

$n^2 - n - 42 = 0$

$(n-7)(n+6) = 0$

$n = 7$

Qus : 222 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 223 MCA NIMCET PYQ

If the mean deviation 1, 1+d, 1+2d, … , 1+100d from their mean is 255, then d is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 224 MCA NIMCET PYQ

If $\overline{X_1}$ and $\overline{X_2}$ are the means of two distributions such that $\overline{X_1} < \overline{X_2}$ and $\overline{X}$ is the mean of the combined distribution, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 225 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 226 MCA NIMCET PYQ

and

, then a possible value of n is among the following is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 227 MCA NIMCET PYQ

The area enclosed within the curve $|X|+|Y| = 1$ (in square units) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 228 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 229 MCA NIMCET PYQ

Let S be the set $\{a\in Z^+:a\leq100\}$.If the equation $[tan^2 x]-tan x - a = 0$ has real roots (where [ . ] is the greatest integer function), then the number of elements is S is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 230 MCA NIMCET PYQ

Let $f(x)$ be a polynomial function of second degree and $f(1) = f(–1)$. If $a, b, c$ are in A.P. then $f'(a), f'(b), f'(c)$ are in.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 231 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 232 MCA NIMCET PYQ

The solution set of the inequality

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 233 MCA NIMCET PYQ

Find the point at which, the tangent to the curve $y=\sqrt{4x-3}-1$ as its slope $\frac{2}{3}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Qus : 237 MCA NIMCET PYQ

Not Available

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 238 MCA NIMCET PYQ

A man observes the angle of elevation of the top of mountain to be 30^o. He walks 1000 feet nearer and finds the angle of elevation to be $45^{o}$. What is the distance of the first point of observation from the foot of the mountain?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Qus : 242 MCA NIMCET PYQ

Force $3\hat{i}+2\hat{j}+5\hat{k}$ and $2\hat{i}+\hat{j}-3\hat{k}$ are acting on a particle and displace it from the point $2\hat{i}-\hat{j}-3\hat{k}$ to $4\hat{i}-3\hat{j}+7\hat{k}$ the point then the work done by the force is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Qus : 258 MCA NIMCET PYQ

If C is the midpoint of AB and P is any point outside AB, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 259 MCA NIMCET PYQ

The number if non –negative integers less than 1000 that contain the digit 1 are.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 260 MCA NIMCET PYQ

Let $\vec{a}, \vec{b}, \vec{c} $ be distinct non-negative numbers. If the vectors $a\hat{i}+a\hat{j}+c\hat{k}$ , $\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ lie in a plane, then c is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

$\vec{a}=a\hat{i}+a\hat{j}+c\hat{k}\, ,\, \vec{b}=\hat{i}+\hat{k}\, \&\, \vec{c}=c\hat{i}+c\hat{j}+b\hat{k}$ are coplanar.

$\Rightarrow\begin{vmatrix}{a} & {a} & {c} \\ {1} & {0} & {1} \\ {c} & {c} & {b}\end{vmatrix}=0$

$\Rightarrow-ac-ab+ac+{c}^2=0$

$\Rightarrow{c}^2=ab$

Qus : 261 MCA NIMCET PYQ

A particle P starts from the point z₀=1+2i, where i=√−1 . It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z₁. From z₁ the particle moves √2 units in the direction of the vector $\hat{i}+\hat{j}$ and then it moves through an angle $\dfrac{\pi}{2}$ in anticlockwise direction on a circle with centre at origin, to reach a point z₂. The point z₂ is given by

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 262 MCA NIMCET PYQ

The average marks of boys in class is 52 and that of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 263 MCA NIMCET PYQ

The lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangent to the same circle. The radius of the this circle is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 264 MCA NIMCET PYQ

The correct expression for $cos^{-1} (-x)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

You should learn it as an important formula.

Qus : 265 MCA NIMCET PYQ

If (4, 3) and (12, 5) are the two foci of an ellipse passing through the origin, then the eccentricity of the ellipse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

For any point $ P $ on the ellipse: $ PF_1 + PF_2 = 2a $

$$OF_1 = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{25} = 5$$

$$\boxed{e = \dfrac{\sqrt{17}}{9}}$$

Qus : 266 MCA NIMCET PYQ

If $\Delta=a^2-(b-c)^2$, where $\Delta$ is the are of the triangle ABC, then $tanA=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 267 MCA NIMCET PYQ

The number of 5 people groups that can be selected from 9 people when two particular persons are not to be in the same group is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 268 MCA NIMCET PYQ

The area of the parallelogram whose diagonals are $\vec{a}=3\hat{i}+\hat{j}-2\hat{k}$ and $\vec{b}=\hat{i}-3\hat{j}+4\hat{k}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 269 MCA NIMCET PYQ

The number of one - one functions f: {1,2,3} → {a,b,c,d,e} is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3) = 5 \times 4 \times 3 = 60$

✅ Final Answer: $\boxed{60}$

Qus : 270 MCA NIMCET PYQ

Two numbers $a$ and $b$ are chosen are random from a set of the first 30 natural numbers, then the probability that $a^2 - b^2$ is divisible by 3 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 271 MCA NIMCET PYQ

The solution set of equation log_x2 log_2x 2 = log_4x 2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 272 MCA NIMCET PYQ

If $sin x + a cos x = b$, then what is the expression for $|a sin x – cos x|$ in terms of $a$ and $b$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 273 MCA NIMCET PYQ

Suppose that the temperature at a point (x,y), on a metal plate is $T(x,y)=4x^2-4xy+y^2$, An ant, walking on the plate, traverses a circle of radius 5 centered at the origin. What is the highest temperature encountered by the ant?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 274 MCA NIMCET PYQ

The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

We need to find: $\displaystyle \lim_{x\to 0}\left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}$

Rewrite each term: $k^x = e^{x\ln k}$

So: $1^x + 2^x + 3^x + 4^x = e^{0} + e^{x\ln 2} + e^{x\ln 3} + e^{x\ln 4}$

As $x \to 0$ use expansion: $e^{x\ln k} = 1 + x\ln k + O(x^2)$

So: $1^x + 2^x + 3^x + 4^x$

$= 1 + (1 + x\ln 2) + (1 + x\ln 3) + (1 + x\ln 4)$ $= 4 + x(\ln 2 + \ln 3 + \ln 4) + O(x^2)$

Thus: $\frac{1^x + 2^x + 3^x + 4^x}{4} = 1 + \frac{x(\ln 2 + \ln 3 + \ln 4)}{4} + O(x^2)$

Now apply the standard limit:

$\lim_{x\to 0} (1 + kx)^{1/x} = e^{k}$

Here: $k = \frac{\ln 2 + \ln 3 + \ln 4}{4}$

Hence the limit is: $e^{(\ln 2 + \ln 3 + \ln 4)/4}$

Combine logs: $\ln 2 + \ln 3 + \ln 4 = \ln(2\cdot 3 \cdot 4) = \ln 24$

Final answer: $\boxed{24^{1/4}}$

Qus : 275 MCA NIMCET PYQ

The equation of a circle with diameters are 2x – 3y + 12 = 0 and x + 4y – 5 = 0 and area of 154 sq. units is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 276 MCA NIMCET PYQ

If A and B are two events such that $P(A \cup B)=\frac{5}{6}$ , $P(A \cap B)=\frac{1}{3}$ and $P(\overline{B})=\frac{1}{2}$, then the events A and B are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 277 MCA NIMCET PYQ

The 10th and 50th percentiles of the observation 32, 49, 23, 29, 118 respectively are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 278 MCA NIMCET PYQ

The value of m for which volume of the parallelepiped is 4 cubic units whose three edges are represented by a = mi + j + k, b = i – j + k, c = i + 2j –k is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given: Volume of a parallelepiped formed by vectors $\vec{a}, \vec{b}, \vec{c}$ is 4 cubic units.

Vectors:

$\vec{a} = m\hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} - \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$

Step 1: Volume = $|\vec{a} \cdot (\vec{b} \times \vec{c})|$

First compute $\vec{b} \times \vec{c}$:

$ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}((-1)(-1) - (1)(2)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(2) - (-1)(1)) \\ = \hat{i}(1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 + 1) = -\hat{i} + 2\hat{j} + 3\hat{k} $

Step 2: Compute dot product with $\vec{a}$:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = (m)(-1) + (1)(2) + (1)(3) = -m + 2 + 3 = -m + 5$

Step 3: Volume = $| -m + 5 | = 4$

So, $|-m + 5| = 4 \Rightarrow -m + 5 = \pm 4$

Case 1: $-m + 5 = 4 \Rightarrow m = 1$
Case 2: $-m + 5 = -4 \Rightarrow m = 9$

✅ Final Answer: $\boxed{m = 1 \text{ or } 9}$

Qus : 279 MCA NIMCET PYQ

is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 280 MCA NIMCET PYQ

If there vectors $2\hat{i}-\hat{j}+\hat{k}$ , $\hat{i}+2\hat{j}-3\hat{k}$ and $3\hat{i}+\lambda \hat{j}+5\hat{k}$ are coplanar, then $\lambda$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 281 MCA NIMCET PYQ

Angles of elevation of the top of a tower from three points (collinear) A, B and C on a road leading to the foot of the tower are 30°, 45° and 60° respectively. The ratio of AB and BC is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 282 MCA NIMCET PYQ

The number of distinct real values of $\lambda$ for which the vectors ${\lambda}^2\hat{i}+\hat{j}+\hat{k},\, \hat{i}+{\lambda}^2\hat{j}+j$ and $\hat{i}+\hat{j}+{\lambda}^2\hat{k}$ are coplanar is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given: Vectors:

$\vec{a} = \lambda^2 \hat{i} + \hat{j} + \hat{k}$
$\vec{b} = \hat{i} + \lambda^2 \hat{j} + \hat{k}$
$\vec{c} = \hat{i} + \hat{j} + \lambda^2 \hat{k}$

Condition: Vectors are coplanar ⟹ Scalar triple product = 0

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Step 1: Use determinant:

$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \lambda^2 & 1 & 1 \\ 1 & \lambda^2 & 1 \\ 1 & 1 & \lambda^2 \end{vmatrix} $

Step 2: Expand the determinant:

$ = \lambda^2(\lambda^2 \cdot \lambda^2 - 1 \cdot 1) - 1(1 \cdot \lambda^2 - 1 \cdot 1) + 1(1 \cdot 1 - \lambda^2 \cdot 1) \\ = \lambda^2(\lambda^4 - 1) - (\lambda^2 - 1) + (1 - \lambda^2) $

Simplify:

$= \lambda^6 - \lambda^2 - \lambda^2 + 1 + 1 - \lambda^2 = \lambda^6 - 3\lambda^2 + 2$

Step 3: Set scalar triple product to 0:

$\lambda^6 - 3\lambda^2 + 2 = 0$

Step 4: Let $x = \lambda^2$, then:

$x^3 - 3x + 2 = 0$

Factor:

$x^3 - 3x + 2 = (x - 1)^2(x + 2)$

So, $\lambda^2 = 1$ (double root), or $\lambda^2 = -2$ (discard as it's not real)

Thus, real values of $\lambda$ are: $\lambda = \pm1$

✅ Final Answer: $\boxed{2}$ distinct real values

Qus : 283 MCA NIMCET PYQ

, and

are vectors of magnitude α then the magnitude of the vector

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 284 MCA NIMCET PYQ

The equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, –1). The length of the side of the triangle is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 285 MCA NIMCET PYQ

If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ are coincide, then the value of $b^2$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 286 MCA NIMCET PYQ

There are 9 bottle labelled 1, 2, 3, ... , 9 and 9 boxes labelled 1, 2, 3,....9. The number of ways one can put these bottles in the boxes so that each box gets one bottle and exactly 5 bottles go in their corresponding numbered boxes is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Total bottles and boxes: 9 each, labeled 1 to 9.

We are asked to count permutations of bottles such that exactly 5 bottles go into their own numbered boxes.

Step 1: Choose 5 positions to be fixed points (i.e., bottle number matches box number).

Number of ways = $\binom{9}{5}$

Step 2: Remaining 4 positions must be a derangement (no bottle goes into its matching box).

Let $D_4$ be the number of derangements of 4 items.

$D_4 = 9$

Step 3: Total ways = $\binom{9}{5} \times D_4 = 126 \times 9 = 1134$

✅ Final Answer: $\boxed{1134}$

Qus : 287 MCA NIMCET PYQ

A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap. If four caps are picked at random, the probability that none of them is green is

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Solution

Qus : 288 MCA NIMCET PYQ

The total number of numbers that can be formed using the digits 3,5 and 7 only if no repetitions are allowed, is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Step 4: Use point-slope form for perpendicular bisector:

$ y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right) $

Step 5: Find y-intercept (put $ x = 0 $)

$ y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right) $

$ y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right) $

Given: y-intercept = -4, so:

$ \dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4 $

Multiply both sides by 2:

$ 7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8 $

$ \Rightarrow k^2 = 16 \Rightarrow k = \pm4 $

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

Qus : 291 MCA NIMCET PYQ

The line 3x + 5y = k touches the ellipse 16 x2 + 25 y2 = 400, if k is

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Solution

Qus : 292 MCA NIMCET PYQ

If x = a cos t, y = b sin t, then $\frac{d^{2}y}{dx^{2}}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 293 MCA NIMCET PYQ

If $a{\lt}b$ then $\int ^b_a\Bigg{(}|x-a|+|x-b|\Bigg{)}dx$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 294 MCA NIMCET PYQ

Let C denote the set of all tuples (x,y) which satisfy $x^2 -2^y=0$ where x and y are natural numbers. What is the cardinality of C?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 295 MCA NIMCET PYQ

A random variable X has the distribution law as given below:

X	1	2	3
P(X=x)	0.3	0.4	0.3

The variance of the distribution is

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Solution

Qus : 296 MCA NIMCET PYQ

The domain of the function $f(x)=\frac{{\cos }^{-1}x}{[x]}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 297 MCA NIMCET PYQ

If $x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then ${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of $ a = 2^{1/6} $

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity $ a^6 = 2 \Rightarrow a^5 = \frac{2}{a} $

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$

Qus : 298 MCA NIMCET PYQ

The value of $\tan\theta+2\tan2\theta + 4\tan4\theta + 8\cot8\theta$ is

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Solution

Qus : 299 MCA NIMCET PYQ

If the volume of the parallelepiped whose adjacent edges are $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}+\alpha \hat{k}$ is 15, then $\alpha$ is equal to

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Solution

Qus : 300 MCA NIMCET PYQ

The number of solutions of ${5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25$ for $x\in(-\mathrm{\pi},\mathrm{\pi})$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Step 1: Recognize the series

This is a geometric series with first term $ a = 1 $, common ratio $ r = |\sin x| \in [0,1] $, so: $$ \text{Sum} = \frac{1}{1 - |\sin x|} $$

Step 2: Rewrite the equation

$$ 5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2 $$

Equating exponents: $$ \frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2} $$

Step 3: Solve for $ x \in (-\pi, \pi) $

We want all $ x \in (-\pi, \pi) $ such that $ |\sin x| = \frac{1}{2} $

So $ \sin x = \pm \frac{1}{2} $. Within $ (-\pi, \pi) $, the values of $ x $ satisfying this are:

$x = \frac{\pi}{6}$
$x = \frac{5\pi}{6}$
$x = -\frac{\pi}{6}$
$x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Qus : 301 MCA NIMCET PYQ

The sum of integers between 200 and 400, that are multiples of 7 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Step 1: Find the first multiple of 7 after 200
$200 \div 7 = 28.57...$ so first multiple $= 29 \times 7 = 203$

Step 2: Find the last multiple of 7 before 400
$400 \div 7 = 57.14...$ so last multiple $= 57 \times 7 = 399$

Step 3: Identify the AP
First term $a = 203$
Last term $l = 399$
Common difference $d = 7$

Step 4: Find number of terms $n$
$l = a + (n-1)d$
$399 = 203 + (n-1) \times 7$
$399 - 203 = (n-1) \times 7$
$196 = (n-1) \times 7$
$n - 1 = 28$
$n = 29$

Step 5: Find the Sum
$S_n = \dfrac{n}{2}(a + l)$

$S_{29} = \dfrac{29}{2}(203 + 399)$

$S_{29} = \dfrac{29}{2} \times 602$

$S_{29} = 29 \times 301$

$S_{29} = 8729$

Answer: The sum of all multiples of 7 between 200 and 400 is $\boxed{8729}$

Qus : 302 MCA NIMCET PYQ

Let $a$ be the distance between the lines $−2x + y = 2$ and $2x − y = 2$, and $b$ be the distance between the lines $4x − 3y= 5$ and $6y − 8x = 1$, then

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Solution

Qus : 303 MCA NIMCET PYQ

The system of equations $x+2y+2z=5$, $x+2y+3z=6$, $x+2y+\lambda z=\mu$ has infinitely many solutions if

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Solution

Given System of Equations:

$x + 2y + 2z = 5$
$x + 2y + 3z = 6$
$x + 2y + \lambda z = \mu$

Goal: Find values of $\lambda$ and $\mu$ such that the system has infinitely many solutions

Step 1: Write Augmented Matrix

$ [A|B] = \begin{bmatrix} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \end{bmatrix} $

Step 2: Row operations: Subtract $R_1$ from $R_2$ and $R_3$

$ \Rightarrow \begin{bmatrix} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \end{bmatrix} $

Step 3: For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < number of variables (3)

This happens when the third row becomes all zeros:

$ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 $

$\Rightarrow \lambda = 2,\quad \mu = 5$

✅ Final Answer: $\boxed{\lambda = 2,\ \mu = 5}$

Now count outcomes for each prime sum:

Sum = 2 → $(1,1)$ → 1 way

Sum = 3 → $(1,2),(2,1)$ → 2 ways

Sum = 5 → $(1,4),(2,3),(3,2),(4,1)$ → 4 ways

Sum = 7 → $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ → 6 ways

Sum = 11 → $(5,6),(6,5)$ → 2 ways

Total favourable outcomes: $1 + 2 + 4 + 6 + 2 = 15$

Total outcomes when 2 dice are rolled: $36$

Therefore, $P(\text{prime sum}) = \frac{15}{36} = \frac{5}{12}$

Final Answer: $\frac{5}{12}$

Qus : 308 MCA NIMCET PYQ

The solution of the equation ${4\cos }^2x+6{\sin }^2x=5$ are

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Solution

Qus : 309 MCA NIMCET PYQ

If F|= 40N (Newtons), |D| = 3m, and $\theta={60^{\circ}}$, then the work done by F acting

from P to Q is

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Solution

Formula for work done:

\[ W = |F| \cdot |D| \cdot \cos\theta \]

Given:

$ |F| = 40 \, \text{N} $
$ |D| = 3 \, \text{m} $
$ \theta = 60^\circ $

Step 1: Plug in the values:

\[ W = 40 \cdot 3 \cdot \cos(60^\circ) \]

Step 2: Use $ \cos(60^\circ) = \frac{1}{2} $

\[ W = 40 \cdot 3 \cdot \frac{1}{2} = 60 \, \text{J} \]

✅ Final Answer: $\boxed{60 \, \text{J}}$

Qus : 310 MCA NIMCET PYQ

Find the equation of the circle which passes through (–1, 1) and (2, 1), and having centre on the line x + 2y + 3 = 0 .

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 311 MCA NIMCET PYQ

The function $f(x)=\begin{cases}{{(1+2x)}^{1/x}} & {,x\ne0} \\ {{e}^2} & {,x=0}\end{cases}$, is

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Solution

Qus : 312 MCA NIMCET PYQ

A committee of 5 is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Total people: 9

Married couple: 2 specific people among them

Total ways to choose 5 people from 9:

\[ \text{Total} = \binom{9}{5} = 126 \]

✅ Case 1: Both are selected

We fix the married couple (2 people), then choose 3 more from remaining 7:

\[ \binom{7}{3} = 35 \]

✅ Case 2: Both are NOT selected

Solution

We are given:

\[ \sin(4x) = \frac{1}{2}, \quad x \in (-9\pi,\ 3\pi) \]

General solutions for $ \sin(θ) = \frac{1}{2} $

\[ θ = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad θ = \frac{5\pi}{6} + 2n\pi \]

Let $ θ = 4x $, so we get:

$ x = \frac{\pi}{24} + \frac{n\pi}{2} $
$ x = \frac{5\pi}{24} + \frac{n\pi}{2} $

Count how many such $ x $ fall in the interval $ (-9\pi, 3\pi) $

By checking all possible $ n $ values, we find:

For $ x = \frac{\pi}{24} + \frac{n\pi}{2} $: 24 valid values
For $ x = \frac{5\pi}{24} + \frac{n\pi}{2} $: 24 valid values

Total distinct values = 24 + 24 = 48

✅ Final Answer: $\boxed{48}$

Qus : 316 MCA NIMCET PYQ

The sum of the focal distances of any point on the ellipse $\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1$ with eccentricity $e$ is given by

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Solution

Qus : 317 MCA NIMCET PYQ

Which term of the series $\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is $\frac{\sqrt{5}}{13}$ ?

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Solution

Step 1: Rewrite all terms with same numerator
$\dfrac{\sqrt{5}}{3},\ \dfrac{\sqrt{5}}{4},\ \dfrac{\sqrt{5}}{5},\ \ldots$

(Since $\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$)

Step 2: Identify the AP in denominators
Denominators: $3,\ 4,\ 5,\ \ldots$
First term $a = 3$
Common difference $d = 1$

Step 3: Set up the equation
The $n^{th}$ term of the series $= \dfrac{\sqrt{5}}{a + (n-1)d}$

Given $n^{th}$ term $= \dfrac{\sqrt{5}}{13}$

So denominator of $n^{th}$ term $= 13$

Step 4: Solve for $n$
$a + (n-1)d = 13$
$3 + (n-1) \times 1 = 13$
$n - 1 = 10$
$n = 11$

Answer: $\dfrac{\sqrt{5}}{13}$ is the $\boxed{11^{th}}$ term of the series.

Qus : 318 MCA NIMCET PYQ

At how many points the following curves intersect $\frac{{y}^2}{9}-\frac{{x}^2}{16}=1$ and $\frac{{x}^2}{4}+\frac{{(y-4)}^2}{16}=1$

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Solution

Qus : 319 MCA NIMCET PYQ

If $\sin x + \sin^{2}x = 1$, then $\cos^{2}x + \cos^{4}x$ is equal to.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 320 MCA NIMCET PYQ

The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are

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Solution

Qus : 321 MCA NIMCET PYQ

If for non-zero x, $cf(x)+df\Bigg{(}\frac{1}{x}\Bigg{)}=|\log |x||+3,$ where $c\ne 0$, then $\int ^e_1f(x)dx=$

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Solution

Qus : 322 MCA NIMCET PYQ

An experiment succeeds twice often as it fails. The probability that in the next six trials there will be at least four successes is.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

An experiment succeeds twice as often as it fails.

Let success probability = $p$

Then failure probability = $1-p$

Given: $p = 2(1-p)$ $\Rightarrow p = 2 - 2p$ $\Rightarrow 3p = 2$ $\Rightarrow p = \frac{2}{3}$

So failure probability = $\frac{1}{3}$.

We perform 6 independent trials.

Let $X$ = number of successes.

We need: $P(X \ge 4)$.

$P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$

Use binomial formula: $P(X=k) = {6 \choose k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{6-k}$

1. $P(X=4)$ $= {6 \choose 4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^2$ $= 15 \cdot \frac{16}{81} \cdot \frac{1}{9}$ $= 15 \cdot \frac{16}{729}$ $= \frac{240}{729}$

2. $P(X=5)$ $= {6 \choose 5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)$ $= 6 \cdot \frac{32}{243} \cdot \frac{1}{3}$ $= 6 \cdot \frac{32}{729}$ $= \frac{192}{729}$

3. $P(X=6)$ $= {6 \choose 6} \left(\frac{2}{3}\right)^6$ $= 1 \cdot \frac{64}{729}$ $= \frac{64}{729}$

Add them:

$P(X \ge 4) = \frac{240}{729} + \frac{192}{729} + \frac{64}{729}$

$= \frac{496}{729}$

Final Answer: $\frac{496}{729}$

Qus : 323 MCA NIMCET PYQ

A survey is done among a population of 200 people who like either tea or coffee. It is found that 60% of the pop lation like tea and 72% of the population like coffee. Let $x$ be the number of people who like both tea & coffee. Let $m{\leq x\leq n}$, then choose the correct option.

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Solution

Total people = 200
People who like tea = $60\% \times 200 = 120$
People who like coffee = $72\% \times 200 = 144$

Using the set formula: \[ |T \cup C| = |T| + |C| - |T \cap C| \] \[ 200 = 120 + 144 - x \quad \Rightarrow \quad x = 64 \]

Minimum possible intersection: \[ x \geq |T| + |C| - 200 = 64 \] Maximum possible intersection: \[ x \leq \min(120,144) = 120 \]

✅ Final Answer

The range of $x$ is: 64 ≤ x ≤ 120 Hence, $m = 64, \; n = 120$.

Qus : 324 MCA NIMCET PYQ

A critical orthopedic surgery is performed on 3 patients. The probability of recovering a patient is 0.6. Then the probability that after surgery, exactly two of them will recover is

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Solution

Given:

Number of patients (n): 3
Probability of recovery (p): 0.6
Probability of failure (q): 0.4
We are asked to find: Probability that exactly 2 out of 3 patients will recover.

Solution: We will use the Binomial Probability Formula to solve this.

Step 1: Binomial Probability Formula:

The formula is:
$$ P(X = r) = \binom{n}{r} p^r (1 - p)^{n - r} $$ Where:
$ n = 3 $, $ r = 2 $, $ p = 0.6 $, $ 1 - p = 0.4 $

Step 2: Calculate the binomial coefficient $ \binom{3}{2} $:

$$ \binom{3}{2} = \frac{3!}{2!(3-2)!} = 3 $$

Step 3: Substitute values into the formula:

$$ P(X = 2) = 3 \times (0.6)^2 \times (0.4)^1 = 3 \times 0.36 \times 0.4 = 0.432 $$

✅ Final Answer: 0.432

Qus : 325 MCA NIMCET PYQ

Sum of 20 terms of the series $–1^{2} + 2^{2} –3^{2} + 4^{2} – …$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Step 1: Group terms in pairs
$(-1^2 + 2^2) + (-3^2 + 4^2) + (-5^2 + 6^2) + \ldots$

Since 20 terms $\Rightarrow$ there are $\dfrac{20}{2} = 10$ pairs.

Step 2: Simplify each pair
Each pair is of the form $-(2k-1)^2 + (2k)^2$ where $k = 1, 2, 3, \ldots, 10$

$= (2k)^2 - (2k-1)^2$

Using $a^2 - b^2 = (a+b)(a-b)$:

$= (2k + 2k - 1)(2k - 2k + 1)$

$= (4k - 1)(1)$

$= 4k - 1$

Step 3: Sum all 10 pairs
$S = \displaystyle\sum_{k=1}^{10} (4k - 1)$

$= 4\displaystyle\sum_{k=1}^{10} k \ - \ \displaystyle\sum_{k=1}^{10} 1$

$= 4 \times \dfrac{10 \times 11}{2} - 10$

$= 4 \times 55 - 10$

$= 220 - 10$

$= 210$

Answer: Sum of 20 terms $= \boxed{210}$

Qus : 326 MCA NIMCET PYQ

The value of $\cot \Bigg{(}{cosec}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\Bigg{)}$ is

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Solution

Qus : 327 MCA NIMCET PYQ

The value of $\tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)}$ is

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Solution

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 328 MCA NIMCET PYQ

If $tan\alpha=\frac{m}{m+1}$ and $tan\beta=\frac{1}{2m+1}$ then $\alpha+\beta$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 329 MCA NIMCET PYQ

If $0{\lt}P(A){\lt}1$ and $0{\lt}P(B){\lt}1$ and $P(A\cap B)=P(A)P(B)$, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 330 MCA NIMCET PYQ

If $\sin x=\sin y$ and $\cos x=\cos y$, then the value of x-y is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given:

\[ \sin x = \sin y \quad \text{and} \quad \cos x = \cos y \]

✳ Step 1: Use the identity for sine

\[ \sin x = \sin y \Rightarrow x = y + 2n\pi \quad \text{or} \quad x = \pi - y + 2n\pi \]

✳ Step 2: Use the identity for cosine

\[ \cos x = \cos y \Rightarrow x = y + 2m\pi \quad \text{or} \quad x = -y + 2m\pi \]

? Combine both conditions

For both $ \sin x = \sin y $ and $ \cos x = \cos y $ to be true, the only consistent solution is:

\[ x = y + 2n\pi \Rightarrow x - y = 2n\pi \]

✅ Final Answer:

\[ \boxed{x - y = 2n\pi \quad \text{for } n \in \mathbb{Z}} \]

Qus : 331 MCA NIMCET PYQ

Which of the following is NOT true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 332 MCA NIMCET PYQ

For an invertible matrix A, which of the following is not always true:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 333 MCA NIMCET PYQ

$f(x)=x+|x|$ is continuous for

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 334 MCA NIMCET PYQ

For what values of $\lambda$ does the equation $6x^2 - xy + \lambda y^2 = 0$ represents two perpendicular lines and two lines inclined at an angle of $\pi/4$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 335 MCA NIMCET PYQ

If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is any positive integer, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

$$ \begin{aligned} &\textbf{To prove: } \quad n\sum_{i=1}^{n} a_i^2 \;\ge\; \left(\sum_{i=1}^{n} a_i\right)^2 \\[6pt] &\text{(RMS}\ge\text{AM)}:\quad \sqrt{\frac{\sum_{i=1}^{n} a_i^2}{\,n\,}} \;\ge\; \frac{\sum_{i=1}^{n} a_i}{\,n\,} \\[6pt] &\Rightarrow\; \frac{\sum_{i=1}^{n} a_i^2}{\,n\,} \;\ge\; \left(\frac{\sum_{i=1}^{n} a_i}{\,n\,}\right)^2 \;\Rightarrow\; n\sum_{i=1}^{n} a_i^2 \;\ge\; \left(\sum_{i=1}^{n} a_i\right)^2. \\[6pt] &\text{Equality iff } a_1=a_2=\cdots=a_n. \end{aligned} $$

Qus : 336 MCA NIMCET PYQ

A speaks truth in 40% and B in 50% of the cases. The probability that they contradict each other while narrating some incident is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

A speaks the truth in 40% of the cases and B in 50% of the cases.

What is the probability that they contradict each other while narrating an incident?

? Let’s Define:

$ P(A_T) = 0.4 $ → A tells the truth
$ P(A_L) = 0.6 $ → A lies
$ P(B_T) = 0.5 $ → B tells the truth
$ P(B_L) = 0.5 $ → B lies

? Contradiction happens in two cases:

A tells the truth, B lies → $ 0.4 \times 0.5 = 0.2 $
A lies, B tells the truth → $ 0.6 \times 0.5 = 0.3 $

Total probability of contradiction: \[ P(\text{Contradiction}) = 0.2 + 0.3 = \boxed{0.5} \]

✅ Final Answer:

\[ \boxed{\frac{1}{2}} \]

Qus : 337 MCA NIMCET PYQ

In a reality show, two judges independently provided marks base do the performance of the participants. If the marks provided by the second judge are given by Y = 10.5 + 2x, where X is the marks provided by the first judge. If the variance of the marks provided by the second judge is 100, then the variance of the marks provided by the first judge is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Statistics Puzzle: Variance under Linear Transformation

Given:

Y = 10.5 + 2X
Var(Y) = 100

Formula: If Y = a + bX, then Var(Y) = b² × Var(X)

Apply:

100 = 2² × Var(X)
100 = 4 × Var(X)
Var(X) = 100 / 4 = 25

✅ Final Answer: The variance of the marks given by the first judge is 25.

Qus : 338 MCA NIMCET PYQ

If $D={\begin{vmatrix}{1} & 1 & {1} \\ 1 & {2+x} & {1} \\ {1} & {1} & {2+y}\end{vmatrix}}\, for\, x\ne0,\, y\ne0$ then D is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 339 MCA NIMCET PYQ

The two parabolas $y^2 = 4a(x + c)$ and $y^2 = 4bx, a > b > 0$ cannot have a common normal unless

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 340 MCA NIMCET PYQ

Area of the parallelogram formed by the lines y=4x, y=4x+1, x+y=0 and x+y=1

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 341 MCA NIMCET PYQ

A man starts at the origin O and walks a distance of 3 units in the north- east direction and then walks a distance of 4 units in the north-west direction to reach the point P. then $\vec{OP}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

A man starts at the origin $ O $, walks 3 units in the north-east direction, then 4 units in the north-west direction to reach point $ P $. Find the displacement vector $ \vec{OP} $.

? Solution:

North-East (45°): \[ \vec{A} = 3 \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \right) \]
North-West (135°): \[ \vec{B} = 4 \cdot \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \left( -\frac{4}{\sqrt{2}}, \frac{4}{\sqrt{2}} \right) \]
Total Displacement: \[ \vec{OP} = \vec{A} + \vec{B} = \left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right) \]

✅ Final Answer:

\[ \boxed{ \vec{OP} = \left( \frac{-1}{\sqrt{2}},\ \frac{7}{\sqrt{2}} \right) } \]

Qus : 342 MCA NIMCET PYQ

A four-digit number is formed using the digits 1, 2, 3, 4, 5 without repetition. The probability that is divisible by 3 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 343 MCA NIMCET PYQ

Among the given numbers below, the smallest number which will be divided by 9, 10, 15 and 20, leaves the remainders 4, 5, 10, and 15, respectively

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Find the smallest number which when divided by 9, 10, 15 and 20 leaves remainders 4, 5, 10 and 15 respectively.

✅ Solution:

Let the number be $ x $.

$ x \equiv 4 \mod 9 \Rightarrow x - 4 $ divisible by 9
$ x \equiv 5 \mod 10 \Rightarrow x - 5 $ divisible by 10
$ x \equiv 10 \mod 15 \Rightarrow x - 10 $ divisible by 15
$ x \equiv 15 \mod 20 \Rightarrow x - 15 $ divisible by 20

So, $ x + 5 $ is divisible by LCM of 9, 10, 15, 20

LCM = $ 2^2 \cdot 3^2 \cdot 5 = 180 $

$ x + 5 = 180 \times 2 = 360 \Rightarrow x = 355 $

Final Answer: $ \boxed{355} $

Qus : 344 MCA NIMCET PYQ

For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)+(na+2)+\ldots+(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 345 MCA NIMCET PYQ

The value of $\sum ^n_{r=1}\frac{{{{}^nP}}_r}{r!}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Question: Find the value of:

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} $$

Solution:

We know: $ nP_r = \frac{n!}{(n - r)!} \Rightarrow \frac{nP_r}{r!} = \frac{n!}{(n - r)! \cdot r!} = \binom{n}{r} $

Therefore,

$$ \sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \binom{n}{r} = 2^n - 1 $$

Final Answer: $$ \boxed{2^n - 1} $$

Qus : 346 MCA NIMCET PYQ

If $\vec{a}=\lambda \hat{i}+\hat{j}-2\hat{k}$ , $\vec{b}=\hat{i}+\lambda \hat{j}-2\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ and $\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}} \end{bmatrix}=7$, then the values of the $\lambda$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 347 MCA NIMCET PYQ

Let A and B be two events defined on a sample space $\Omega$. Suppose $A^C$ denotes the complement of A relative to the sample space $\Omega$. Then the probability $P\Bigg{(}(A\cap{B}^C)\cup({A}^C\cap B)\Bigg{)}$ equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given: Two events $ A $ and $ B $ defined on sample space $ \Omega $. We are to find the probability:

$$ P\left((A \cap B^c) \cup (A^c \cap B)\right) $$

Step 1: This is the probability of events that are in exactly one of A or B (but not both), i.e., symmetric difference of A and B:

$$ (A \cap B^c) \cup (A^c \cap B) = A \Delta B $$

Step 2: So, we use:

$$ P(A \Delta B) = P(A) + P(B) - 2P(A \cap B) $$

Final Answer:

$$ \boxed{P(A) + P(B) - 2P(A \cap B)} $$

Qus : 348 MCA NIMCET PYQ

The function $f(x)=\log (x+\sqrt[]{{x}^2+1})$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 349 MCA NIMCET PYQ

Let Z be the set of all integers, and consider the sets $X=\{(x,y)\colon{x}^2+2{y}^2=3,\, x,y\in Z\}$ and $Y=\{(x,y)\colon x{\gt}y,\, x,y\in Z\}$. Then the number of elements in $X\cap Y$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given: $$x^2 + 2y^2 = 3 \text{ and } x > y \text{ with } x, y \in \mathbb{Z}$$

Solutions to the equation are: $$\{(1,1), (1,-1), (-1,1), (-1,-1)\}$$

Among them, only $ (1, -1) $ satisfies $ x > y $.

Answer: $$\boxed{1}$$

Qus : 350 MCA NIMCET PYQ

The mean of 25 observations was found to be 38. It was later discovered that 23 and 38 were misread as 25 and 36, then the mean is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 351 MCA NIMCET PYQ

The value of $f(1)$ for $f\Bigg{(}\frac{1-x}{1+x}\Bigg{)}=x+2$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given:
$$f\left(\frac{1 - x}{1 + x}\right) = x + 2$$

To Find: $ f(1) $

Let $ \frac{1 - x}{1 + x} = 1 \Rightarrow x = 0 $

Then, $ f(1) = f\left(\frac{1 - 0}{1 + 0}\right) = 0 + 2 = 2 $

Answer: $$\boxed{2}$$

Qus : 352 MCA NIMCET PYQ

The area enclosed within the curve |x|+|y|=2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 353 MCA NIMCET PYQ

Given a set A with median $m_1 = 2$ and set B with median $m_2 = 4$

What can we say about the median of the combined set?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given two sets:

Set $ A $ has median $ m_1 = 2 $
Set $ B $ has median $ m_2 = 4 $

What can we say about the median of the combined set $ A \cup B $?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

\[ \text{Combined Median} \in [2, 4] \]

So, the exact median cannot be determined with the given data.

Qus : 354 MCA NIMCET PYQ

If ${\Bigg{(}\frac{x}{a}\Bigg{)}}^2+{\Bigg{(}\frac{y}{b}\Bigg{)}}^2=1$, $(a{\gt}b)$ and ${x}^2-{y}^2={c}^2$ cut at right angles, then

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Solution

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ are orthogonal.

Then

$a^2-b^2=c^2-d^2$

Similarly

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=c^2$ are orthogonal.

It means

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{-c^2}=1$ are orthogonal

Then

$a^2-b^2=c^2-(-c^2)$

$a^2-b^2=2c^2$

Qus : 355 MCA NIMCET PYQ

Let $f(x)=\begin{cases}{{x}^2\sin \frac{1}{x}} & {,\, x\ne0} \\ {0} & {,x=0}\end{cases}$

Then which of the follwoing is true

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 356 MCA NIMCET PYQ

If $\alpha , \beta$ are the roots of $x^2-x-1=0$ and $A_n=\alpha^n+\beta^n$, the Arithmetic mean of $A_{n-1}$ and $A_n$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 357 MCA NIMCET PYQ

A coin is thrown 8 number of times. What is the probability of getting a head in an odd number of throw?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Total outcomes = $ 2^8 = 256 $

Favorable outcomes (odd heads):

$ \binom{8}{1} = 8 $
$ \binom{8}{3} = 56 $
$ \binom{8}{5} = 56 $
$ \binom{8}{7} = 8 $

Total favorable = $ 8 + 56 + 56 + 8 = 128 $

So, Probability = $ \frac{128}{256} = \boxed{\frac{1}{2}} $

? Final Answer: $ \boxed{\frac{1}{2}} $

Qus : 358 MCA NIMCET PYQ

If $a_1, a_2, a_3,...a_n$, are in Arithmetic Progression with common difference d, then the sum $(sind) (cosec a_1 . cosec a_2+cosec a_2.cosec a_2+...+cosec a_{n-1}.cosec a_n)$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 359 MCA NIMCET PYQ

Consider the function $f(x)={x}^{2/3}{(6-x)}^{1/3}$. Which of the following statement is false?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given Function: $$f(x) = x^{2/3}(6 - x)^{1/3}$$

f is increasing in (0, 4): ✅ True
f has a point of inflection at x = 0: ✅ True
f has a point of inflection at x = 6: ✅ True
f is decreasing in (6, ∞): ❌ False (function not defined there)

Qus : 365 MCA NIMCET PYQ

If one AM (Arithmetic mean) 'a' and two GM's (Geometric means) p and q be inserted between any two positive numbers, the value of p^3+q^3 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Problem:

If one Arithmetic Mean (AM) $ a $ and two Geometric Means $ p $ and $ q $ are inserted between any two positive numbers, find the value of: \[ p^3 + q^3 \]

Given:

Let two positive numbers be $ A $ and $ B $.
One AM: $ a = \frac{A + B}{2} $
Two GMs inserted: so the four terms in G.P. are: \[ A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B \]

Now calculate:

\[ pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB \]
\[ p^3 = A^2B, \quad q^3 = AB^2 \]
\[ p^3 + q^3 = A^2B + AB^2 = AB(A + B) \]

Also,

\[ 2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B) \]

✅ Therefore,

$ \boxed{p^3 + q^3 = 2apq} $

Qus : 366 MCA NIMCET PYQ

A straight line through the point (4, 5) is such that its intercept between the axes is bisected at A, then its equation is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 367 MCA NIMCET PYQ

The equation $3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0$ represents

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Rule for Classifying Conics Using Discriminant

Given the equation: $ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $

Compute: $ \Delta = B^2 - 4AC $

? Based on value of $ \Delta $:

Ellipse: $ \Delta < 0 $ and $ A \ne C $, $ B \ne 0 $ → tilted ellipse
Circle: $ \Delta < 0 $ and $ A = C $, $ B = 0 $
Parabola: $ \Delta = 0 $
Hyperbola: $ \Delta > 0 $

Example:

For the equation: $ 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 $

$ A = 3 $, $ B = 10 $, $ C = 11 $ →
$ \Delta = 10^2 - 4(3)(11) = 100 - 132 = -32 $

Since $ \Delta < 0 $, it represents an ellipse.

Qus : 368 MCA NIMCET PYQ

The value of $\int \frac{({x}^2-1)}{{x}^3\sqrt[]{2{x}^4-2{x}^2+1}}dx$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 369 MCA NIMCET PYQ

The points (1,1/2) and (3,-1/2) are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given:

Points: $ A = (1, \frac{1}{2}) $, $ B = (3, -\frac{1}{2}) $

Line: $ 2x + 3y = k $

Step 1: Evaluate $ 2x + 3y $

For A: $ 2(1) + 3\left(\frac{1}{2}\right) = \frac{7}{2} $
For B: $ 2(3) + 3\left(-\frac{1}{2}\right) = \frac{9}{2} $

✅ Option-wise Check:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $: ✔️ True since $ \frac{7}{2}, \frac{9}{2} \in (-6, 6) $
On the same side of $ 2x + 3y = 6 $: ✔️ True, both values are less than 6
On the same side of $ 2x + 3y = -6 $: ✔️ True, both values are greater than -6
On the opposite side of $ 2x + 3y = -6 $: ❌ False, both are on the same side

✅ Final Answer:

The correct statements are:

In between the lines $ 2x + 3y = -6 $ and $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = 6 $
On the same side of the line $ 2x + 3y = -6 $

Qus : 370 MCA NIMCET PYQ

Coordinate of the focus of the parabola $4y^2+12x-20y+67=0$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 371 MCA NIMCET PYQ

How much work does it take to slide a crate for a distance of 25m along a loading dock by pulling on it with a 180 N force where the dock is at an angle of $45°$ from the horizontal?

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Solution

Work Done Problem:

A crate is pulled 25 m along a dock with a force of 180 N at an angle of 45°.

✅ Formula Used:

\[ \text{Work} = F \cdot d \cdot \cos(\theta) \]

✅ Substituting Values:

\[ W = 180 \times 25 \times \cos(45^\circ) = 180 \times 25 \times 0.70710678118 = 3181.98052\, \text{J} \]

✅ Final Answer (to 5 decimal places):

\[ \boxed{3.181\times 10^3 \, \text{Joules}} \]

Qus : 372 MCA NIMCET PYQ

There are two circles in xy −plane whose equations are $x^2+y^2-2y=0$ and $x^2+y^2-2y-3=0$. A point $(x,y)$ is chosen at random inside the larger circle. Then the probability that the point has been taken from smaller circle is

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Solution

Qus : 373 MCA NIMCET PYQ

The vector $\vec{A}=(2x+1)\hat{i}+(x^2-6y)\hat{j}+(xy^2+3z)\hat{k}$ is a

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Vector Field:

\[ \vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k} \]

Divergence:

\[ \nabla \cdot \vec{A} = 2 - 6 + 3 = -1 \neq 0 \]

Not solenoidal ❌

Curl:

\[ \nabla \times \vec{A} = (2xy)\hat{i} - (y^2)\hat{j} + (2x)\hat{k} \neq \vec{0} \]

Not conservative ❌

Final Answer:

$ \vec{A} $ is neither conservative nor solenoidal.

Vector Sink Field Analysis

Given vector field:

\[ \vec{A} = (2x + 1)\hat{i} + (x^2 - 6y)\hat{j} + (xy^2 + 3z)\hat{k} \]

Divergence:

\[ \nabla \cdot \vec{A} = 2 - 6 + 3 = -1 \]

✅ Conclusion:

The divergence is negative at every point, so $ \vec{A} $ is a sink field.

Qus : 374 MCA NIMCET PYQ

In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is

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Solution

Qus : 375 MCA NIMCET PYQ

Region R is defined as region in first quadrant satisfying the condition $x^2 + y^2 < 4$. Given that a point P=(r,s) lies in R, what is the probability that r>s?

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Solution

Probability that $ r > s $ in Region $ R $

Given: $ R = \{ (x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4 \} $ in the first quadrant

Area of region $ R $ in first quadrant: \[ A = \frac{1}{4} \pi (2)^2 = \pi \]

Region where $ r > s $ (i.e., below line $ x = y $) occupies half of that quarter-circle: \[ A_{\text{favorable}} = \frac{1}{2} \pi \]

Therefore, the required probability is:

\[ \text{Probability} = \frac{\frac{1}{2} \pi}{\pi} = \boxed{\frac{1}{2}} \]

Qus : 376 MCA NIMCET PYQ

If $x^my^n$=$(x+y)^{m+n}$, then $\frac{dy}{dx}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 377 MCA NIMCET PYQ

Lines $L_1, L_2, .., L_10 $are distinct among which the lines $L_2, L_4, L_6, L_8, L_{10}$ are parallel to each other and the lines $L_1, L_3, L_5, L_7, L_9$ pass through a given point C. The number of point of intersection of pairs of lines from the complete set $L_1, L_2, L_3, ..., L_{10}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Total Number of Intersection Points

Given:

10 distinct lines: $ L_1, L_2, \ldots, L_{10} $
$ L_2, L_4, L_6, L_8, L_{10} $: parallel (no intersections among them)
$ L_1, L_3, L_5, L_7, L_9 $: concurrent at point $ C $ (intersect at one point)

? Calculation:

\[ \text{Total line pairs: } \binom{10}{2} = 45 \]

\[ \text{Subtract parallel pairs: } \binom{5}{2} = 10 \Rightarrow 45 - 10 = 35 \]

\[ \text{Concurrent at one point: reduce } 10 \text{ pairs to 1 point} \Rightarrow 35 - 9 = \boxed{26} \]

✅ Final Answer: $\boxed{26}$ unique points of intersection

Qus : 378 MCA NIMCET PYQ

If $cos^{-1} \frac{x}{2}+cos^{-1} \frac{y}{3}=\phi$, then $9x^2-12xy cos\phi+4y^2$ is

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Solution

Qus : 379 MCA NIMCET PYQ

If the line $a^2 x + ay +1=0$, for some real number $a$, is normal to the curve $xy=1$ then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Problem:

The line \[ a^2x + ay + 1 = 0 \] is normal to the curve \[ xy = 1 \]. Find possible values of $ a \in \mathbb{R} $.

Step 1: Slope of Line

Rewrite: \[ y = -a x - \frac{1}{a} \] → slope = $ -a $

Step 2: Curve Derivative

\[ xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \] Slope of normal = $ \frac{x}{y} $

Match Slopes

\[ -a = \frac{x}{y} \Rightarrow x = -a y \]

Plug into Curve

\[ xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a} \]

For real $ y $, we need $ a < 0 $

✅ Final Answer:

\[ \boxed{a < 0} \]

Qus : 380 MCA NIMCET PYQ

The value of ${3}^{3-\log _35}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 381 MCA NIMCET PYQ

Out of a group of 50 students taking examinations in Mathematics, Physics, and Chemistry, 37 students passed Mathematics, 24 passed Physics, and 43 passed Chemistry. Additionally, no more than 19 students passed both Mathematics and Physics, no more than 29 passed both Mathematics and Chemistry, and no more than 20 passed both Physics and Chemistry. What is the maximum number of students who could have passed all three examinations?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Maximum Students Passing All Three Exams

Given:

Total students = 50
$ |M| = 37 $, $ |P| = 24 $, $ |C| = 43 $
$ |M \cap P| \leq 19 $, $ |M \cap C| \leq 29 $, $ |P \cap C| \leq 20 $

We use the inclusion-exclusion principle:

\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C| \]

Let $ x = |M \cap P \cap C| $. Then:

\[ 50 \geq 37 + 24 + 43 - 19 - 29 - 20 + x \Rightarrow 50 \geq 36 + x \Rightarrow x \leq 14 \]

✅ Final Answer: $\boxed{14}$

Qus : 382 MCA NIMCET PYQ

If $f(x)$ is a polynomial satisfying $f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(3)=28$, then $f(4)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

We know that $f(x) = 1 + x^n$

Finding $n$ using $f(3) = 28$:

$f(3) = 1 + 3^n = 28$

$3^n = 27$

$\therefore n = 3$

Thus: $f(x) = 1 + x^3$

Finding $f(4)$:

$f(4) = 1 + 4^3 = 1 + 64$

$$\boxed{f(4) = 65}$$

Qus : 383 MCA NIMCET PYQ

There are two sets A and B with |A| = m and |B| = n. If |P(A)| − |P(B)| = 112 then choose the wrong option (where |A| denotes the cardinality of A, and P(A) denotes the power set of A)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

$|P(A)| - |P(B)| = 112$

$2^m - 2^n = 112$

$2^n(2^{m-n} - 1) = 112$

$112 = 2^4 \cdot 7$

$2^n = 2^4 \Rightarrow n = 4$

$2^{m-n} - 1 = 7 \Rightarrow 2^{m-n} = 8$

$m - n = 3 \Rightarrow m = 7$

$m + n = 7 + 4 = 11 ; (\text{True})$

$2m - n = 14 - 4 = 10 \ne 1 ; (\text{False})$

$2n - m = 8 - 7 = 1 ; (\text{True})$

$3n - m = 12 - 7 = 5 ; (\text{True})$

Qus : 384 MCA NIMCET PYQ

Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be a function such that $f(0)=\frac{1}{\pi}$ and $f(x)=\frac{x}{e^{\pi x}-1}$ for $x\ne0$, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Analysis of Continuity and Differentiability

Function:

\[ f(x) = \begin{cases} \dfrac{x}{e^{\pi x} - 1}, & x \neq 0 \\ \dfrac{1}{\pi}, & x = 0 \end{cases} \]

✅ Continuity at $ x = 0 $:

\[ \lim_{x \to 0} f(x) = \frac{1}{\pi} = f(0) \quad \Rightarrow \quad \text{Function is continuous at } x = 0 \]

✏️ Differentiability at $ x = 0 $:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = -\frac{1}{2} \]

✅ Final Result:

Function is continuous at $ x = 0 $
Function is differentiable at $ x = 0 $
$ f'(0) = \boxed{-\frac{1}{2}} $

Qus : 385 MCA NIMCET PYQ

Suppose $P_1,P_2,\dots,P_{30}$ are thirty sets each having $5$ elements and $Q_1,Q_2,\dots,Q_n$ are $n$ sets with $3$ elements each. Let $\bigcup_{i=1}^{30}P_i=\bigcup_{j=1}^{n}Q_j=S$ and each element of $S$ belongs to exactly $10$ of the $P$’s and exactly $9$ of the $Q$’s. Then $n$ equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Total elements counted in $P$ sets: $30 \times 5 = 150$

Each element counted $10$ times: $|S| = \frac{150}{10} = 15$

Total elements in $Q$ sets: $15 \times 9 = 135$

Each $Q$ has $3$ elements: $n = \frac{135}{3} = 45$

Qus : 386 MCA NIMCET PYQ

The eccentricity of an ellipse, with its center at the origin is $\frac{1}{3}$ . If one of the directrices is $x=9$, then the equation of ellipse is:

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Solution

Qus : 387 MCA NIMCET PYQ

If f(x)=cos[$\pi$^2]x+cos[-$\pi$^2]x, where [.] stands for greatest integer function, then $f(\pi/2)$=

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Solution

? Function with Greatest Integer and Cosine

Given:

\[ f(x) = \cos\left([\pi^2]x\right) + \cos\left([-\pi^2]x\right) \]

Find: \[ f\left(\frac{\pi}{2}\right) \]

Step 1: Estimate Floor Values

\[ \pi^2 \approx 9.8696 \Rightarrow [\pi^2] = 9,\quad [-\pi^2] = -10 \]

Step 2: Plug into the Function

\[ f\left(\frac{\pi}{2}\right) = \cos\left(9 \cdot \frac{\pi}{2}\right) + \cos\left(-10 \cdot \frac{\pi}{2}\right) = \cos\left(\frac{9\pi}{2}\right) + \cos(-5\pi) \]

Step 3: Simplify

\[ \cos\left(\frac{9\pi}{2}\right) = 0,\quad \cos(-5\pi) = -1 \]

✅ Final Answer:

\[ \boxed{-1} \]

Qus : 388 MCA NIMCET PYQ

The number of functions $f$ from $A={0,1,2}$ into $B={0,1,2,3,4,5,6,7}$ such that $f(i) \le f(j)$ for $i
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Number of non-decreasing functions = combinations with repetition $= \binom{8+3-1}{3}=\binom{10}{3}$

Qus : 393 MCA NIMCET PYQ

It is given that the mean, median and mode of a data set is $1, 3^x$ and $9^x$ respectively. The possible values of the mode is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Mean, Median, and Mode Relation

Given:

Mean = 1
Median = $ 3^x $
Mode = $ 9^x $

Use empirical formula:

\[ \text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean} \]

\[ 9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2 \]

Let $ y = 3^x $, then:

\[ y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0 \]

So, $ y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4 $

✅ Final Answer: $\boxed{1 \text{ or } 4}$

Qus : 394 MCA NIMCET PYQ

The integer $n$ for which $\displaystyle \lim_{x\to0}\frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is finite and non-zero

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$\cos x-1 \sim -\frac{x^2}{2}$ $\cos x-e^x \sim -x$ Numerator $\sim x^3$ Hence $n=3$

Qus : 395 MCA NIMCET PYQ

The value of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Given the infinite series:

\[ S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]

This is a known convergent series, and its sum is:

\[ \boxed{e^{-1}} \]

✅ Final Answer: $\boxed{e^{-1}}$

Qus : 396 MCA NIMCET PYQ

The area of the plane bounded by $y=\sqrt{x},\ x\in[0,1]$, $y=x^2,\ x\in[1,2]$, $y=-x^2+2x+4,\ x\in[0,2]$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Area = $\displaystyle \int_0^1[( -x^2+2x+4)-\sqrt{x}]dx + \int_1^2[( -x^2+2x+4)-x^2]dx$ Evaluating gives $\boxed{\frac{19}{3}}$

Qus : 397 MCA NIMCET PYQ

The function $f(x)=2\sin x+\sin 2x,\ x\in[0,2\pi]$ has absolute maximum and minimum at

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$f'(x)=2\cos x+2\cos 2x$

$=2(2\cos^2x+\cos x-1)$

Critical points:

$\cos x=\frac{1}{2},-1$

$x=\frac{\pi}{3},\pi,\frac{5\pi}{3}$

Check values → max at $\frac{\pi}{3}$, min at $\frac{5\pi}{3}$

Qus : 410 MCA NIMCET PYQ

Identify the wrong statement.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Check (2):

Left side = $(A-B)-C = A - (B \cup C)$

Right side = $(A-C)-(B-C)$ = $A - (B \cap C)$

These are not equal in general.

Qus : 411 MCA NIMCET PYQ

Suppose $a,b,c$ are in A.P. with common difference $d$. Then $e^{1/c},e^{1/b},e^{1/a}$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in H.P. Exponentials of H.P. form G.P. Answer: $\boxed{\text{G.P.}}$

Qus : 412 MCA NIMCET PYQ

Two person A and B agree to meet 20 april 2018 between 6pm to 7pm with understanding that they will wait no longer than 20 minutes for the other. What is the probability that they meet?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 413 MCA NIMCET PYQ

Survey: $63%$ like cheese, $76%$ like apples. If $x%$ like both?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Using inclusion–exclusion:

$63 + 76 - x \le 100$

$139 - x \le 100$

$x \ge 39$

Max value of $x$ cannot exceed minimum of both percentages

→ $x \le 63$

Thus $39 \le x \le 63$.

Qus : 414 MCA NIMCET PYQ

Let $\alpha$ and $\beta$ be the roots of $x^2+x+1=0$. The equation whose roots are $\alpha^{19}$ and $\beta^{19}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$\alpha,\beta$ are complex cube roots of unity. $\alpha^{19}=\alpha,\ \beta^{19}=\beta$ Same equation remains. Answer: $\boxed{x^2+x+1=0}$

Qus : 415 MCA NIMCET PYQ

Three numbers a,b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that $a^3+b^2+c^2-3abc$ is divisible by 3 is,

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

The total number of ways in which numbers can be choosed =25 x 25=625 The number of ways in which either players can choose same numbers = 25

Probability that they win a prize = 25/625 = 1/25

Thus, the probability that they will not win a prize in a single trial = 1 - 1/25 = 24/25

Qus : 419 MCA NIMCET PYQ

The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Find the number of common terms in:
Series 1: $17, 21, 25, \ldots, 817$
Series 2: $16, 21, 26, \ldots, 851$

Step 1: Identify both APs
Series 1: $a_1 = 17,\ d_1 = 4$
Series 2: $a_2 = 16,\ d_2 = 5$

Step 2: First common term
By observation, $21$ is the first common term (appears in both series).

Step 3: Find common difference of new AP
Common terms will themselves form an AP.

Common difference $= \text{LCM}(d_1, d_2) = \text{LCM}(4, 5) = 20$

So common terms form AP: $21, 41, 61, \ldots$
with $a = 21,\ d = 20$

Step 4: Find the last common term
Last term must be $\leq$ smaller of the two last terms:
$\min(817, 851) = 817$

$T_n = a + (n-1)d \leq 817$
$21 + (n-1) \times 20 \leq 817$
$(n-1) \times 20 \leq 796$
$n - 1 \leq 39.8$
$n \leq 40.8$

So $n = 40$

Qus : 420 MCA NIMCET PYQ

If $(1+x)^n = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$, then $\left(1+\frac{a_1}{a_0}\right)\left(1+\frac{a_2}{a_1}\right)\left(1+\frac{a_3}{a_2}\right)\dots\left(1+\frac{a_n}{a_{n-1}}\right)$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Using $a_k = \binom{n}{k}$: $1 + \frac{a_k}{a_{k-1}} = \frac{n+1}{k}$ Product = $\frac{(n+1)^n}{n!}$

Qus : 421 MCA NIMCET PYQ

The value of $y=0.36\log_{0.25}\left(\dfrac13+\dfrac1{3^2}+\cdots\right)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

The series is a G.P. with first term $\dfrac13$ and ratio $\dfrac13$. Sum $=\dfrac{\frac13}{1-\frac13}=\dfrac12$ So $y=0.36\log_{0.25}\left(\dfrac12\right)$ $\log_{0.25}\left(\dfrac12\right)=\dfrac{\log(1/2)}{\log(1/4)}=\dfrac{-1}{-2}=\dfrac12$ Hence $y=0.36\times\dfrac12=0.18$ Answer: $\boxed{0.18}$

Qus : 422 MCA NIMCET PYQ

The quadratic equation whose roots are $\sin^218^o$ and $\cos^236^o$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 423 MCA NIMCET PYQ

India plays 4 matches. Probabilities of scoring $0,1,2$ points: $0.45,\ 0.05,\ 0.50$. Find probability of at least $7$ points.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$P(8)=0.5^4=0.0625$

$P(7)=4 \cdot 0.05 \cdot 0.5^3 = 0.025$

Total $= 0.0625 + 0.025 = 0.0875$

Qus : 424 MCA NIMCET PYQ

If $H_1,H_2,\ldots,H_n$ are $n$ harmonic means between $a$ and $b$, $a\ne b$, then the value of $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Step 1: Convert HP to AP
Since $a, H_1, H_2, \ldots, H_n, b$ is in HP,
$\dfrac{1}{a},\ \dfrac{1}{H_1},\ \ldots,\ \dfrac{1}{H_n},\ \dfrac{1}{b}$ is in AP with $n+2$ terms.

Step 2: Find common difference $d$
$d = \dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{n+1} = \dfrac{a-b}{ab(n+1)}$

Step 3: Find $H_1$ and $H_n$
$\dfrac{1}{H_1} = \dfrac{1}{a} + d = \dfrac{a+bn}{ab(n+1)}$ $\Rightarrow H_1 = \dfrac{ab(n+1)}{a+bn}$

$\dfrac{1}{H_n} = \dfrac{1}{b} - d = \dfrac{an+b}{ab(n+1)}$ $\Rightarrow H_n = \dfrac{ab(n+1)}{an+b}$

Step 4: Evaluate $\dfrac{H_1+a}{H_1-a}$ using Componendo-Dividendo
$\dfrac{H_1}{a} = \dfrac{b(n+1)}{a+bn}$

Applying componendo-dividendo:
$\dfrac{H_1+a}{H_1-a} = \dfrac{b(n+1)+(a+bn)}{b(n+1)-(a+bn)} $

$= \dfrac{a+b(2n+1)}{b-a} \quad \cdots(1)$

Step 5: Evaluate $\dfrac{H_n+b}{H_n-b}$ using Componendo-Dividendo
$\dfrac{H_n}{b} = \dfrac{a(n+1)}{an+b}$

Applying componendo-dividendo:
$\dfrac{H_n+b}{H_n-b} = \dfrac{a(n+1)+(an+b)}{a(n+1)-(an+b)} $

$= \dfrac{b+a(2n+1)}{a-b} \quad \cdots(2)$

Step 6: Add (1) and (2)
$\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} $

$= \dfrac{a+b(2n+1)}{b-a} + \dfrac{b+a(2n+1)}{a-b}$

$= \dfrac{a+b(2n+1) - b - a(2n+1)}{b-a}$

$= \dfrac{(a-b) + (2n+1)(b-a)}{b-a}$

$= \dfrac{(b-a)(2n+1)-(b-a)}{b-a}$

$= (2n+1) - 1$

$= 2n$

Answer: $\dfrac{H_1+a}{H_1-a}+\dfrac{H_n+b}{H_n-b} = \boxed{2n}$

Qus : 425 MCA NIMCET PYQ

Sum to infinity of a geometric is twice the sum of the first two terms. Then what are the possible values of common ratio?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Let first term $= a$ and common ratio $= r$, where $|r| < 1$

Sum to infinity: $S_{\infty} = \dfrac{a}{1-r}$

Sum of first two terms: $S_2 = a + ar = a(1+r)$

Step 2: Set up the equation
Given: $S_{\infty} = 2 \times S_2$

$\dfrac{a}{1-r} = 2 \cdot a(1+r)$

Step 3: Simplify
Dividing both sides by $a$ (since $a \ne 0$):

$\dfrac{1}{1-r} = 2(1+r)$

$1 = 2(1+r)(1-r)$

$1 = 2(1 - r^2)$

$1 = 2 - 2r^2$

$2r^2 = 1$

$r^2 = \dfrac{1}{2}$

$r = \pm\dfrac{1}{\sqrt{2}}$

Qus : 426 MCA NIMCET PYQ

The area of the region bounded by x-axis and the curves defined by $y=tanx$, $-\frac{\pi}{3}\leq x\leq \frac{\pi}{3}$ and $y=cotx$, $-\frac{\pi}{6}\leq x\leq \frac{3\pi}{2}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 427 MCA NIMCET PYQ

A coin is tossed three times. Probability of getting heads and tails alternately is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Alternating sequences = $HTH,\ THT$

Probability $= 2/8 = 1/4$

Qus : 428 MCA NIMCET PYQ

For $a>0,\ a\ne1$, the number of values of $x$ satisfying $2\log_x a+\log_{ax} a+3\log_{a^2x} a=0$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Let $\log_x a=t$ Then $\log_{ax}a=\dfrac{t}{1+t}$, $\log_{a^2x}a=\dfrac{t}{2+t}$ Equation becomes $2t+\dfrac{t}{1+t}+3\dfrac{t}{2+t}=0$ Solving gives $t=0,-1,-2$ All give valid $x$ values. Number of solutions $=3$ Answer: $\boxed{3}$

Qus : 429 MCA NIMCET PYQ

Suppose that m and n are fixed numbers such that the mth term of an HP is equal to n and the nth term is equal to m, (m ≠ n). Then the (m + n)th term is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Given: The sequence is in H.P. with $a_m = n$ and $a_n = m$ $(m \ne n)$.

In an H.P., reciprocals of terms form an A.P. So, $\tfrac{1}{a_m} = \tfrac{1}{n}$ and $\tfrac{1}{a_n} = \tfrac{1}{m}$.

This A.P. leads to common difference $d = \tfrac{1}{mn}$ and first term $A = \tfrac{1}{mn}$.

Thus, the reciprocal of the $(m+n)$th term is:

$b_{m+n} = \tfrac{m+n}{mn}$

Hence, the $(m+n)$th term of H.P. is:

$a_{m+n} = \dfrac{mn}{m+n}$

Final Answer: $ \dfrac{mn}{m+n} $

Qus : 430 MCA NIMCET PYQ

If $\log (1-x+x^2)={{a}}_1x+{{a}}_{2{}^{{}^{}}}{x}^2+{{{}{{a}}_3{x}^3+.\ldots.}}^{}$ then ${{a}}_3+{{a}}_6+{{a}}_9+.\ldots.$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 431 MCA NIMCET PYQ

One hundred identical coins, each with probability $p$ of showing up a head, are tossed. If $0
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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Condition: $\binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49}$ Divide both sides: $\frac{1}{\frac{100-50}{51}} \cdot \frac{1-p}{p} = 1$ $\frac{51}{50} \cdot \frac{1-p}{p} = 1$ $\frac{1-p}{p} = \frac{50}{51}$ $51 - 51p = 50p$ $51 = 101p$ $p = \frac{51}{101}$

Qus : 432 MCA NIMCET PYQ

An eight digit number divisible by $9$ is to be formed by using $8$ digits out of the digits $0,1,\ldots,9$ without replacement. The number of ways in which this can be done is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Sum of digits must be divisible by $9$. Total sum of digits $0$ to $9$ is $45$. Choose $8$ digits such that their sum is divisible by $9$. Possible digit-exclusions give $4$ valid cases. Arrangements of remaining $8$ digits excluding leading zero restriction gives $4\times7!$ Answer: $\boxed{4(7!)}$

Qus : 433 MCA NIMCET PYQ

If A is an invertible skew-symmetric matrix, then

is a

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 434 MCA NIMCET PYQ

$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 435 MCA NIMCET PYQ

The length of the projection of $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ on $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$, is equal to:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$

The length of projection of $\vec{a}$ on $\vec{b}$ is:

$\displaystyle \text{Proj length} = \frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

Compute dot product: $\vec{a}\cdot\vec{b} = (2)(-2) + (3)(1) + (1)(2)$ $= -4 + 3 + 2 = 1$

Compute $|\vec{b}|$: $|\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2}$ $= 3$

Thus projection length: $\displaystyle \frac{1}{3}$

Qus : 436 MCA NIMCET PYQ

n a Poisson distribution if $P[X=3] = \frac{1}{4} P[X=4]$ then $P[X=5] = k P[X=7]$ where $k$ equals to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Poisson PMF:

$P[X=x] = e^{-\lambda} \frac{\lambda^x}{x!}$

Given: $P(3) = \frac{1}{4} P(4)$

$\frac{\lambda^3/3!}{\lambda^4/4!} = \frac{1}{4}$

$\frac{4}{\lambda} = \frac{1}{4}$

$\lambda = 16$

Now find $k$:

$\frac{P(5)}{P(7)} = \frac{\lambda^5/5!}{\lambda^7/7!} = \frac{7 \cdot 6}{\lambda^2}$

Substitute $\lambda = 16$:

$\frac{42}{256} = \frac{21}{128}$

So, $k = \frac{21}{128}$

Qus : 437 MCA NIMCET PYQ

The number of ordered pairs $(m,n)$, $m,n\in{1,2,\ldots,100}$ such that $7^m+7^n$ is divisible by $5$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$7\equiv2\pmod5$ So $7^k\equiv2^k\pmod5$ $2^k\pmod5$ cycles as $2,4,3,1$ (period $4$). $7^m+7^n\equiv0\pmod5$ when residues are complementary. Total valid ordered pairs $=2500$ Answer: $\boxed{2500}$

Qus : 438 MCA NIMCET PYQ

If the mean of the squares of first n natural numbers be 11, then n is equal to?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Given: Mean of squares of first n natural numbers is 11.

Use the identity for sum of squares: $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$

Mean $=$ (sum) $ \div n \Rightarrow $ $$\frac{1}{n}\sum_{k=1}^{n}k^2=\frac{(n+1)(2n+1)}{6}=11$$

Solve for $n$: $$(n+1)(2n+1)=66 $$ $$\;\Rightarrow\; 2n^2+3n+1=66 \;\Rightarrow\; $$ $$n=\frac{-3\pm\sqrt{9+520}}{4}=\frac{-3\pm 23}{4}$$

Only positive integer solution: $$n=\frac{20}{4}=5$$

Answer: $n=5$.

Qus : 439 MCA NIMCET PYQ

A and B are independent witness in a case. The chance that A speaks truth is x and B speaks truth is y. If A and B agree on certain statement, the probability that the statement is true is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

P(A speaks truth) = x

P(B speaks truth) = y

Since, both A and B agree on certain statement.

Hence, Total Probability =P(A)P(B)+P(A')P(B')

=xy + (1-x)(1-y)

If statement is true then it means both A and B speaks truth.

∴ Required Probability = $\dfrac{xy}{xy+(1-x)(1-y)}$

Qus : 440 MCA NIMCET PYQ

The number of common tangents to the circle $x^2+y^2=4$ and $x^2+y^2-6x-8y=24$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 441 MCA NIMCET PYQ

If $8^{x-1}=(1/4)^{x}$, then the value of $\frac{1}{\log_{x+1}4-\log_{x+1}5}+\frac{1}{\log_{1-x}4-\log_{1-x}5}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: $8^{x-1} = (1/4)^{x}$

Rewrite both sides with base 2: $8 = 2^3$ and $1/4 = 2^{-2}$

So: $(2^3)^{x-1} = (2^{-2})^x$ $\Rightarrow 2^{3(x-1)} = 2^{-2x}$

Equate powers: $3(x - 1) = -2x$

$3x - 3 = -2x$

$5x = 3$

$\Rightarrow x = \frac{3}{5}$

We need the value of:

$\displaystyle \frac{1}{\log_{x+1} 4 - \log_{x+1} 5} + \frac{1}{\log_{1-x} 4 - \log_{1-x} 5}$

Use property: $\log_a m - \log_a n = \log_a \left(\frac{m}{n}\right)$

So the expression becomes: $\displaystyle \frac{1}{\log_{x+1} \left(\frac{4}{5}\right)} + \frac{1}{\log_{1-x} \left(\frac{4}{5}\right)}$

Now use: $\displaystyle \frac{1}{\log_a b} = \log_b a$

So expression becomes: $\log_{4/5}(x+1) + \log_{4/5}(1-x)$

Use product property: $\log_{4/5}[(x+1)(1-x)]$

Compute: $(x+1)(1-x) = 1 - x^2$

Substitute $x = \frac{3}{5}$:

$1 - x^2 = 1 - \frac{9}{25} = \frac{16}{25}$

Thus value = $\log_{4/5}\left(\frac{16}{25}\right)$

Rewrite: $\frac{16}{25} = \left(\frac{4}{5}\right)^2$

Therefore: $\log_{4/5}\left( (4/5)^2 \right) = 2$

Qus : 442 MCA NIMCET PYQ

The average marks per student in a class of 30 students were 45. On rechecking it was found that marks had been entered wrongly in two cases. After correction these marks were increased by 24 and 34 in the two cases. The correct average marks per student are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Increase in total marks $= 24 + 34 = 58$ New average $45 + \frac{58}{30} = 45 + 1.933 = 46.933 \approx 47$

Qus : 443 MCA NIMCET PYQ

If $a,b,c$ are the roots of the equation $x^3-3px^2+3qx-1=0$, then the centroid of the triangle with vertices $\left(a,\frac1a\right),\left(b,\frac1b\right),\left(c,\frac1c\right)$ is the point

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Centroid $=\left(\dfrac{a+b+c}{3},\dfrac{\frac1a+\frac1b+\frac1c}{3}\right)$ From the equation, $a+b+c=3p$ Also $\dfrac1a+\dfrac1b+\dfrac1c=\dfrac{ab+bc+ca}{abc}=\dfrac{3q}{1}=3q$ Hence centroid $=(p,q)$ Answer: $\boxed{(p,q)}$

Qus : 444 MCA NIMCET PYQ

The set of points, where $f(x)=\frac{x}{1+|x|}$ is differentiable in

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Qus : 451 MCA NIMCET PYQ

In an entrance test there are multiple choice questions, with four possible answer to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If the student gets the correct answer to a question, then the probability that he as guessing is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 452 MCA NIMCET PYQ

Let $\vec{a}=2\widehat{i}\, +\widehat{j}\, +2\widehat{k}$ , $\vec{b}=\widehat{i}-\widehat{j}+2\widehat{k}$ and $\vec{c}=\widehat{i}+\widehat{j}-2\widehat{k}$ are are three vectors. Then, a vector in the plane of $\vec{a}$ and $\vec{c}$ whose projection on $\vec{b}$ is of magnitude $\frac{1}{\sqrt{6}}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 453 MCA NIMCET PYQ

The curve $y=\frac{x}{1+x\tan x}$ attains maxima

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given the curve

$y = \dfrac{x}{1 + x \tan x}$

Differentiate using quotient rule:

$y' = \dfrac{(1 + x \tan x) - x(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}$

Simplify the numerator:

$N = 1 + x \tan x - x \tan x - x^2 \sec^2 x$

$N = 1 - x^2 \sec^2 x$

Set $N = 0$ for maxima/minima:

$1 - x^2 \sec^2 x = 0$

$x^2 \sec^2 x = 1$

$x \sec x = \pm 1$

Hence the curve attains maxima when:

$x \sec x = 1$

$x=\cos x$

Qus : 454 MCA NIMCET PYQ

The value of $X^4 + 9X^3 + 35X^2 - X + 4$ for $X = -5 + 2\sqrt{-4}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$X = -5 + 4i$ Compute modulus: $|X|^2 = (-5)^2 + 4^2 = 25 + 16 = 41$ Because polynomial is symmetric to complex conjugates, evaluate: $X^4 + 9X^3 + 35X^2 - X + 4 = -160$

Qus : 455 MCA NIMCET PYQ

The number of roots of the equation $|x^2-x-6|=x+2$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$x^2-x-6=(x-3)(x+2)$ Case 1: $x^2-x-6\ge0$ $|x^2-x-6|=x^2-x-6=x+2$ $x^2-2x-8=0 \Rightarrow x=4,-2$ Case 2: $x^2-x-6<0$ $-(x^2-x-6)=x+2$ $x^2=4 \Rightarrow x=\pm2$ Valid roots: $-2,2,4$ Total roots $=3$

Qus : 456 MCA NIMCET PYQ

Let $f: R \rightarrow R$ be defined by $f(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ |x - 2| & \text{if } x \geq 0 \end{cases}$. Find $\int_{-2}^{3} f(x)\, dx$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 457 MCA NIMCET PYQ

A man is known to speak the truth 2 out of 3 times. He threw a dice cube with 1 to 6 on its faces and reports that it is 1. Then the probability that it is actually 1 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 458 MCA NIMCET PYQ

For what value of p, the polynomial $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 459 MCA NIMCET PYQ

The scores of students in a national level examination are normally distributed with a mean of 500 and a standard deviation of 100. If the value of the cumulative distribution of the standard normal random variable at 0.5 is 0.691, then the probability that a randomly selected student scored between 450 and 500 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given:

Scores are normally distributed with mean $ \mu = 500 $,

standard deviation $ \sigma = 100 $

We want the probability that a student scored between 450 and 500:

$P(450 < X < 500)$

Convert to $Z$-scores:

For $X = 450$: $Z = \dfrac{450 - 500}{100} = -0.5$

For $X = 500$: $Z = \dfrac{500 - 500}{100} = 0$

Thus:

$P(450 < X < 500) = P(-0.5 < Z < 0)$

Given: $P(Z < 0.5) = 0.691$

Using symmetry of the standard normal curve:

$P(-0.5 < Z < 0.5) = 2(0.691 - 0.5) = 2(0.191) = 0.382$

So: $P(-0.5 < Z < 0) = \dfrac{0.382}{2} = 0.191$

Final answer: 0.191

Qus : 460 MCA NIMCET PYQ

If $y = a \log x + b x^2 + x$ has its extremum value at $x = -1$ and $x = 2$, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$y' = \frac{a}{x} + 2bx + 1$ Since $y'=0$ at $x=-1$ and $2$: At $x= -1$: $-!a + (-2b) + 1 = 0$ $\Rightarrow -a -2b + 1 = 0$ … (i) At $x= 2$: $\frac{a}{2} + 4b + 1 = 0$ $\Rightarrow a + 8b + 2 = 0$ … (ii) Solve (i) and (ii): From (i): $a = 1 - 2b$ Put into (ii): $(1 - 2b) + 8b + 2 = 0$ $1 - 2b + 8b + 2 = 0$ $6b + 3 = 0$ $b = -\frac12$ Then $a = 1 - 2b = 1 - 2(-\frac12) = 1 + 1 = 2$ So $a = 2,\ b = -\frac12$

Qus : 461 MCA NIMCET PYQ

A pair of unbiased dice is rolled together till a sum of either $5$ or $7$ is obtained. The probability that $5$ comes before $7$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$P(5)=\dfrac4{36}$, $P(7)=\dfrac6{36}$ Required probability $=\dfrac{P(5)}{P(5)+P(7)}=\dfrac4{10}=\dfrac25$

Qus : 462 MCA NIMCET PYQ

The slope of two-lines $6x^2-xy-2y^2=0$ differ by

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Equation:

$6x^2 - xy - 2y^2 = 0$

Compare with: $ ax^2 + 2hxy + by^2 = 0 $

$a = 6$

$2h = -1 \;\;\Rightarrow\;\; h = -\tfrac{1}{2}$

$b = -2$

Slopes satisfy:

$bm^2 + 2hm + a = 0$

$-2m^2 - m + 6 = 0$

$\Rightarrow 2m^2 + m - 6 = 0$

Solving quadratic:

$m = \dfrac{-1 \pm \sqrt{49}}{4}$

$m = \dfrac{-1 \pm 7}{4}$

Therefore:

$m_1 = \tfrac{3}{2}, \quad m_2 = -2$

Answer: $ \tfrac{7}{2} $

Qus : 463 MCA NIMCET PYQ

Let A and B be two events such that

and

where

stands for the complement of event A. Then the events A and B are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 464 MCA NIMCET PYQ

If $F(\theta)=\begin{bmatrix}{\cos \theta} & {-\sin \theta} & {0} \\ {\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{bmatrix}$ , then $F(\theta)F(\alpha)$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 465 MCA NIMCET PYQ

Number of permutations of the letters of the word BANGLORE such that the string ANGLE appears together in all permutations, is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Treat the fixed string ANGLE as one block. Remaining letters from BANGLORE are B, O, R. So we arrange 4 distinct items: {ANGLE, B, O, R}.

Number of permutations =4! = 24.

Qus : 466 MCA NIMCET PYQ

If $a, b, c$ are in A.P., $p, q, r$ are in H.P. and $ap, bq, cr$ in G.P.$,$ then $\frac{p}{r} + \frac{r}{p}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$a, b, c$ in A.P.

⇒ $b = \frac{a+c}{2}$

$p, q, r$ in H.P.

⇒ $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ in A.P.

Given $ap, bq, cr$ in G.P.

$\frac{bq}{ap} = \frac{cr}{bq}$

Substitute $b = \frac{a+c}{2}$ and simplify: $\frac{p}{r} + \frac{r}{p} = \frac{a}{c} + \frac{c}{a}$

Qus : 467 MCA NIMCET PYQ

A letter is taken at random from the letters of the word STATISTICS and another letter is taken at random from the letters of the word ASSISTANT. The probability that they are the same letter is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

STATISTICS: $S(3),T(3),A(1),I(2)$ ASSISTANT: $S(3),T(2),A(2),I(1),N(1)$ Probability $=\dfrac{3\cdot3+3\cdot2+1\cdot2+2\cdot1}{10\cdot9}$ $=\dfrac{19}{90}$ Answer: $\boxed{\dfrac{19}{90}}$

Qus : 468 MCA NIMCET PYQ

If the radius of the circle changes at the rate of $-\dfrac{2}{\pi}$ m/s, at what rate does the circle's area change when the radius is 10m?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 469 MCA NIMCET PYQ

The mean and variance of a random variable X having binomial distribution are 4 and 2 respectively. The P(X = 1) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

np = 4

npq = 2

q = 1/2, p = 1/2, n = 8

p(X = 1) = 8C1 (1/2)(1/2)7

Qus : 470 MCA NIMCET PYQ

If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio 2:1, then the value of n is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 471 MCA NIMCET PYQ

Let A and B be two square matrices of same order satisfying $A^2+5A+5I =0$ and $B^2+3B+I=0$ repectively. Where I is the identity matrix. Then the inverse of the matrix $C= BA+2B+2A+4I$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: $A^2 + 5A + 5I = 0 \quad\Rightarrow\quad A^2 = -5A - 5I$

$B^2 + 3B + I = 0 \quad\Rightarrow\quad B^2 = -3B - I$

Matrix: $C = BA + 2B + 2A + 4I$

We want $C^{-1}$.

Step 1: Try a linear expression as inverse Assume: $C^{-1} = \alpha AB + \beta A + \gamma B + \delta I$

Test using the fact that quadratic equations give inverses as linear polynomials.

Check option (2):

$AB + A + 3B + 3I$ Multiply with $C$:

$C(AB + A + 3B + 3I)$

$= (BA + 2B + 2A + 4I)(AB + A + 3B + 3I)$

Using reductions:

$A^2 = -5A - 5I$ $B^2 = -3B - I$

Everything simplifies to: $I$

Hence: $C^{-1} = AB + A + 3B + 3I$

Qus : 472 MCA NIMCET PYQ

If $a \ne p$, $b \ne q$, $c \ne r$ and $\left|\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}\right| = 0$, then the value of $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Solution: Given $\left|\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}\right| = 0,$ the rows are linearly dependent. Using the determinant identity, we get $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 1.$

Solution

Solution: Using $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0,$ simplify the entries. After row/column reduction and applying cube root identities, the determinant becomes $\omega^2.$

Qus : 479 MCA NIMCET PYQ

The value of $\lambda$ for which the volume of the parallelepiped formed by the vectors $\vec i+\lambda\vec j+\vec k,\ \vec j+\lambda\vec k,\ \lambda\vec i+\vec k$ is minimum is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Volume $=|\det|$ $V=|\lambda^3-3\lambda|$ Minimum when derivative $=0$ $\Rightarrow 3\lambda^2-3=0$ $\Rightarrow \lambda=\pm\dfrac1{\sqrt3}$ Minimum positive value at $\lambda=\dfrac1{\sqrt3}$ Answer: $\boxed{\dfrac1{\sqrt3}}$

Qus : 480 MCA NIMCET PYQ

The number of solutions of the equation sinx + sin5x = sin3x lying in the interval $[0, \pi]$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 481 MCA NIMCET PYQ

If E₁ and E₂ are two events associated with a random experiment such that P (E₂) = 0.35, P (E₁ or E₂) = 0.85 and P (E₁ & E₂) = 0.15 then P(E₁) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 482 MCA NIMCET PYQ

If the position vector of A and B relative to O be $\widehat{i}\, -4\widehat{j}+3\widehat{k}$ and $-\widehat{i}\, +2\widehat{j}-\widehat{k}$ respectively, then the median through O of ΔABC is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 483 MCA NIMCET PYQ

The value of $\frac{d}{dx}\int ^{2\sin x}_{\sin {x}^2}{e}^{{t}^2}dt$ at $x=\pi$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We want to find $\dfrac{d}{dx}\left(\int_{\sin^2 x}^{2\sin x} e^{t^2}, dt\right)$ at $x=\pi$.

Using Leibniz rule: $\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)} f(t) dt\right) = f(b(x)) b'(x) ;-; f(a(x)) a'(x)$

Here $a(x)=\sin^2 x$, $b(x)=2\sin x$, $f(t)=e^{t^2}$.

Compute derivatives:

$a'(x)=2\sin x\cos x$

$b'(x)=2\cos x$

So the derivative is:

$e^{(2\sin x)^2}(2\cos x)-e^{(\sin^2 x)^2}(2\sin x\cos x)$

Now evaluate at $x=\pi$

$\sin\pi=0,\quad \cos\pi=-1$

Thus: $b'(\pi)=2(-1)=-2$

$a'(\pi)=0$

Therefore: $e^{0}(-2)-e^{0}(0)=-2$

So the final answer is: $-2$

Qus : 484 MCA NIMCET PYQ

The point $(4,1)$ undergoes the following transformations successively: (i) Reflection about the line $y=x$ (ii) Translation through a distance $2$ units along the positive $x$-axis (iii) Rotation by an angle $\frac{\pi}{4}$ anticlockwise about the origin The final position of the point is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Solution: Step 1: Reflect (4,1) about y=x → (1,4) Step 2: Translate 2 units in +x direction → (1+2, 4) = (3,4) Step 3: Rotate (3,4) by π/4 anticlockwise: New x = (3 - 4)/√2 = -1/√2 New y = (3 + 4)/√2 = 7/√2 Final point = $\left(\frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

Qus : 485 MCA NIMCET PYQ

A six-faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Let $P(\text{even})=p$ and $P(\text{odd})=3p$ $p+3p=1 \Rightarrow p=\dfrac14$ So $P(\text{even})=\dfrac14,\quad P(\text{odd})=\dfrac34$ Sum is even when both outcomes are even or both are odd. $P=\left(\dfrac14\right)^2+\left(\dfrac34\right)^2=\dfrac1{16}+\dfrac9{16}=\dfrac{10}{16}=\dfrac58$ Answer: $\boxed{\dfrac58}$

Qus : 486 MCA NIMCET PYQ

In an acute-angled ΔABC the least value of secA+secB+secC is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 487 MCA NIMCET PYQ

Find a matrix X such that 2A + B + X = 0, whose A =

and B =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 488 MCA NIMCET PYQ

The general value of $\theta$, satisfying the equation $\sin \theta=\frac{-1}{2},\, \tan \theta=\frac{1}{\sqrt[]{3}}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 489 MCA NIMCET PYQ

There are two coins, say blue and red. For blue coin, probability of getting head is 0.99 and for red coin, it is 0.01. One coin is chosen randomly and is tossed. The probability of getting head is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

One coin is chosen at random, so each has probability $\tfrac{1}{2}$ of being selected.

Let $P(H)$ = probability of getting head.

Using the law of total probability:

$P(H) = \tfrac{1}{2} \times 0.99 + \tfrac{1}{2} \times 0.01 $$ = \tfrac{1}{2}(1.00) = 0.50$

✅ Final Answer: ${0.5}$

Qus : 490 MCA NIMCET PYQ

If the two pair of lines $X^2 - 2mXY - Y^2 = 0$ and $X^2 - 2nXY - Y^2 = 0$ are such that one represents the bisector of the angles between the other, then:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Slopes of first pair: m ± √(m² + 1) Slopes of second pair: n ± √(n² + 1) Condition for angle bisector of two homogeneous pair of lines: mn + 1 = 0

Qus : 491 MCA NIMCET PYQ

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

TATANAGAR has $2$ occurrences of TA CALCUTTA has $3$ occurrences of TA Total occurrences $=5$ Required probability $=\dfrac{3}{5}$ This is not listed. Answer: $\boxed{\text{None of these}}$

Qus : 492 MCA NIMCET PYQ

Let $P = \{\theta : \sin\theta - \cos\theta = \sqrt{2}\cos\theta \}$ and $Q = \{\theta : \sin\theta + \cos\theta = \sqrt{2}\sin\theta \}$ be two sets. Then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 493 MCA NIMCET PYQ

If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in HP, then sin A, sin B, sin C are in

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

If altitudes of a triangle are in HP then its side will be in AP because sides are inverse proportion to height as area is constant. a, b, c are sides of triangle.

a, b, c are in A.P.

sin A, sin B, sin C are in A.P.

Qus : 494 MCA NIMCET PYQ

The area of the triangle formed by the vertices whose position vectors are $3\widehat{i}+\widehat{j}$ , $5\widehat{i}+2\widehat{j}+\widehat{k}$ , $\widehat{i}-2\widehat{j}+3\widehat{k}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 495 MCA NIMCET PYQ

The number of all even integers between 99 and 999 which are not multiple of 3 and 5 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Solution:

Even numbers from 100 to 998: count $= \frac{998-100}{2}+1=450$.

Exclude evens divisible by $3$ or $5$ using inclusion–exclusion:

Multiples of $6$ in $[100,998]$: $\lfloor 998/6 \rfloor-\lfloor 99/6 \rfloor = 166-16=150$.
Multiples of $10$ in $[100,998]$: $\lfloor 998/10 \rfloor-\lfloor 99/10 \rfloor = 99-9=90$.
Multiples of $30$ in $[100,998]$: $\lfloor 998/30 \rfloor-\lfloor 99/30 \rfloor = 33-3=30$.

Forbidden $=150+90-30=210$ ⇒ Allowed $=450-210={240}$.

Qus : 496 MCA NIMCET PYQ

The circle $x^2 + y^2 = 9$ is contained in the circle $x^2 + y^2 - 6x - 8y + 25 = c^2$ if

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Given circle 1: center (0,0), radius 3 Circle 2 (rewrite): (x-3)² + (y-4)² = c² Distance between centers = √(3² + 4²) = 5 Condition: larger radius ≥ smaller radius + distance c ≥ 3 + 5 = 8 Given options → smallest suitable c is 10

Qus : 497 MCA NIMCET PYQ

If $\cos\alpha+\cos\beta=a$, $\sin\alpha+\sin\beta=b$ and $\theta$ is the arithmetic mean between $\alpha$ and $\beta$, then $\sin2\theta+\cos2\theta$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$a^2+b^2=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$ $=2+2\cos(\alpha-\beta)=4\cos^2\frac{\alpha-\beta}{2}$ $\Rightarrow \cos(\alpha-\beta)=\dfrac{a^2+b^2}{2}-1$ Since $\theta=\dfrac{\alpha+\beta}{2}$, $\sin2\theta+\cos2\theta=\dfrac{a^2-b^2}{a^2+b^2}$ Answer: $\boxed{\dfrac{a^2-b^2}{a^2+b^2}}$

Qus : 498 MCA NIMCET PYQ

If $\dfrac{\tan x}{2} = \dfrac{\tan y}{3} = \dfrac{\tan z}{5}$ and $x + y + z = \pi$, then the value of $\tan^2 x + \tan^2 y + \tan^2 z$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 499 MCA NIMCET PYQ

α, β are the roots of the an equation $x^2- 2x cosθ + 1 = 0$, then the equation having roots αn and βn is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 500 MCA NIMCET PYQ

The standard deviation of 20 numbers is 30. If each of the numbers is increased by 4, then the new standard deviation will be

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Concept: Standard deviation does not change when the same constant is added to every observation.

For data $x_i$ and constant $k$: $$\operatorname{SD}(x_i+k)=\operatorname{SD}(x_i).$$

Given: Original SD = 30, each number increased by 4.

New standard deviation: $30$.

Qus : 501 MCA NIMCET PYQ

Let A = {1,2,3, ... , 20}. Let $R\subseteq A\times A$ such that R = {(x,y): y = 2x - 7}. Then the number of elements in R, is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given relation: $R = \{(x,y) : y = 2x - 7\}$ where $A = \{1,2,3,\dots,20\}$.

We need both $x, y \in A$.

So, $1 \le y = 2x - 7 \le 20$

⇒ $1 \le 2x - 7 \le 20$

Add 7: $8 \le 2x \le 27$

Divide by 2: $4 \le x \le 13.5$

Hence, integer values of $x$ are $4,5,6,7,8,9,10,11,12,13$ → total $10$ values.

✅ Number of elements in R = 10

Qus : 502 MCA NIMCET PYQ

If any tangent to the ellipse $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$ intercepts equal length $l$ on both axes, then $l =$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Tangent intercept form: x/A + y/B = 1 Equal intercepts → A = B = l Condition for tangent to ellipse: 1 = a²/A² + b²/B² = a²/l² + b²/l² ⇒ l² = a² + b² ⇒ l = √(a² + b²)

Qus : 503 MCA NIMCET PYQ

If $(1+\tan1^\circ)(1+\tan2^\circ)\cdots(1+\tan45^\circ)=2^n$, then the value of $n$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Using identity $(1+\tan\theta)(1+\tan(45^\circ-\theta))=2$ There are $22$ such pairs from $1^\circ$ to $44^\circ$ and $(1+\tan45^\circ)=2$ So $2^{22}\times2=2^{23}$ Hence $n=23$ Answer: $\boxed{23}$

Qus : 504 MCA NIMCET PYQ

The circles whose equations are $x^2+y^2+c^2=2ax$ and $x^2+y^2+x^2-2by=0$ will touch one another externally, if

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 505 MCA NIMCET PYQ

The equation (x-a)³+(x-b)³+(x-c)³ = 0 has

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let f(x) = (x – a)³ + (x – b)³ + (x – c)³.

Then f'(x) = 3{(x – a)² + (x – b)² + (x –c)²}

clearly , f'(x) > 0 for all x.

so, f'(x) = 0 has no real roots.

Hence, f(x) = 0 has two imaginary and one real root

Qus : 506 MCA NIMCET PYQ

The function $f(x)=\frac{x}{1+x\tan x}$ , $0\leq x\leq\frac{\pi}{2}$ is maximum when

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 507 MCA NIMCET PYQ

If $\vec{a}, \vec{b}$ and $\vec{c} $ are three vectors such that $\vec{a} \times \vec{b}=\vec{c}$ , $\vec{a}.\vec{c} = 2$ and $\vec{b}.\vec{c} = 1$. If $|\vec{b}| = 1$, then the value of $|\vec{a}| $ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given conditions: $\vec{a} \times \vec{b} = \vec{c}$

$\vec{a} \cdot \vec{c} = 2$

$\vec{b} \cdot \vec{c} = 1$

$|\vec{b}| = 1$

Since $\vec{c} = \vec{a} \times \vec{b}$,

we have: $|\vec{c}| = |\vec{a}||\vec{b}| \sin\theta = |\vec{a}|\sin\theta$ (because $|\vec{b}| = 1$)

Also: $\vec{a} \cdot \vec{c} = \vec{a} \cdot (\vec{a} \times \vec{b}) = 0$

But given: $\vec{a} \cdot \vec{c} = 2$

This is possible only if $\vec{c}$ is not perpendicular to $\vec{a}$, meaning $\vec{c}$ is not just $\vec{a} \times \vec{b}$ but also has a component along $\vec{a}$.

Use the identity: $(\vec{a} \times \vec{b}) \cdot \vec{c} = \det(\vec{a},\vec{b},\vec{c})$

But we need magnitudes. Take dot product of $\vec{b}$ with $\vec{c}$:

$\vec{b} \cdot \vec{c} = \vec{b} \cdot (\vec{a} \times \vec{b}) = 0$

But given $\vec{b} \cdot \vec{c} = 1$

So again $\vec{c}$ has components outside the perpendicular direction $\vec{c}$ is independent.

Use the vector triple product identity: $\vec{c} = \vec{a} \times \vec{b}$

Take magnitude squared:

$|\vec{c}|^2 = |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

Since $|\vec{b}| = 1$: $|\vec{c}|^2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$

Now use the given dot products.

We know: $\vec{a} \cdot \vec{c} = 2$ $\vec{b} \cdot \vec{c} = 1$

Take magnitude squared of $\vec{c}$: $|\vec{c}|^2 = (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b})$

But also: $|\vec{c}|^2 = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) = 2 \cdot 1 = 2$

So: $|\vec{c}|^2 = 2$

Thus: $2 = |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2$

Also: $\vec{a} \cdot \vec{b} = \dfrac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$

Therefore: $(\vec{a} \cdot \vec{b})^2 = 2$

Substitute: $2 = |\vec{a}|^2 - 2$

So: $|\vec{a}|^2 = 4$

Hence: $|\vec{a}| = 2$

Qus : 508 MCA NIMCET PYQ

The angle between the asymptotes of the hyperbola $27x^2 - 9y^2 = 24$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Qus : 517 MCA NIMCET PYQ

, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Parametric form of parabola: x = 2at, y = at² Focal chord endpoints: t and –1/t Slope of tangent at t: 1/t Equation of tangent meets the tangent at –1/t Product of x-intercepts = a²

Qus : 521 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$|\vec A\times\vec B|=\sqrt{(2,-2,1)\cdot(2,-2,1)}=3$ Using $|(\vec A\times\vec B)\times\vec C|=|\vec A\times\vec B||\vec C|\sin30^\circ$ From given conditions, $|\vec C|=1$ Hence $=3\times1\times\dfrac12=\dfrac32$ Answer: $\boxed{\dfrac32}$

Qus : 522 MCA NIMCET PYQ

The area enclosed between the curves $y^2=x$ and $y=|x|$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 523 MCA NIMCET PYQ

The value of A that satisfies the equation asinA + bcosA = c is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let

$R = \sqrt{a^2 + b^2}$

Choose $\theta$ such that

$\cos \theta = \frac{b}{R}, \quad \sin \theta = \frac{a}{R}$

Then

$a\sin A + b\cos A = R(\sin A \sin \theta + \cos A \cos \theta) = R\cos(A - \theta)$

Given that

$a\sin A + b\cos A = c,$

we get

$R\cos(A - \theta) = c \quad \Rightarrow \quad \cos(A - \theta) = \frac{c}{\sqrt{a^2 + b^2}}$

Hence,

$A - \theta = \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right) + 2\pi n, \quad n \in \mathbb{Z}$

and since $\tan \theta = \dfrac{a}{b}$, we have $\theta = \tan^{-1}\!\left(\dfrac{a}{b}\right)$

Therefore,

$\boxed{A = \tan^{-1}\!\left(\frac{a}{b}\right) \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right)}$

Qus : 524 MCA NIMCET PYQ

The probability of occurrence of two events E and F are 0.25 and 0.50, respectively. the probability of their simultaneous occurrence is 0.14. the probability that neither E nor F occur is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 525 MCA NIMCET PYQ

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be unit vectors such that the angle between them is ${\cos }^{-1}\Bigg{\{}\frac{1}{4}\Bigg{\}}$. If $\vec{b}=2\vec{c}+\lambda \vec{a}$, where $\lambda$ > 0 and $\vec{b}=4$, then $\lambda$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and the angle between any pair is:

$\cos^{-1}\left(\frac{1}{4}\right)$

Thus, $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=\frac{1}{4}$.

We are given: $\vec{b}=2\vec{c}+\lambda \vec{a}$ and the magnitude: $|\vec{b}|=4$.

Since $\vec{b}$ is not a unit vector here, we use:

$|\vec{b}|^2 = (2\vec{c}+\lambda \vec{a})\cdot(2\vec{c}+\lambda \vec{a})$.

Expand: $|\vec{b}|^2 = 4|\vec{c}|^2 + \lambda^2|\vec{a}|^2 + 4\lambda(\vec{c}\cdot\vec{a})$

Since all are unit vectors: $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$ and $\vec{c}\cdot\vec{a}=\frac14$.

So: $|\vec{b}|^2 = 4 + \lambda^2 + 4\lambda\left(\frac14\right)$

Simplify: $|\vec{b}|^2 = 4 + \lambda^2 + \lambda$

Given: $|\vec{b}| = 4 \Rightarrow |\vec{b}|^2 = 16$

Thus: $\lambda^2 + \lambda + 4 = 16$ $\lambda^2 + \lambda - 12 = 0$

Solve quadratic: $\lambda = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}$

So: $\lambda = 3$ (positive root, since $\lambda>0$)

Final answer: $\boxed{3}$

Qus : 526 MCA NIMCET PYQ

The value of the integral $\displaystyle \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}, dx$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Use substitution: x → 9 – x I = ∫ √x / (√(9–x) + √x) dx I = ∫ √(9–x) / (√x + √(9–x)) dx Add both expressions: 2I = ∫₃⁶ 1 dx = 3 ⇒ I = 3/2

Qus : 527 MCA NIMCET PYQ

A rigid body is rotating at the rate of $3$ radians per second about an axis $AB$, where $A(1,-2,1)$ and $B(3,-4,2)$. The velocity of the point $P(5,-1,-1)$ of the body is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Direction vector of axis $\vec{AB}=(2,-2,1)$ Unit vector along axis $\hat n=\dfrac{(2,-2,1)}{3}$ Angular velocity $\vec\omega=3\hat n=(2,-2,1)$ Position vector of $P$ relative to $A$: $\vec r=(4,1,-2)$ Velocity $\vec v=\vec\omega\times\vec r=3\vec i+8\vec j+10\vec k$

Qus : 528 MCA NIMCET PYQ

Equation of the line perpendicular to x-2y=1 and passing through (1,1) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Qus : 532 MCA NIMCET PYQ

The value of the integral $\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + \sin 2x}, dx$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

sin2x = 2 sinx cosx Rewrite numerator: sinx + cosx = √2 sin(x + π/4) Denominator: 3 + sin2x = 3 + 2 sinx cosx = 1 + (sinx + cosx)² Let t = sinx + cosx dt/dx = cosx – sinx → use identity to convert dx Integral evaluates to (1/4) log 3

Qus : 533 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

$\vec C=-(\vec A+\vec B)$ $|\vec C|^2=|\vec A|^2+|\vec B|^2+2\vec A\cdot\vec B$ $49=9+25+2(3)(5)\cos\theta$ $49=34+30\cos\theta$ $\cos\theta=\dfrac12 \Rightarrow \theta=\dfrac{\pi}{3}$

Qus : 534 MCA NIMCET PYQ

If $A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$ and $f(x) = 1 + x + x^2 + \cdots + x^{16}$, then $f(A) =$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 535 MCA NIMCET PYQ

Find the principal value of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 536 MCA NIMCET PYQ

The value of $\tan 9{^{\circ}}-\tan 27{^{\circ}}-\tan 63{^{\circ}}+\tan 81{^{\circ}}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 537 MCA NIMCET PYQ

If $\vec{a}$ and $\vec{b}$ are twp vectors such that |$\vec{a}$|=3, |$\vec{b}$|=4 and |$\vec{a}+\vec{b}$|=1, then the value of $|\vec{a}-\vec{b}|$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{a} + \vec{b}| = 1$.

Use the identity: $|\vec{a} + \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} + 2\vec{a}\cdot\vec{b}$

Substitute values: $1^{2} = 3^{2} + 4^{2} + 2\vec{a}\cdot\vec{b}$

$1 = 9 + 16 + 2\vec{a}\cdot\vec{b}$

$1 = 25 + 2\vec{a}\cdot\vec{b}$

$2\vec{a}\cdot\vec{b} = -24$

$\vec{a}\cdot\vec{b} = -12$

Now compute $|\vec{a} - \vec{b}|$:

$|\vec{a} - \vec{b}|^{2} = |\vec{a}|^{2} + |\vec{b}|^{2} - 2\vec{a}\cdot\vec{b}$

$= 9 + 16 - 2(-12)$

$= 25 + 24$ $= 49$

So: $|\vec{a} - \vec{b}| = 7$

Qus : 538 MCA NIMCET PYQ

If $ f(x)= \begin{cases} x \sin\left(\frac{1}{x}\right), & x \ne 0 \\ 0, & x = 0 \end{cases} $ then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Solution: lim_{x→0} x·sin(1/x) = 0 ⇒ f is continuous at 0. f'(0) = lim_{x→0} [x sin(1/x)]/x = sin(1/x) But sin(1/x) has no limit as x→0⁺ or x→0⁻. ⇒ Both f'(0+) and f'(0-) do not exist.

Qus : 539 MCA NIMCET PYQ

9 balls are to be placed in 9 boxes and 5 of the balls cannot fit into 3 small boxes. The number of ways of arranging one ball in each of the boxes is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

First off all select 5 boxes out 6 boxes in which 5 big ball can fit then arrange these ball in these 5 boxes and then put remaining 4 ball in any remaining box.

So Ans is [(6C5)5!](4!) = 6!4! = 17280

Qus : 540 MCA NIMCET PYQ

If cosθ = 4/5 and cosϕ = 12/13, θ and ϕ both in the fourth quadrant, the value of cos( θ + ϕ )is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 541 MCA NIMCET PYQ

In a triangle, if the sum of two sides is x and their product is y such that (x+z)(x-z)=y, where z is the third side of the triangle , then triangle is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 542 MCA NIMCET PYQ

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{c}=\hat{i}-2\hat{j}+\hat{k}$, then a vector of magnitude $\sqrt{22}$ which is parallel to $2\vec{a}-\vec{b}+3\vec{c}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$

$\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$

$\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$

Compute: $2\vec{a} - \vec{b} + 3\vec{c}$

First:

$2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}$

$-\vec{b} = -2\hat{i} + \hat{j} - 3\hat{k}$

$3\vec{c} = 3\hat{i} - 6\hat{j} + 3\hat{k}$

Add components:

$i$–component: $2 - 2 + 3 = 3$

$j$–component: $2 + 1 - 6 = -3$

$k$–component: $2 - 3 + 3 = 2$

So the vector: $\vec{v} = 3\hat{i} - 3\hat{j} + 2\hat{k}$

Its magnitude: $|\vec{v}| = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22}$

We need a vector parallel to $\vec{v}$ with magnitude $\sqrt{22}$.

Since $|\vec{v}| = \sqrt{22}$, the required vector is simply: $\boxed{3\hat{i} - 3\hat{j} + 2\hat{k}}$

Qus : 543 MCA NIMCET PYQ

$\int \log_{10} x , dx$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Use change of base: $\log_{10} x = \dfrac{\ln x}{\ln 10}$ $\int \log_{10} x , dx = \dfrac{1}{\ln 10} \int x(\ln x)' dx = \dfrac{1}{\ln 10} (x\ln x - x)$ Rewrite: $= \log 10 \cdot x \log e\left(\frac{x}{e}\right) + c$

Qus : 544 MCA NIMCET PYQ

Which of the following function is the inverse of itself?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Qus : 552 MCA NIMCET PYQ

Let the line $\frac{x}{4}+\frac{y}{2}=1$ meets the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line x + y = 1. Let the point P lie on the line x + y = 1 such that the $\Delta$ABP is an isosceles triangle with AP = BP. Then the distance between M' and P is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

A = $(4,0)$, B = $(0,2)$

Midpoint of AB: $M=\left(\frac{4+0}{2},\frac{0+2}{2}\right)=(2,1)$

Image of $M(2,1)$ across line $x+y=1$:

$M'=(0,-1)$

Since $AP=BP$, point $P$ lies on perpendicular bisector of AB.

Perpendicular bisector of AB passes through $M(2,1)$ and has slope $2$:

$y-1=2(x-2)$

$y=2x-3$

Also $P$ lies on: $x+y=1$

So, $x+2x-3=1$ $3x=4$

$x=\frac{4}{3}$ $y=1-\frac{4}{3}=-\frac{1}{3}$

So, $P=\left(\frac{4}{3},-\frac{1}{3}\right)$

Distance between $M'(0,-1)$ and $P$:

$d=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(-\frac{1}{3}+1\right)^2}$

$d=\sqrt{\frac{16}{9}+\frac{4}{9}}$

$d=\sqrt{\frac{20}{9}}$

$d=\frac{2\sqrt5}{3}$

Final Answer: $\boxed{\frac{2\sqrt5}{3}}$

Qus : 553 MCA NIMCET PYQ

Area between $y = 2 - x^2$ and $y = x^2$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Solve intersection: $2 - x^2 = x^2$ $2x^2 = 2$ $x^2 = 1$ → $x = -1,\ 1$ Area = $\int_{-1}^{1} [(2 - x^2) - x^2] dx$ $= \int_{-1}^{1} (2 - 2x^2) dx$ $= 2\int_{-1}^{1} (1 - x^2) dx$ Compute: $\int_{-1}^{1} 1 dx = 2$ $\int_{-1}^{1} x^2 dx = \frac{2}{3}$ Area = $2(2 - \frac{2}{3}) = \frac{8}{3}$

Qus : 554 MCA NIMCET PYQ

In a survey where 100 students reported which subject they like, 32 students in total liked Mathematics, 38 students liked Business and 30 students liked Literature. Moreover, 7 students liked both Mathematics and Literature, 10 students liked both Mathematics and Business. 8 students like both Business and Literature, 5 students liked all three subjects. Then the number of people who liked exactly one subject is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Given totals: $|M|=32,\ |B|=38,\ |L|=30$

Intersections (inclusive): $|M\cap L|=7,\ |M\cap B|=10,\ |B\cap L|=8,\ |M\cap B\cap L|=5$.

Compute “only” counts:

$\displaystyle |M\ \text{only}|=|M|-|M\cap B|-|M\cap L|+|M\cap B\cap L|=32-10-7+5=20$
$\displaystyle |B\ \text{only}|=|B|-|M\cap B|-|B\cap L|+|M\cap B\cap L|=38-10-8+5=25$
$\displaystyle |L\ \text{only}|=|L|-|M\cap L|-|B\cap L|+|M\cap B\cap L|=30-7-8+5=20$

Exactly one subject: $20+25+20=\boxed{65}$.

Qus : 555 MCA NIMCET PYQ

If non-zero numbers a, b, c are in A.P., then the straight line ax + by + c = 0 always passes through a fixed point, then the point is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Since a, b and c are in A. P. 2b = a + c

a –2b + c =0

The line passes through (1, –2).

Qus : 556 MCA NIMCET PYQ

If $32\, \tan ^8\theta=2\cos ^2\alpha-3\cos \alpha$ and $3\, \cos \, 2\theta=1$, then the general value of $\alpha$ =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 557 MCA NIMCET PYQ

Which one of the following is NOT a correct statement?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Question: Which one of the following is NOT a correct statement?

The value of standard deviation changes by a change of scale.
The standard deviation is greater than or equal to the mean deviation (about mean).
The sum of squares of deviations is minimum when taken from the mean.
The variance is expressed in the same units as the units of observation.

Answer: Option 4

Why: Variance has squared units, not the same units as the data (e.g., if data are in cm, variance is in cm²). Standard deviation (the square root of variance) has the same units as the data.

Notes:

Change of scale: if each value is multiplied by $k$, then $\text{SD}$ becomes $|k|\cdot \text{SD}$ and $\text{Var}$ becomes $k^2\cdot \text{Var}$.
$\text{SD} \ge \text{Mean Deviation (about mean)}$ is a standard inequality.
$\sum (x_i - a)^2$ is minimized at $a = \bar{x}$ (the mean).

Qus : 558 MCA NIMCET PYQ

A vector $\vec{a}$ has components $2p$ and $1$. After rotation, it becomes $(p+1,\ 1)$. Find $p$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Length of vector does not change under rotation: $\sqrt{(2p)^2 + 1^2} = \sqrt{(p+1)^2 + 1^2}$ Square both sides: $4p^2 + 1 = (p+1)^2 + 1$ $4p^2 + 1 = p^2 + 2p + 2$ $3p^2 - 2p - 1 = 0$ Solve: $p = 1$ or $p = -\frac{1}{3}$ But the option lists $\frac13$ (positive), so valid match is:

Qus : 559 MCA NIMCET PYQ

If A and B are two events and $P(A \cup B) = \dfrac{5}{6}$, $P(A \cap B) = \dfrac{1}{2}$, the A and B are two events which are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 560 MCA NIMCET PYQ

If the lines x + (a – 1)y + 1 = 0 and 2x + a²y – 1 = 0 are perpendicular, then the condition satisfies by a is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 561 MCA NIMCET PYQ

In a triangle ABC, $a\cos ^2\frac{C}{2}+\, c\, \, {\cos }^2\frac{A}{2}=\frac{3b}{2}$ then the sides of the triangle are in

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 562 MCA NIMCET PYQ

An equilateral triangle is inscribed in the parabola $y^2 = x$. One vertex of the triangle is at the vertex of the parabola. The centroid of triangle is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given parabola $y^{2} = x. $ $ A(0,0) $ is one vertex of the equilateral triangle.

Let the other two vertices be $ B(t^{2}, t) $ and $ C(t^{2}, -t) $, since they lie on the parabola and are symmetric.

Distance: $ BC = \sqrt{(t^{2}-t^{2})^{2} + (t - (-t))^{2}} = 2t. $

And $ AB = \sqrt{(t^{2}-0)^{2} + (t-0)^{2}} = \sqrt{t^{4}+t^{2}} = t\sqrt{t^{2}+1}. $

For equilateral triangle: $ AB = BC $

$ t\sqrt{t^{2}+1} = 2t $ $ \sqrt{t^{2}+1} = 2 $

$ t^{2}+1 = 4 $

$ t^{2} = 3. $

So the points become:

$ B = (3,\sqrt{3}), \quad C = (3,-\sqrt{3}). $

Centroid: $ G = \left( \frac{0+3+3}{3},; \frac{0+\sqrt3-\sqrt3}{3} \right) = (2,0). $

Qus : 563 MCA NIMCET PYQ

The vectors $\vec{a},\vec{b},\vec{c}$ are equal in length and taken pairwise make equal angles. If $\vec{a}=\hat{i}+\hat{j}$, $\vec{b}=\hat{j}+\hat{k}$ and $\vec{c}$ makes an obtuse angle with $\hat{i}$, then $\vec{c}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Equal lengths and equal pairwise angles ⇒ vectors form a symmetric set. Dot-products must satisfy: $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}$ Also $\vec{c}$ must make obtuse angle with $\hat{i}$ ⇒ its $i$-component < 0. Solving gives: $\vec{c}=\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$

Qus : 564 MCA NIMCET PYQ

If $a_1, a_2,...a_n$ are positive real numbers whose product is a fixed number c, then the minimum of $a_1, a_2, ....2a_n$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Key Concept: AM-GM Inequality
$\dfrac{a_1 + a_2 + \cdots + a_n}{n} \geq (a_1 \cdot a_2 \cdots a_n)^{1/n}$

Step 1: Rewrite the sum
$a_1 + a_2 + \cdots + a_{n-1} + 2a_n$

This has $n$ terms: $(a_1, a_2, \ldots, a_{n-1}, 2a_n)$

Step 2: Apply AM-GM
$\dfrac{a_1 + a_2 + \cdots + a_{n-1} + 2a_n}{n} \geq (a_1 \cdot a_2 \cdots a_{n-1} \cdot 2a_n)^{1/n}$

$\geq (2 \cdot a_1 a_2 \cdots a_n)^{1/n}$

$\geq (2c)^{1/n}$

Step 3: Find minimum
$a_1 + a_2 + \cdots + 2a_n \geq n(2c)^{1/n}$

Minimum value $= n(2c)^{1/n}$

Equality holds when $a_1 = a_2 = \cdots = a_{n-1} = 2a_n$

Answer: Minimum value $= \boxed{n(2c)^{1/n}}$

Qus : 565 MCA NIMCET PYQ

In a triangle ABC, let angle C = π/2. If R is the inradius and R is circumradius of the triangle ABC, then 2(r + R) equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 566 MCA NIMCET PYQ

If |k|=5 and 0° ≤ θ ≤ 360°, then the number of distinct solutions of 3cos⁡θ + 4sin⁡θ = k is

NIMCET 2021

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 567 MCA NIMCET PYQ

If θ is acute angle between the pair of lines $x^2-7xy+12y^2=0$, then $\frac{2\cos \theta+3\sin \theta}{4\sin \theta+5\cos \theta}=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 568 MCA NIMCET PYQ

The angles of depression of the top and bottom of an 8m tall building from the top of a multi storied building are 30° and 45°, respectively. What is the height of the multistoried building and the distance between the two buildings?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

The height of the multistoried building is $H$ and the horizontal distance between the buildings is $D$.

The angles of depression to the top and bottom of the $8\text{ m}$ building are $30^\circ$ and $45^\circ$ respectively.

From the $45^\circ$ angle:

$\tan 45^\circ = \dfrac{H}{D}$

$1 = \dfrac{H}{D}$

$H = D$

From the $30^\circ$ angle:

$\tan 30^\circ = \dfrac{H - 8}{D}$

$\dfrac{1}{\sqrt{3}} = \dfrac{H - 8}{H}$

$H - 8 = \dfrac{H}{\sqrt{3}}$

$H\left(1 - \dfrac{1}{\sqrt{3}}\right) = 8$

Solve for $H$:

$H = \dfrac{8}{1 - \dfrac{1}{\sqrt{3}}}$

$H = 4(3 + \sqrt{3})$

Since $H = D$:

$D = 4(3 + \sqrt{3})$

Final answers:

$\boxed{H = 4(3 + \sqrt{3})\ \text{m}}$

$\boxed{D = 4(3 + \sqrt{3})\ \text{m}}$

Qus : 569 MCA NIMCET PYQ

The position vectors of $A,B,C,D$ are $\hat{i}+\hat{j}+\hat{k}$, $2\hat{i}+5\hat{j}$, $3\hat{i}+2\hat{j}-3\hat{k}$, $\hat{i}-6\hat{j}-\hat{k}$ Angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Compute vectors: $\overrightarrow{AB} = (2-1)\hat{i}+(5-1)\hat{j}+(0-1)\hat{k}=\hat{i}+4\hat{j}-\hat{k}$ $\overrightarrow{CD} = (1-3)\hat{i}+(-6-2)\hat{j}+(-1+3)\hat{k}=-2\hat{i}-8\hat{j}+2\hat{k}$ They are scalar multiples: $\overrightarrow{CD} = -2(\overrightarrow{AB})$ ⇒ Angle = $\pi$

Qus : 570 MCA NIMCET PYQ

If $a, b, c$ are the roots of the equation $x^3 - 3x^2 + 3x + 7 = 0$, then the value of\[\begin{vmatrix}2bc - a^2 & c^2 & b^2 \\c^2 & 2ac - b^2 & a^2 \\b^2 & a^2 & 2ab - c^2\end{vmatrix}\]is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 571 MCA NIMCET PYQ

If x² + 3xy + 2y² – x – 4y – 6 = 0 represents a pair of straight lines, their point of intersection is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 572 MCA NIMCET PYQ

The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as end with T is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 573 MCA NIMCET PYQ

The four geometric means between 2 and 64 are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Step 1: Set up the GP
The sequence is: $2,\ G_1,\ G_2,\ G_3,\ G_4,\ 64$
Total terms $= 6$, so $n = 6$

Step 2: Find common ratio $r$
$a = 2,\ a_6 = 64$

$a_6 = a \cdot r^{n-1}$
$64 = 2 \cdot r^5$
$r^5 = 32$
$r^5 = 2^5$
$r = 2$

Step 3: Find the four geometric means
$G_1 = ar = 2 \times 2 = 4$
$G_2 = ar^2 = 2 \times 4 = 8$
$G_3 = ar^3 = 2 \times 8 = 16$
$G_4 = ar^4 = 2 \times 16 = 32$

Answer: The four geometric means are $\boxed{4, 8, 16, 32}$

Qus : 574 MCA NIMCET PYQ

The number of accidents per week in a town follows Poisson distribution with mean 3 (In Exam Given 2, which is incorrect). If the probability that there are three accidents in two weeks time is $ke^{-6}$, then the value of k is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Mean accidents per week = $3$.

Mean accidents in two weeks = $6$.

For a Poisson distribution,

$P(X=3) = \dfrac{6^{3}}{3!} e^{-6}$.

Given $P(X=3) = k e^{-6}$,

so $k = \dfrac{6^{3}}{3!}$.

$k = \dfrac{216}{6} = 36$.

Qus : 575 MCA NIMCET PYQ

Let $\vec{a},\vec{b},\vec{c}$ be three non-zero vectors, no two collinear. If $\vec{a}+\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+\vec{b}+\vec{c}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Conditions: $\vec{a}+\vec{b}=\lambda\vec{c}$ $\vec{b}+\vec{c}=\mu\vec{a}$ Solving gives: $\vec{a}+\vec{b}+\vec{c}=0$ Which is none of the given vectors.

$B_{\min} = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{3}{4}$.

Maximum occurs at $t = 0$ or $t = 1$:

$B_{\max} = 1$.

Hence, $B \in \left[\dfrac{3}{4},, 1\right]$.

Qus : 581 MCA NIMCET PYQ

If $C$ is midpoint of $AB$ and $P$ is any point outside $AB$, then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

$\vec{C}=\frac{\vec{A}+\vec{B}}{2}$ $\overrightarrow{PA} = \vec{A}-\vec{P}$ $\overrightarrow{PB} = \vec{B}-\vec{P}$ Add: $\overrightarrow{PA}+\overrightarrow{PB} = \vec{A}+\vec{B}-2\vec{P}$ But $\overrightarrow{PC}=\frac{\vec{A}+\vec{B}}{2} - \vec{P}$ ⇒ $2\overrightarrow{PC} = \vec{A}+\vec{B}-2\vec{P}$ Thus: $\overrightarrow{PA}+\overrightarrow{PB} = 2\overrightarrow{PC}$

Qus : 582 MCA NIMCET PYQ

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - px + r = 0$ and $\dfrac{\alpha}{2}, \beta$ are the roots of the equation $x^2 - ax + r = 0$, then the value of $r$,

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 583 MCA NIMCET PYQ

If the graph of y = (x – 2)² – 3 is shifted by 5 units up along y-axis and 2 units to the right along the x-axis, then the equation of the resultant graph is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

When y= f (x) is shifted by k units to the right along x

– axis, it become y= f (x - k )

Hence, new equation of

graph is y = (x - 4)² + 2

Qus : 584 MCA NIMCET PYQ

Let P(E) denote the probability of event E. Given P(A) = 1, P(B) =$\frac{1}{2}$ the value of P(A|B) and P(B|A) respectively are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 585 MCA NIMCET PYQ

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k}$ , $(a,b,c\ne1)$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 586 MCA NIMCET PYQ

Let $F_1, F_2$ be foci of hyperbola $\frac{{x}^2}{{a}^2}-\frac{{y}^2}{{b}^2}=1$, a>0, b>0, and let O be the origin. Let M be an arbitrary point on curve C and above X-axis and H be a point on $MF_1$ such that $MF_2\perp{{F}}_1{{F}}_2$, $MF_1\perp{{O}}{{H}}$, $|OH|=\lambda |OF_2|$ with $\lambda \in(2/5, 3/5)$, then the range of the eccentricity $e$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

If the system of equations $3x-y+4z=3$ , $x+2y-3z=-2$ , $6x+5y+λz=-3 $ has atleast one solution, then $λ=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 597 MCA NIMCET PYQ

If $\alpha$ and $\beta$ are the two roots of the quadratic equation $x^2 + ax + b = 0, (ab \ne 0)$ then the quadratic roots whose roots $\frac{1}{\alpha^3+\alpha}$ and $\frac{1}{\beta^3+\beta}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given $x^{2} + ax + b = 0$ with roots $\alpha,\beta$,

we use $\alpha + \beta = -a$ and $\alpha\beta = b$.

Required new roots are $\dfrac{1}{\alpha^{3}+\alpha}$ and $\dfrac{1}{\beta^{3}+\beta}$.

Since $\alpha^{3}+\alpha = \alpha(\alpha^{2}+1)$ and using $\alpha^{2}=-a\alpha-b$ (and same for $\beta$), after simplification the sum and product of new roots become:

$u+v = \dfrac{a^{3}+a-3ab}{b(b^{2}+1+a^{2}-2b)}$

$uv = \dfrac{1}{b(b^{2}+1+a^{2}-2b)}$

So the required quadratic is:

$b(b^{2}+1+a^{2}-2b)x^{2} - (a^{3}+a-3ab)x + 1 = 0$

Qus : 598 MCA NIMCET PYQ

In $\triangle ABC$, $R$ is circumradius and $8R^2=a^2+b^2+c^2$. Then $\triangle ABC$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

In any triangle: $a^2+b^2+c^2 = 8R^2$ ⇔ right-angled triangle identity.

Qus : 599 MCA NIMCET PYQ

We use the identity

$\sin A + \sin B = 2 \sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2}$.

So, $\sin x + \sin(x+1) = 2 \sin\left(x+\dfrac12\right)\cos\dfrac12$.

Maximum value of $\sin(x+\tfrac12)$ is $1$.

Therefore maximum of the expression is: $2 \cos\dfrac12$.

Given maximum $= k \cos\dfrac12$, so $k = 2$.

Qus : 604 MCA NIMCET PYQ

The rate of increase of length of the shadow of a man $2$ meters high, due to a lamp at $10$ meters height, when he is moving away from it at $2 \text{ m/sec}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Let $x =$ distance of man from lamp

Let $y =$ length of shadow

By similar triangles: $\dfrac{10}{x+y} = \dfrac{2}{y}$ Cross-multiply:

$10y = 2(x+y)$

$10y = 2x + 2y$

$8y = 2x$

$y = \dfrac{x}{4}$

Differentiate w.r.t time $t$:

$\dfrac{dy}{dt} = \dfrac{1}{4}\dfrac{dx}{dt}$

Given $\dfrac{dx}{dt} = 2$ m/sec

: $\dfrac{dy}{dt} = \dfrac{1}{4} \times 2 = \dfrac{1}{2}$

Qus : 605 MCA NIMCET PYQ

and

, (a ≠ b ≠ c ≠ 1) are co-planar, then the value of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 606 MCA NIMCET PYQ

A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1, 30% are brand 2 and 20% are brand 3. Each manufacturer offers one year warranty on parts and labor. It is known that 25% of brand 1 DVD players require warranty repair work whereas the corresponding percentage for brands 2 and 3 are 20% and 10% respectively. The probability that a randomly selected purchaser has a DVD player that will need repair while under warranty, is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

This is a Total Probability Problem.
Given Information:
$P(B_1) = 0.50, $

$ P(B_2) = 0.30, $

$P(B_3) = 0.20$

$P(R|B_1) = 0.25$

$P(R|B_2) = 0.20$

$P(R|B_3) = 0.10$

Law of Total Probability:

$P(R) = \sum_{i=1}^{3} P(R|B_i) \cdot P(B_i)$

$P(R) = P(R|B_1)\cdot P(B_1) + P(R|B_2)\cdot P(B_2) + P(R|B_3)\cdot P(B_3)$

Substituting Values:

$P(R) = (0.25)(0.50) + (0.20)(0.30) + (0.10)(0.20)$
$P(R) = 0.125 + 0.060 + 0.020$

Final Answer:

$\boxed{P(R) = 0.205}$
$\therefore $

The probability that a randomly selected purchaser will need warranty repair is $ \mathbf{20.5\%}$

Qus : 607 MCA NIMCET PYQ

If A={1,2,3,4} and B={3,4,5}, then the number of elements in (A∪B)×(A∩B)×(AΔB)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Given: $A=\{1,2,3,4\}$, $B=\{3,4,5\}$

$A\cup B=\{1,2,3,4,5\}\Rightarrow |A\cup B|=5$
$A\cap B=\{3,4\}\Rightarrow |A\cap B|=2$
$A\triangle B=(A\cup B)\setminus(A\cap B)=\{1,2,5\}\Rightarrow |A\triangle B|=3$

Size of Cartesian product: $|(A\cup B)\times(A\cap B)\times(A\triangle B)| = 5\times 2\times 3 = \mathbf{30}$.

Qus : 608 MCA NIMCET PYQ

Let $\mathbb{R}\rightarrow\mathbb{R}$ be any function defined as $f(x)=\begin{cases}{{x}^{\alpha}\sin \frac{1}{{x}^{\beta}}} & {,x\ne0} \\ {0} & {,x=0}\end{cases}$, $\alpha , \beta \in \mathbb{R}$. Which of the following is true? ($\mathbb{R}$ denotes the set of all real numbers)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

The function is $f(x) = x^{\alpha}\sin\left(\dfrac{1}{x^{\beta}}\right)$ for $x \ne 0$, and $f(0)=0$.

To check continuity at $x=0$, consider: $\displaystyle \lim_{x\to 0} x^{\alpha}\sin\left(\frac{1}{x^{\beta}}\right)$.

Since $|\sin \theta| \le 1$ for all $\theta$

$|x^{\alpha}\sin(1/x^{\beta})| \le |x^{\alpha}|$.

Now, $\displaystyle \lim_{x\to 0} x^{\alpha} = 0 \quad \text{iff } \alpha > 0$.

Therefore, $\displaystyle \lim_{x\to 0} x^{\alpha}\sin(1/x^{\beta}) = 0 = f(0)$ iff $\alpha > 0$.

The value of $\beta$ does not matter because $\sin(1/x^{\beta})$ is always bounded.

Hence: ${f(x)\text{ is continuous at }x=0\text{ for all }\alpha>0\text{ and }\beta\in\mathbb{R}.}$

Qus : 609 MCA NIMCET PYQ

A person stands at a point $A$ due south of a tower and observes elevation $60^\circ$. He walks west to $B$, elevation becomes $45^\circ$. At point $C$ on $AB$ extended, elevation becomes $30^\circ$. Find $\dfrac{AB}{BC}$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Let height of tower = $h$

From A (angle $60^\circ$):

$\tan 60^\circ = \frac{h}{OA} \Rightarrow OA = \frac{h}{\sqrt{3}}$

From B (angle $45^\circ$):

$\tan 45^\circ = \frac{h}{OB} \Rightarrow OB = h$

$AB = OB - OA = h - \frac{h}{\sqrt{3}}$

From C (angle $30^\circ$):

$\tan 30^\circ = \frac{h}{OC} \Rightarrow OC = \sqrt{3}h$

$BC = OC - OB = \sqrt{3}h - h$

Now ratio:

$\frac{AB}{BC} = \frac{h - \frac{h}{\sqrt{3}}}{\sqrt{3}h - h}$

$= \frac{h\left(1 - \frac{1}{\sqrt{3}}\right)}{h(\sqrt{3} - 1)}$

$= \frac{1 - \frac{1}{\sqrt{3}}}{\sqrt{3} - 1}$

Multiply numerator & denominator by $\sqrt{3}$:

$= \frac{\sqrt{3} - 1}{\sqrt{3}(\sqrt{3} - 1)}$

$= \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{1} = 1$

Final Answer: $$\boxed{1}$$

Qus : 610 MCA NIMCET PYQ

Differential coefficient of $\log_{10} x$ with respect to $\log_{x} 10$ to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 611 MCA NIMCET PYQ

Let a, b and c be three vectors having magnitudes 1, 1 and 2 respectively. If a x (a x c) - b = 0, then the acute angle between a and c is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 612 MCA NIMCET PYQ

The locus of the intersection of the two lines $\sqrt{3} x-y=4k\sqrt{3}$ and $k(\sqrt{3}x+y)=4\sqrt{3}$, for different values of k, is a hyperbola. The eccentricity of the hyperbola is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 613 MCA NIMCET PYQ

If n is an integer between 0 to 21, then find a value of n for which the value of $n!(21-n)!$ is minimum

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 614 MCA NIMCET PYQ

The number of 3-digit integers that are multiple of 6 which can be formed by using the digits 1,2,3,4,5,6 without repetition is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

To be a multiple of $6$:

• last digit must be even → ${2,4,6}$

• sum of digits must be divisible by $3$.

For each even last digit, exactly 8 valid choices of the first two digits satisfy divisibility by $3$ (checked by mod – 3 pairing).

So total numbers: 3×8=24

Qus : 615 MCA NIMCET PYQ

Distance between the parallel lines $y = 2x + 4$ and $6x = 3y + 5$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Rewrite second line: $6x - 3y - 5 = 0$ Divide by 3: $2x - y - \dfrac{5}{3} = 0$ First line: $y = 2x + 4 \Rightarrow 2x - y + 4 = 0$ Distance between parallel lines: $d = \dfrac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$ Here: $c_1 = 4$, $c_2 = -\dfrac{5}{3}$, $a=2,\ b=-1$ Compute: $d = \dfrac{\left|4 - \left(-\dfrac{5}{3}\right)\right|}{\sqrt{4+1}} = \dfrac{\left|\dfrac{12}{3}+\dfrac{5}{3}\right|}{\sqrt{5}} = \dfrac{17/3}{\sqrt{5}} = \dfrac{17\sqrt{5}}{15}$

Qus : 616 MCA NIMCET PYQ

$f(x) = x + |x|$ is continuous for

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 617 MCA NIMCET PYQ

Let

and

be three vector such that |

| = 2, |

| = 3, |

| = 5 and

= 0. The value of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 618 MCA NIMCET PYQ

Constant forces $\vec{P}= 2\hat{i} - 5\hat{j} + 6\hat{k} $ and $\vec{Q}= -\hat{i} + 2\hat{j}- \hat{k}$ act on a particle. The work done when the particle is displaced from A whose position vector is $4\hat{i} - 3\hat{j} - 2\hat{k} $, to B whose position vector is $6\hat{i} + \hat{j} - 3k\hat{k}$ , is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 619 MCA NIMCET PYQ

Suppose $A_1,A_2,\ldots,A_{30}$ are 30 sets each with five elements and $B_1,B_2,B_3,\ldots,B_n$ are n sets (each with three elements) such that $\bigcup ^{30}_{i=1}{{A}}_i={{\bigcup }}^n_{j=1}{{B}}_i=S\, $ and each element of S belongs to exactly ten of the $A_i$'s and exactly 9 of the $B^{\prime}_j$'s. Then $n=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Let $|S|=m$.

Thus $n\times 3=135 \Rightarrow n=\dfrac{135}{3}=\boxed{45}.$

Qus : 620 MCA NIMCET PYQ

The circle $x^2 + y^2+ \alpha x+ \beta y+ \gamma=0$ is the image of the circle $x^2 + y^2- 6x- 10y+ 30=0$ across the line 3x + y = 2. The value of $[\alpha+ \beta+ \gamma]$ is (where [.] represents the floor function.)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given circle:

$x^{2}+y^{2}-6x-10y+30=0$

Center and radius: $C(3,5), \quad r=2$

We reflect $C(3,5)$ across the line $3x + y - 2 = 0$.

Compute reflection factor: $k = \dfrac{2(3\cdot 3 + 5 - 2)}{3^{2} + 1^{2}} = \dfrac{24}{10} = \dfrac{12}{5}$

Reflected center:

$x' = 3 - 3k = 3 - \dfrac{36}{5} = -\dfrac{21}{5}$

$y' = 5 - k = 5 - \dfrac{12}{5} = \dfrac{13}{5}$

So new center is:

$C'\left(-\dfrac{21}{5},\dfrac{13}{5}\right)$

The image circle is $x^{2}+y^{2}+\alpha x+\beta y+\gamma = 0$

Using center formula: $-\dfrac{\alpha}{2} = -\dfrac{21}{5} \Rightarrow \alpha = \dfrac{42}{5}$

$-\dfrac{\beta}{2} = \dfrac{13}{5} \Rightarrow \beta = -\dfrac{26}{5}$

Using radius $r=2$: $\left(-\dfrac{21}{5}\right)^{2} + \left(\dfrac{13}{5}\right)^{2} - \gamma = 4$

$\dfrac{610}{25} - \gamma = 4$

$\gamma = \dfrac{102}{5}$

Now compute:

$\alpha + \beta + \gamma = \dfrac{42}{5} - \dfrac{26}{5} + \dfrac{102}{5} = \dfrac{118}{5} = 23.6$

Thus, $\boxed{23}$

Qus : 621 MCA NIMCET PYQ

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors, then $|\mathbf{a} - \mathbf{b}|^2 + |\mathbf{b} - \mathbf{c}|^2 + |\mathbf{c} - \mathbf{a}|^2$ does not exceeds

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 622 MCA NIMCET PYQ

= (i + 2j - 3k) and

=(3i -j + 2k), then the angle between (

) and (

)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 623 MCA NIMCET PYQ

The value of $\int \sqrt{x} e^{\sqrt{x}} dx$ is equal to:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 624 MCA NIMCET PYQ

The eccentric angle of the extremities of latus-rectum of the ellipse $\frac{{x}^2}{{a}^2}^{}+\frac{{y}^2}{{b}^2}^{}=1$ are given by

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 625 MCA NIMCET PYQ

Let $\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-\hat{k}$ and $\vec{c}=3\hat{i}+\hat{j}+\lambda \hat{k}$ be the co-terminal edges

of a parallelopiped whose volume is 5 units. Then the value of $\lambda$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given $\vec a = \langle 2,-3,4\rangle,\; \vec b = \langle 1,2,-1\rangle,\; \vec c = \langle 3,1,\lambda\rangle$

Volume condition: $|\vec a \cdot (\vec b \times \vec c)| = 5$

Compute cross product using determinant: \[ \vec b \times \vec c = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 3 & 1 & \lambda \end{vmatrix} \] \[ = (2\lambda + 1)\mathbf{i} - (\lambda + 3)\mathbf{j} - 5\mathbf{k} \] So $\vec b \times \vec c = \langle 2\lambda + 1,\; -(\lambda + 3),\; -5\rangle$

Scalar triple product: \[ \vec a \cdot (\vec b \times \vec c) = 2(2\lambda + 1) + (-3)(-(\lambda + 3)) + 4(-5) \] Simplifying: \[ = 4\lambda + 2 + 3(\lambda + 3) - 20 = 7\lambda - 9 \] Volume equation: \[ |7\lambda - 9| = 5 \] Solve: \[ 7\lambda - 9 = 5 \Rightarrow \lambda = 2 \] \[ 7\lambda - 9 = -5 \Rightarrow \lambda = \frac{4}{7} \] Final answer: \[ \boxed{\lambda = 2 \text{ or } \frac{4}{7}} \]

Qus : 626 MCA NIMCET PYQ

The vector $\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}$ lies in the plane of the vector $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which of the following gives possible values of $\alpha$ and $\beta$? \begin{enumerate}

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 627 MCA NIMCET PYQ

The number of elements in the power set P(S) of the set S = {2, (1, 4)} is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Given: $ S = \{2, (1,4)\} $

Here, the set $S$ has two elements:

The number $2$
The ordered pair $(1,4)$

Hence, $|S| = 2$.

Formula: The number of elements in the power set of a set having $n$ elements is $2^n$.

\[ |P(S)| = 2^{|S|} = 2^2 = 4 \]

✅ Therefore, the number of elements in $P(S)$ is 4.

Qus : 628 MCA NIMCET PYQ

For the vectors $\vec{a}=-4\hat{i}+2\hat{j}, \vec{b}=2\hat{i}+\hat{j}$ and $\vec{c}=2\hat{i}+3\hat{j}$, if $\vec{c}=m\vec{a}+n\vec{b}$ then the value of m + n is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 629 MCA NIMCET PYQ

If α≠β and $\alpha^2=5\alpha-3,\beta^2=5\beta-3$, then the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 630 MCA NIMCET PYQ

The area enclosed between the curve y = sin x, y = cosx, $0\leq x\leq\frac{\pi}{2}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 637 MCA NIMCET PYQ

m distinct animals of a circus have to be placed in m cages, one is each cage. There are n small cages and p large animal (n < p < m). The large animals are so large that they do not fit in small cage. However, small animals can be put in any cage. The number of putting the animals into cage is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 638 MCA NIMCET PYQ

The number of ways in which 5 days can be chosen in each of the 12 months of a non-leap year, is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 639 MCA NIMCET PYQ

Angle between $\vec{a}$ and $\vec{b}$ is $120{^{\circ}}$. If $|\vec{b}|=2|\vec{a}|$ and the vectors , $\vec{a}+x\vec{b}$ , $\vec{a}-\vec{b}$ are at right angle, then $x=$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 640 MCA NIMCET PYQ

Number of three digit numbers that can be formed using 0, 1, 2, 3 and 5 where these digits are allowed to repeat any number of times, is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Digits available: 0, 1, 2, 3, 5 (total 5 digits).

Repetition is allowed.

To form a three-digit number:

1) The first digit cannot be 0.

Possible choices = 1, 2, 3, 5 → 4 choices.

2) The second digit can be any of the 5 digits → 5 choices.

3) The third digit can also be any of the 5 digits → 5 choices.

Total three-digit numbers = $4 \times 5 \times 5 = 100$.

Final Answer: 100

Qus : 641 MCA NIMCET PYQ

The value of $\cot\!\left(cosec^{-1}\dfrac{5}{3} + \tan^{-1}\dfrac{2}{3}\right)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 642 MCA NIMCET PYQ

Let A and B two sets containing four and two elements respectively. The number of subsets of the A × B, each having at least three elements is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$|A \times B| = 4 \times 2 = 8$

Total subsets $= 2^8 = 256$

Subsets with fewer than 3 elements: $^8C_0 + ^8C_1 + ^8C_2 = 1 + 8 + 28 = 37$

Subsets with at least 3 elements $= 256 - 37 = 219$

Answer: $\boxed{219}$ ✅

Qus : 643 MCA NIMCET PYQ

If [x] represents the greatest integer not exceeding x, then $\int_{0}^{9} [x] dx $ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 644 MCA NIMCET PYQ

If a number x is selected at random from natural numbers 1,2,…,100, then the probability for $x+\frac{100}{x}{\gt}29$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 645 MCA NIMCET PYQ

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ and $h:\mathbb{R}\rightarrow \mathbb{R}$, be two functions such that $h(x) = sgn(g(x))$. Then select which of the following is not true?( $\mathbb{R}$ denotes the set of all real numbers, sgn stands for signum function)

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let $g:\mathbb{R}\to\mathbb{R}$ and $h:\mathbb{R}\to\mathbb{R}$ be such that $h(x) = \operatorname{sgn}(g(x))$.

Recall: $\operatorname{sgn}(t) = \begin{cases} 1, & t>0\\ 0, & t=0\\ -1, & t<0 \end{cases}$

Check each statement:

1) "The domain of $h(x)$ is the same as the domain of $g(x)$." $\Rightarrow$ True, because $\operatorname{sgn}(g(x))$ is defined for every $x$ where $g(x)$ is defined.

2) "The domain of continuity of $h(x)$ equals the domain of continuity of $g(x) - \{x\in\mathbb{R} : g(x)=0\}$."

At points where $g(x)\neq 0$, $h(x)$ is locally constant ($1$ or $-1$), hence continuous there (provided $g$ itself is continuous).

At points where $g(x)=0$, $h(x)$ jumps from $-1$ to $1$, so it is discontinuous. $\Rightarrow$

This statement is true.

3) "The domain of $h(x)$ is different from the domain of $g(x)$ at the same point."

Since for every $x$ in the domain of $g$, $h(x)=\operatorname{sgn}(g(x))$ is defined, the domains are exactly the same; they never differ.

$\Rightarrow$ This statement is false.

4) " $h(x)$ is discontinuous at $g(x)=0$."

At any $x_0$ where $g(x_0)=0$, the left and right limits of $h(x)$ are $-1$ and $1$, not equal to $h(x_0)=0$.

$\Rightarrow$ $h$ is discontinuous there, so this statement is true.

Therefore, the statement which is **not true** is: $\boxed{\text{Option 3}}$

Qus : 646 MCA NIMCET PYQ

If $\sin\theta = 3\sin(\theta + 2\alpha)$, then value of $\tan(\theta + \alpha) + 2\tan\alpha$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 647 MCA NIMCET PYQ

The slope of the function \[ f(x) = \begin{cases} x^2 \sin\!\left(\dfrac{1}{x}\right), & \text{if } x \ne 0, \\[8pt] 0, & \text{if } x = 0 \end{cases} \]

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$

$=\lim_{x\to0}\dfrac{x^2\sin(1/x)}{x}$

Let first term $= a$, common difference $= d$, so $t_n = a + (n-1)d$

Step 1: Expand $\displaystyle\sum_{l=0}^{18} t_{3l+1} = 1197$
Terms: $t_1, t_4, t_7, \ldots, t_{55}$ (19 terms, $l = 0$ to $18$)

$t_{3l+1} = a + 3ld$

$\displaystyle\sum_{l=0}^{18}(a + 3ld) $

$= 19a + 3d\cdot\dfrac{18 \times 19}{2} $

$= 19a + 513d = 1197$

$\Rightarrow a + 27d = 63 \quad \cdots(1)$

Step 2: Use $t_7 + 3t_{22} = 174$
$t_7 = a + 6d,\quad t_{22} = a + 21d$

$(a + 6d) + 3(a + 21d) = 174$
$4a + 69d = 174 \quad \cdots(2)$

Step 3: Solve (1) and (2)
From (1): $a = 63 - 27d$

Substitute in (2):
$4(63 - 27d) + 69d = 174$
$252 - 108d + 69d = 174$
$-39d = -78$
$d = 2$

$a = 63 - 54 = 9$

Step 4: Find $\displaystyle\sum_{l=1}^{9} t_l^2$
$t_l = 9 + (l-1) \times 2 = 7 + 2l$

$\displaystyle\sum_{l=1}^{9} t_l^2 = \sum_{l=1}^{9}(7+2l)^2 = \sum_{l=1}^{9}(49 + 28l + 4l^2)$

$= 49(9) + 28\cdot\dfrac{9 \times 10}{2} + 4\cdot\dfrac{9 \times 10 \times 19}{6}$

$= 441 + 1260 + 1140$

$= 2841$

Step 5: Find $b$
$947b = 2841$

$b = \dfrac{2841}{947} = 3$

Answer: $b = \boxed{3}$

Qus : 656 MCA NIMCET PYQ

Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 657 MCA NIMCET PYQ

The value of $\int_{0}^{\pi}x^3 \sin x dx$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let $I=\displaystyle\int_{0}^{\pi}x^{3}\sin x\,dx$.

Using integration by parts: let $u=x^{3},\; dv=\sin x\,dx$ $\Rightarrow du=3x^{2}dx,\; v=-\cos x$

$I = [-x^{3}\cos x]_0^{\pi} + \displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx$

Now, let $J=\displaystyle\int_{0}^{\pi}x^{2}\cos x\,dx$

Again, by parts: $u=x^{2},\; dv=\cos x\,dx \Rightarrow du=2x\,dx,\; v=\sin x$

$J = [x^{2}\sin x]_0^{\pi} - \displaystyle\int_{0}^{\pi}2x\sin x\,dx$

Let $K=\displaystyle\int_{0}^{\pi}x\sin x\,dx$ By parts: $u=x,\; dv=\sin x\,dx \Rightarrow du=dx,\; v=-\cos x$

$K=[-x\cos x]_0^{\pi}+\displaystyle\int_{0}^{\pi}\cos x\,dx = \pi$

So $J=0-2K=-2\pi$

Then $\displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx = 3J = -6\pi$

$I = [-x^{3}\cos x]_0^{\pi} + 3J = (-\pi^{3}\cos\pi - 0) - 6\pi = \pi^{3} - 6\pi$

Answer: $\boxed{\pi^{3} - 6\pi}$ ✅

Qus : 658 MCA NIMCET PYQ

If $A=\begin{bmatrix} a &b &c \\ b & c & a\\ c& a &b \end{bmatrix}$ , where $a, b, c$ are real positive numbers such that $abc = 1$ and $A^{T}A=I$ then the equation that not holds true among the following is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 659 MCA NIMCET PYQ

If the matrix $ \begin{bmatrix} -1 & 3 & 2 \\ 1& k &-3 \\ 1 & 4 & 5\\ \end{bmatrix}$ has an inverse matrix, then the value of K is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 660 MCA NIMCET PYQ

If $\vec{e_1}=(1,1,1)$ and $\vec{e_2}=(1,1,-1)$ and $\vec{a}$ and $\vec{b}$ and two vectors such that $\vec{e_2}=\vec{a}+2\vec{b}$ , then angle between $\vec{a}$ and $\vec{b}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 661 MCA NIMCET PYQ

Let E and F be two events such that P(E) > 0 and P(F) > 0. Which one of the following is NOT equivalent to the condition that $P(E) =P(E|F)$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We are given the condition: $P(E) = P(E|F)$.

Since $P(E|F) = \dfrac{P(E \cap F)}{P(F)}$, the condition becomes: $P(E)P(F) = P(E \cap F)$, which is exactly the definition of independence of $E$ and $F$.

Now check each option:

1) "E and F are independent" → This is exactly equivalent to $P(E)=P(E|F)$ (TRUE).

3) $P(F) = P(F|E)$ → Also true under independence (TRUE).

4) $E^c$ and $F$ are independent → Independence is preserved under complements (TRUE).

2) $2P(E^c)P(F^c) \ne P(E \cap F^c)$ → This statement has no relation to $P(E)=P(E|F)$ and does NOT follow from independence (NOT equivalent).

Therefore, the option that is NOT equivalent is: Option 2.

Qus : 662 MCA NIMCET PYQ

A line passing through (4, 2) meets the x and y-axis at P and Q respectively. If O is the origin, then the locus of the centre of the circumcircle of ΔOPQ is -

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 663 MCA NIMCET PYQ

Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If $\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$, then f(2) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$

Given

$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$

$\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3$

$\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3$

For limit to have finite value, value of 'd' and 'e' must be 0

⇒d=0 & e=0

Substituting x=0 in limit

⇒ c+1=3

⇒ c=2

$f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d$

$x=1$ and $x=2$ are extreme values,

⇒$f^{\prime}(1)=0$ and $f^{\prime}(2)=0

⇒ $4a+3b+4=0$ and $32a+12b+8=0$

By solving these equations

we get, $a=\frac{1}{2}$ and $b=-2$

So,

$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$

⇒$f(2)=2^{2}(2-4+2)$

⇒$f(2)=0$

Multiply by $t$:

$ t^2 + 1 = 2t $.

Rearrange:

$ t^2 - 2t + 1 = 0 $.

$ (t - 1)^2 = 0 $.

So: $ t = 1 $.

Thus: $ \tan\theta = 1 $.

General solution: $ \theta = \dfrac{\pi}{4} + n\pi,\; n \in \mathbb{Z}. $

Qus : 668 MCA NIMCET PYQ

The maximum value of 4 sin²x + 3 cos²x + sin(x/2) + cos(x/2) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$4\sin^2x + 3\cos^2x = 3(\sin^2x + \cos^2x) + \sin^2x = 3 + \sin^2x \le 4$

$\sin\frac{x}{2} + \cos\frac{x}{2} \le \sqrt{2}$ with equality when $\dfrac{x}{2}=\dfrac{\pi}{4}+2n\pi$

When $\sin^2x=1$ and $\sin\frac{x}{2}+\cos\frac{x}{2}=\sqrt{2}$, both maxima occur simultaneously.

Therefore, maximum value $= 4 + \sqrt{2}$

Answer: $\boxed{4+\sqrt{2}}$ ✅

Equivalently: $\dfrac{e^{y}}{y+1}=K(1+e^x)$.

Answer : $e^y=k(y+1)(1+e^x)$

Qus : 674 MCA NIMCET PYQ

The value of $\lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 675 MCA NIMCET PYQ

The value of tan 1° tan 2° tan 3° ... tan 89° is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Let $t=1-x \Rightarrow x=1-t,\ dx=-dt$.

As $x:0\to1$,

$t:1\to0$.

\[ \int_{0}^{1} x(1-x)^n\,dx \]

\[ = \int_{1}^{0} (1-t)\,t^n\,(-dt) \]

\[= \int_{0}^{1} \big(t^n - t^{n+1}\big)\,dt \]

\[ = \left[\frac{t^{n+1}}{n+1}-\frac{t^{n+2}}{n+2}\right]_{0}^{1} = \frac{1}{n+1}-\frac{1}{n+2}. \]

Simplify: \[ \frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}. \]

Answer: $\boxed{\dfrac{1}{(n+1)(n+2)}}$ (valid for $n>-1$).

Qus : 679 MCA NIMCET PYQ

, then the values of n and r are:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 680 MCA NIMCET PYQ

The value of $\int_{-\pi/3}^{\pi/3} \frac{x sinx}{cos^{2}x}dx$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 681 MCA NIMCET PYQ

If $\alpha$ and $\beta$ are the roots of the equation $2x^{2}+ 2px + p^{2} = 0$, where $p$ is a non-zero real number, and $\alpha^{4}$ and $\beta^{4}$ are the roots of $x^{2} - rx + s = 0$, then the roots of $2x^{2} - 4p^{2}x + 4p^{4} - 2r = 0$ are:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 682 MCA NIMCET PYQ

$\int {3}^{{3}^{{3}^x}}.{3}^{{3}^x}.{3}^xdx$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 683 MCA NIMCET PYQ

The critical point and nature for the function f(x, y) = x² –2x + 2y² + 4y – 2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Compute partial derivatives:

$f_x=2x-2,\; f_y=4y+4$.

Set $f_x=f_y=0 \Rightarrow 2x-2=0,\; 4y+4=0 \Rightarrow (x,y)=(1,-1)$.

Hessian: $H=\begin{pmatrix}2 & 0\\ 0 & 4\end{pmatrix}$ (positive definite since $2>0$ and $\det=8>0$).

Therefore, $(1,-1)$ is a strict (global) minimum. Also, by completing squares: $f(x,y)=(x-1)^2+2(y+1)^2-5 \Rightarrow f_{\min}=-5$ at $(1,-1)$.

Answer: Critical point $\boxed{(1,-1)}$; nature: $\boxed{\text{minimum}}$; minimum value $\boxed{-5}$.

Qus : 684 MCA NIMCET PYQ

In a class of 50 students, it was found that 30 students read "Hitava", 35 students read "Hindustan" and 10 read neither. How many students read both: "Hitavad" and "Hindustan" newspapers?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Given: Total students = 50, read “Hitava/Hitavad” = 30, read “Hindustan” = 35, read neither = 10.

At least one: $50 - 10 = 40$.

By inclusion–exclusion:

$|H| + |N| - |H \cap N| = |H \cup N| = 40$
$30 + 35 - |H \cap N| = 40 \Rightarrow 65 - |H \cap N| = 40$
$|H \cap N| = 25$

Answer: 25 students read both newspapers.

Qus : 685 MCA NIMCET PYQ

The foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{{81}}=\frac{1}{25}$ coincide, then the value of $b^{2}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 686 MCA NIMCET PYQ

The number of ways to arrange the letters of the English alphabet, so that there are exactly 5 letters between a and b, is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 687 MCA NIMCET PYQ

There are 50 questions in a paper. Find the number of ways in which a student can attempt one or more questions :

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 688 MCA NIMCET PYQ

A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point h meters above the first, the depression of the foot of the tower is 60°. What is the horizontal distance of the tower from the point?

Qus : 692 MCA NIMCET PYQ

Suppose, the system of linear equations

-2x + y + z = l

x - 2y + z = m

x + y - 2z = n

is such that l + m + n = 0, then the system has:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 693 MCA NIMCET PYQ

Consider the following frequency distribution table.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 686 and the median is 42.6, then the value of $f_1$;and $f_2$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 694 MCA NIMCET PYQ

There are 40 female and 20 male students in a class. If the average heights of female and male students are 5.15 feet and 5.66 feet, respectively, then the average height (in feet) of all the students in the class equals

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given: Females = 40 (avg = 5.15 ft), Males = 20 (avg = 5.66 ft). Total students = 60.

Weighted average height = $$\frac{40\times 5.15 + 20\times 5.66}{40+20}$$

Compute: $$\frac{206 + 113.2}{60}=\frac{319.2}{60}=5.32$$

Answer: 5.32 feet

Qus : 695 MCA NIMCET PYQ

The derivative of (x³ + e^x + 3^x + cotx) with respect to x is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$\dfrac{d}{dx}(x^3) = 3x^2$

$\dfrac{d}{dx}(e^x) = e^x$

$\dfrac{d}{dx}(3x) = 3$

$\dfrac{d}{dx}(\cot x) = -cosec^2 x$

Therefore, total derivative $= 3x^2 + e^x + 3 - cosec^2 x$

Answer: $\boxed{3x^2 + e^x + 3 - cosec^2 x}$ ✅

Qus : 696 MCA NIMCET PYQ

If A = { x, y, z }, then the number of subsets in powerset of A is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Given: $A = \{x, y, z\}$

Step 1: Find the number of subsets of $A$:

If a set has $n$ elements, its power set has $2^n$ subsets.

$|A| = 3 \Rightarrow |P(A)| = 2^3 = 8.$

Step 2: Now we need the number of subsets of $P(A)$, i.e. the power set of the power set.

So, $|P(P(A))| = 2^{|P(A)|} = 2^8 = \boxed{256}.$

✅ Final Answer: 256

Qus : 697 MCA NIMCET PYQ

If the mean deviation of the numbers 1, 1 + d, 1 + 2d, ....., 1 + 100d from their mean is 255, then the value of d is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Given AP: $1,\,1+d,\,1+2d,\,\ldots,\,1+100d$ (total $101$ terms). The mean is the middle term $1+50d$.

Mean Deviation (about mean): For $2n+1$ terms $a,a+d,\ldots,a+2nd$, $$\text{MD}=\frac{d\,n(n+1)}{2n+1}.$$

Here $2n+1=101 \Rightarrow n=50$. So $$\text{MD}=\frac{d\cdot 50\cdot 51}{101}=\frac{2550}{101}\,d.$$

Given $\text{MD}=255$. Solve for $d$: $$\frac{2550}{101}\,d=255 \;\Rightarrow\; d=\frac{255\times 101}{2550}=10.1.$$

Answer: $d=10.1$.

Qus : 698 MCA NIMCET PYQ

If $\vec{A}=4\hat{i}+3\hat{j}+\hat{k}$ and $\vec{B}=2\hat{i}-\hat{j}+2\hat{k}$ , then the unit vector $\hat{N}$ perpendicular to the vectors $\vec{A}$ and $\vec{B}$ ,such that $\vec{A}, \vec{B}$ , and $\hat{N}$ form a right handed system, is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 699 MCA NIMCET PYQ

Let $x$ be a positive real number such that $x^{(8\log_5x-24)}=5^{-4}$. Then the product of all possible values of x is =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We have $ x^{8\log_5 x - 24} = 5^{-4} $

Let $ \log_5 x = t $.

Then $ x = 5^t $.

Substitute: $ (5^t)^{8t - 24} = 5^{-4} $

$ 5^{t(8t - 24)} = 5^{-4} $

So: $ t(8t - 24) = -4 $

$ 8t^2 - 24t + 4 = 0 $

Divide by 4:

$ 2t^2 - 6t + 1 = 0 $

Solutions: $ t = \frac{3 \pm \sqrt{7}}{2} $

Thus: $ x_1 = 5^{\frac{3 + \sqrt{7}}{2}},\ x_2 = 5^{\frac{3 - \sqrt{7}}{2}} $

Product: $ x_1 x_2 = 5^{\frac{3 + \sqrt{7}}{2} + \frac{3 - \sqrt{7}}{2}} = 5^{\frac{6}{2}} = 5^3 = 125 $

Qus : 700 MCA NIMCET PYQ

If $X={4^n-3n-1,; n\in N}$ and $Y={9n-9,; n\in N}$, then $X\cup Y$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 701 MCA NIMCET PYQ

The solution of the differential equation $\dfrac{dy}{dx}=e^{x+y}+x^2e^y$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$\dfrac{dy}{dx}=e^y(e^x+x^2)\ \Rightarrow\ e^{-y}dy=(e^x+x^2)dx$

$\int e^{-y}dy=\int (e^x+x^2)dx$ $\Rightarrow\ -e^{-y}=e^x+\dfrac{x^3}{3}+C$

Hence, $e^{-y}+e^x+\dfrac{x^3}{3}=C$ (or $y=-\ln\!\big(C-e^x-\dfrac{x^3}{3}\big)$).

Answer: $\boxed{e^{-y}+e^x+\dfrac{x^3}{3}=C}$ ✅

Qus : 702 MCA NIMCET PYQ

How many words can be formed starting with letter D taking all letters from the word DELHI so that the letters are not repeated:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 703 MCA NIMCET PYQ

If $P=sin^{20} \theta + cos^{48} \theta $ then the inequality that holds for all values of is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 704 MCA NIMCET PYQ

The value of $\int \frac{(x+1)}{x(xe^{x}+1)} dx$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 705 MCA NIMCET PYQ

The obtuse angle between lines 2y = x + 1 and y = 3x + 2 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Given lines:

$2y = x + 1$ and $y = 3x + 2$.

From $2y = x + 1$

we get $y = \dfrac{1}{2}x + \dfrac{1}{2}$,

so slope $m_1 = \dfrac{1}{2}$.

From $y = 3x + 2$,

slope $m_2 = 3$.

Angle between two lines: $ \tan\theta = \left|\dfrac{m_2 - m_1}{1 + m_1 m_2}\right| $

$ \tan\theta = \left|\dfrac{3 - \dfrac{1}{2}}{1 + \dfrac{1}{2}\cdot 3}\right| = \dfrac{\dfrac{5}{2}}{\dfrac{5}{2}} = 1 $

So $\theta = 45^\circ$ or $135^\circ$.

Required obtuse angle $= 135^\circ$.

Qus : 706 MCA NIMCET PYQ

$\int \left(\dfrac{\log x - 1}{1 + (\log x)^2}\right)^2 dx$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 707 MCA NIMCET PYQ

Differentiate [- log(log x), x > 1] with respect to x

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let $y=-\log(\log x)$

$\dfrac{dy}{dx}=-\dfrac{d}{dx}[\log(\log x)]$

$\dfrac{d}{dx}[\log(\log x)]=\dfrac{1}{\log x}\cdot\dfrac{1}{x}$

Therefore, $\dfrac{dy}{dx}=-\dfrac{1}{x\log x}$

Answer: $\boxed{-\dfrac{1}{x\log x}},\; x>1$ ✅

Qus : 708 MCA NIMCET PYQ

Naresh has 10 friends, and he wants to invite 6 of them to a party. How many times will 3 particular friends never attend the party?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

^(10-3)C₆= ⁷C₆ = 7

Qus : 709 MCA NIMCET PYQ

If a, b, c are in geometric progression, then $log_{ax}^{a}, log_{bx}^{a}$ and $log_{cx}^{a}$ are in

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

$ a, b, c $ are in G.P.

$ \Rightarrow b^2 = ac $

Take logs base $a$:

$ \log_a a = 1,\quad \log_a b,\quad \log_a c $

Since $ a, b, c $ are in G.P.

$ \Rightarrow \log_a a,\ \log_a b,\ \log_a c $ are in A.P.

Now given:

$ \log_a(ax) = \log_a a + \log_a x = 1 + \log_a x $

$ \log_a(bx) = \log_a b + \log_a x $

$ \log_a(cx) = \log_a c + \log_a x $

These are: $ (1 + k),\ (\log_a b + k),\ (\log_a c + k) $ where $k = \log_a x$

Adding same constant does not change A.P. nature $\boxed{\text{They are in A.P.}}$

Qus : 710 MCA NIMCET PYQ

The sum of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ such that $|\vec{a}|=|\vec{b}|=|\vec{c}|=2$. Then, the magnitude of $\vec{a}-\vec{b}$ is equal to:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 711 MCA NIMCET PYQ

What is the value of $\lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)}$?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We want to evaluate $ \displaystyle \lim_{x \to \infty} -(x+1)\left(e^{\frac{1}{x+1}} - 1\right). $

Rewrite it as a product: $ -(x+1)\left(e^{\frac{1}{x+1}} - 1\right) = -\dfrac{e^{\frac{1}{x+1}} - 1}{\frac{1}{x+1}}. $

Now let $ t = \frac{1}{x+1} \Rightarrow t \to 0^+ $ as $ x \to \infty $.

The expression becomes: $ -\dfrac{e^{t} - 1}{t}. $

Now apply L'Hospital’s Rule to the limit: $ \displaystyle \lim_{t \to 0} \dfrac{e^{t} - 1}{t} = \lim_{t \to 0} \dfrac{e^{t}}{1} = 1. $

So the original limit is: $ -1. $

Final Answer: $ -1 $.

Step 1: General term
$T_n = \dfrac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$

Step 2: Rationalize by multiplying numerator and denominator by $(n+1)\sqrt{n}-n\sqrt{n+1}$
Denominator becomes:
$[(n+1)\sqrt{n}]^2 - [n\sqrt{n+1}]^2$

$ = n(n+1)^2 - n^2(n+1) $

$= n(n+1)(n+1-n)$

$ = n(n+1)$

So:
$T_n = \dfrac{(n+1)\sqrt{n} - n\sqrt{n+1}}{n(n+1)}$

$= \dfrac{\sqrt{n}}{n} - \dfrac{\sqrt{n+1}}{n+1}$

$= \dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}$

Step 3: Telescoping sum from $n=1$ to $n=24$
$\displaystyle\sum_{n=1}^{24} T_n = \left(\dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{2}}\right) + \left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{3}}\right) + \cdots + \left(\dfrac{1}{\sqrt{24}} - \dfrac{1}{\sqrt{25}}\right)$

$= \dfrac{1}{\sqrt{1}} - \dfrac{1}{\sqrt{25}}$

$= 1 - \dfrac{1}{5}$

$= \dfrac{4}{5}$

Answer: $\boxed{\dfrac{4}{5}}$

Qus : 717 MCA NIMCET PYQ

If x and y are positive real numbers satisfying the system of equations $x^{2}+y\sqrt{xy}=336$ and $y^{2}+x\sqrt{xy}=112$, then x + y is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 718 MCA NIMCET PYQ

The value of $\int ^{\frac{\pi}{2}}_0\frac{(1+2\cos x)}{({2+\cos x)}^2}dx$ lies in the interval

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

We need to evaluate $ \displaystyle \int_{0}^{\pi/2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\, dx $.

Let $ t = 2 + \cos x $. Then $ dt = -\sin x\, dx $.

But the integral has no $\sin x$, so rewrite numerator:

$ 1 + 2\cos x = (2 + \cos x) - 1 = t - 1 $.

Now express $ dx $ using $ \sin^2 x = 1 - \cos^2 x $, but the standard trick is to differentiate:

$ \dfrac{d}{dx}\left(\dfrac{1}{2+\cos x}\right) = -\dfrac{-\sin x}{(2+\cos x)^2} = \dfrac{\sin x}{(2+\cos x)^2}. $ We

use complementary substitution:

Let $ x = \frac{\pi}{2} - y $.

Then $\cos x = \sin y$ and $\sin x = \cos y$.

Integral becomes:

$ I = \int_{0}^{\pi/2} \frac{1 + 2\sin y}{(2 + \sin y)^2}\, dy. $

Average the two forms:

$ I = \frac{1}{2}\int_{0}^{\pi/2} \left[ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} \right] dx. $

Now observe identity:

$ \frac{1 + 2\cos x}{(2 + \cos x)^2} + \frac{1 + 2\sin x}{(2 + \sin x)^2} = \frac{d}{dx}\left(\frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)}\right). $

Thus integral becomes a telescoping form and evaluates to:

$ I = \left[ \frac{\sin x - \cos x}{(2+\cos x)(2+\sin x)} \right]_{0}^{\pi/2}. $

Now compute:

At $ x = \frac{\pi}{2}$: $ \sin x = 1,\;\cos x = 0 $

Expression = $ \dfrac{1 - 0}{(2+0)(2+1)} = \dfrac{1}{6}. $

At $ x = 0$: $ \sin 0 = 0,\;\cos 0 = 1 $

Expression = $ \dfrac{0 - 1}{(2+1)(2+0)} = -\dfrac{1}{6}. $

Therefore: $ I = \dfrac{1}{6} - (-\dfrac{1}{6}) = \dfrac{2}{6} = \dfrac{1}{3}. $

Final Answer: $\dfrac{1}{3} $

Qus : 719 MCA NIMCET PYQ

The volume of the parallelepiped determined by $u = i + 2j - k,; v = -2i + 3k$ and $w = 7j - 4k$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 720 MCA NIMCET PYQ

If $\vec{a}=\hat{i}-\hat{k},\, \vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}$ and $\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}$ , then $[\vec{a} , \vec{b}, \vec{c}]$ depends on

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 721 MCA NIMCET PYQ

From three collinear points A, B and C on a level ground, which are on the same side of a tower, the angles of elevation of the top of the tower are 30°, 45° and 60° respectively. If BC = 60 m, then AB is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 722 MCA NIMCET PYQ

The vector perpendicular to the plane passing through $(1,-1,0),; (2,1,-1)$ and $(-1,1,2)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 723 MCA NIMCET PYQ

A problem in Mathematics is given to 3 students A, B, and C. If the probability of A solving the problem is 1/2 and B not solving it is 1/4 . The whole probability of the problem being solved is 63/64 , then what is the probability of solving it by C?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 724 MCA NIMCET PYQ

If $42 (^nP_2)=(^nP_4)$ then the value of n is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 725 MCA NIMCET PYQ

If $x = 1$ is the directrix of the parabola $y^{2} = kx - 8$, then k is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 726 MCA NIMCET PYQ

If $2x^2 + 7xy + 3y^2 + 8x + 14y + \lambda = 0$ represents a pair of straight lines, the value of $\lambda$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 727 MCA NIMCET PYQ

A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both win a prize. The probability that they will not win a prize in a single trial is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Probability of winning a prize = $\frac{1}{25}$
Thus, probability of not winning = $1-\frac{1}{25}=\frac{24}{25}$

Qus : 728 MCA NIMCET PYQ

The foot of the perpendicular from the point (2, 4) upon $x + y = 1$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 729 MCA NIMCET PYQ

If $sin x + a cos x = b$, then $|a sin x - cos x|$ is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 730 MCA NIMCET PYQ

The area of region bounded by the lines $y = |x - 1|$ and $y = 3 - |x|$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 731 MCA NIMCET PYQ

A, B, C are three sets of values of x:

A: 2,3,7,1,3,2,3

B: 7,5,9,12,5,3,8

C: 4,4,11,7,2,3,4

Select the correct statement among the following:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Given sets

A: 2, 3, 7, 1, 3, 2, 3
B: 7, 5, 9, 12, 5, 3, 8
C: 4, 4, 11, 7, 2, 3, 4

Set A: sum = 21, n = 7 ⇒ Mean = 21/7 = 3.
Sorted A = (1, 2, 2, 3, 3, 3, 7) ⇒ Median = 3.
Mode(A) = most frequent = 3.

Set B: sorted = (3, 5, 5, 7, 8, 9, 12) ⇒ Median(B) = 7.

Set C: sum = 35, n = 7 ⇒ Mean(C) = 5; Mode(C) = 4.

Check options

Mean(A) = 3 vs Mode(C) = 4 → False
Mean(C) = 5 vs Median(B) = 7 → False
Median(B) = 7 vs Mode(A) = 3 → False
Mean(A) = Median(A) = Mode(A) = 3 → True

Correct statement: Option 4

Qus : 732 MCA NIMCET PYQ

The value of k for which the equation $(k-2)x^{2}+8x+k+4=0$ has both real, distinct and negative roots is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 733 MCA NIMCET PYQ

A condition that $x^{3} + ax^{2} + bx + c$ may have no extremum is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 734 MCA NIMCET PYQ

In a triangle $ABC$, $a = 4$, $b = 3$, $\angle BAC = 60^\circ$, then the equation for which $c$ is the root is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 735 MCA NIMCET PYQ

Standard deviation for the following distribution is

Size of item	6	7	8	9	10	11	12
Frequency	3	6	9	13	8	5	4

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Total number of items in the distribution = Σ f_i = 3 + 6 + 9 + 13 + 8 + 5 + 4 = 48.

The Mean (x̅) of the given set = $\rm \dfrac{\sum f_i x_i}{\sum f_i}$.

⇒ x̅ = $\rm \dfrac{6\times3+7\times6+8\times9+9\times13+10\times8+11\times5+12\times4}{48}=\dfrac{432}{48}$ = 9.

Let's calculate the variance using the formula: $\rm \sigma^2 =\dfrac{\sum x_i^2}{n}-\bar x^2$.

$\rm \dfrac{\sum {x_i}^2}{n}=\dfrac{6^2\times3+7^2\times6+8^2\times9+9^2\times13+10^2\times8+11^2\times5+12^2\times4}{48}=\dfrac{4012}{48}$ = 83.58.

∴ σ² = 83.58 - 9² = 83.58 - 81 = 2.58.

And, Standard Deviation (σ) = $\rm \sqrt{\sigma^2}=\sqrt{Variance}=\sqrt{2.58}$ ≈ 1.607.

posses a non-zero solution is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

A homogeneous system has a non-trivial solution $\iff$ the determinant of its coefficient matrix is $0$.
Coefficient matrix $A=\begin{bmatrix}4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1\end{bmatrix}$. Hence, $$ \det(A)= \begin{vmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{vmatrix} =-(k-4)(k-2). $$ Setting $\det(A)=0 \Rightarrow -(k-4)(k-2)=0 \Rightarrow k=2 \text{ or } k=4.$
Therefore, the number of values of $k$ is $\boxed{2}$.

Qus : 748 MCA NIMCET PYQ

The number of bit strings of length 10 that contain either five consecutive 0’s or five consecutive 1’s is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

We count bit strings of length 10 that contain at least one run of five identical bits.

Case 1: Exactly one block of five consecutive 0’s.
The block $00000$ can start at positions 1 to 6, so 6 choices.
The remaining 5 positions can be filled freely with 0 or 1, except they should not create another block of five 0’s.
Valid fillings = $2^5 - 1 = 31$.
So number of strings with exactly one block of five 0’s is
$6 \times 31 = 186$

Case 2: Exactly one block of five consecutive 1’s.
By symmetry, the count is the same.
$186$

Case 3: One block of five 0’s and one block of five 1’s.
This is possible only when the blocks do not overlap.
The only such strings are
$0000011111$ and $1111100000$

So total such strings = $2$.
Using inclusion–exclusion principle:
$186 + 186 + 2 = 222$
Final Answer: $222$

MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

Solution

Step 1: Set up AP
Let first row (front) have $a$ children, common difference $d = -3$

For $n$ rows, sum $= 630$:
$S_n = \dfrac{n}{2}[2a + (n-1)(-3)] = 630$

$n[2a - 3(n-1)] = 1260$

$2a = \dfrac{1260}{n} + 3(n-1)$

$a = \dfrac{630}{n} + \dfrac{3(n-1)}{2}$

Step 2: Conditions for valid solution
$a$ must be a positive integer, and last row $= a + (n-1)(-3) > 0$

$\Rightarrow a > 3(n-1)$

$\Rightarrow \dfrac{630}{n} + \dfrac{3(n-1)}{2} > 3(n-1)$

$\Rightarrow \dfrac{630}{n} > \dfrac{3(n-1)}{2}$

$\Rightarrow 1260 > 3n(n-1)$

$\Rightarrow n(n-1) < 420$

$\Rightarrow n \leq 21$ (since $21 \times 20 = 420$, not $< 420$, so $n \leq 20$)

Step 3: Also $a$ must be a positive integer
$a = \dfrac{630}{n} + \dfrac{3(n-1)}{2}$ must be a positive integer.

For $a$ to be integer: $\dfrac{630}{n}$ and $\dfrac{3(n-1)}{2}$ must together give integer.

Check $n = 6$: $a = \dfrac{630}{6} + \dfrac{3(5)}{2} = 105 + 7.5 = 112.5$ — not integer! ✗

Check $n = 7$: $a = \dfrac{630}{7} + \dfrac{3(6)}{2} = 90 + 9 = 99$ ✓
Check $n = 9$: $a = \dfrac{630}{9} + \dfrac{3(8)}{2} = 70 + 12 = 82$ ✓
Check $n = 14$: $a = \dfrac{630}{14} + \dfrac{3(13)}{2} = 45 + 19.5 = 64.5$ — not integer! ✗

Qus : 767 MCA NIMCET PYQ

The sum of the expression $\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\dfrac{1}{\sqrt{80}+\sqrt{81}}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 768 MCA NIMCET PYQ

If a + b + c = 0, then the value of $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Just Let a=1, b=1 and c=-2 and put in given question
$\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}$

$=-1+4=3$

Qus : 785 MCA NIMCET PYQ

Let $\bar{P}$ and $\bar{Q}$ denote the complements of two sets P and Q. Then the set $(P-Q)\cup (Q-P) \cup (P \cap Q)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Expression: $(P - Q) \cup (Q - P) \cup (P \cap Q)$

Recall: $P - Q = P \cap \bar{Q}$ and $Q - P = Q \cap \bar{P}$.

Thus the union consists of three mutually exclusive parts:

Elements only in $P$: $P \cap \bar{Q}$
Elements only in $Q$: $Q \cap \bar{P}$
Elements in both: $P \cap Q$

The union of these three pieces is precisely all elements that are in $P$ or $Q$:

$(P - Q) \cup (Q - P) \cup (P \cap Q) = P \cup Q$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 803 MCA NIMCET PYQ

A matrix $M_r$ is defined as $M_r=\begin{bmatrix} r &r-1 \\ r-1&r \end{bmatrix} , r \in N$ then the value of $det(M_1) + det(M_2) +...+ det(M_{2015})$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 804 MCA NIMCET PYQ

The value of sin 20° sin 40° sin 80° is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 805 MCA NIMCET PYQ

The perimeter of a $\Delta ABC$ is 6 times the arithmetic mean of the sines of its angles. If the side a is 1, then the angle A is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Quick Solution

Given: Perimeter = 6 × Arithmetic Mean of sin A, sin B, sin C

Using Law of Sines: $ \frac{a}{\sin A} = 2R $, and a = 1 ⇒ $ \sin A = \frac{1}{2R} $

Assume $ A = 30^\circ \Rightarrow \sin A = \frac{1}{2} \Rightarrow 2R = 2 $

⇒ b = 2 sin B, c = 2 sin C

Perimeter = $ 1 + b + c = 1 + 2 \sin B + 2 \sin C $

Mean = $ \frac{\sin A + \sin B + \sin C}{3} $

Check: $ 1 + 2\sin B + 2\sin C = 6 \cdot \frac{1/2 + \sin B + \sin C}{3} $ ✅

✅ Final Answer: 30°

Qus : 806 MCA NIMCET PYQ

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Red = 8
Blue = 7
Green = 6
Total balls = 8 + 7 + 6 = 21

Probability (neither red nor green) = Probability (blue)
= Number of blue balls / Total balls
= 7 / 21
= 1 / 3

✅ Final Answer: 1/3

Qus : 807 MCA NIMCET PYQ

Sum of the roots of the equation $4^x - 3(2^{x+3}) + 128 = 0$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $4^x - 3(2^{x+3}) + 128 = 0$
Writing in terms of $2^x$:
$(2^x)^2 - 3 \cdot 2^x \cdot 2^3 + 128 = 0$
$(2^x)^2 - 24(2^x) + 128 = 0$
Let $t = 2^x$
$\Rightarrow t^2 - 24t + 128 = 0$
$\Rightarrow (t - 8)(t - 16) = 0$
$\Rightarrow t = 8$ or $t = 16$
When $t = 8$:
$2^x = 8 = 2^3 \Rightarrow x = 3$
When $t = 16$:
$2^x = 16 = 2^4 \Rightarrow x = 4$
Sum of roots $= 3 + 4$
$\therefore \boxed{\text{Sum of roots} = 7}$

Qus : 808 MCA NIMCET PYQ

Find the area bounded by the line y = 3 - x, the parabola y = x² - 9 and

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 809 MCA NIMCET PYQ

If $\vec{AC}=2\hat{i}+\hat{j}+\hat{k}$ and $\vec{BD}=-\hat{i}+3\hat{j}+2\hat{k}$ then the area of the quadrilateral ABCD is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

The median AD of ΔABC is bisected at E and BE is produced to meet the side AC at F. Then, AF ∶ FC is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 816 MCA NIMCET PYQ

For a group of 100 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on, it was found that the scores 25 and 35 were misread as 52 and 53 respectively. Then the corrected mean and standard deviation corresponding to the corrected figures are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

$ \mu' = \frac{3955}{100} = \boxed{39.55} $

Corrected Standard Deviation:

$ \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} $

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96

Qus : 817 MCA NIMCET PYQ

The system of equations

x + y + 2z = a

x + z = b

2x + y + 3z = c

has a solution if

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given system of equations:
$x + y + 2z = a$
$x + z = b$
$2x + y + 3z = c$
Writing in matrix form $[A|B]$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{array}\right]$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & -1 & -1 & c-2a \end{array}\right]$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-2a-(b-a) \end{array}\right]$
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-a-b \end{array}\right]$
For system to have a solution:
$\rho(A) = \rho(A|B)$
$\Rightarrow c - a - b = 0$
$\therefore \boxed{a + b = c}$

Qus : 818 MCA NIMCET PYQ

Two forces F₁ and F₂ are used to pull a car, which met an accident. The angle between the two forces is θ . Find the values of θ for which the resultant force is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 819 MCA NIMCET PYQ

If two circles $x^{2}+y^{2}+2gx+2fy=0$ and $x^{2}+y^{2}+2g'x+2f'y=0$ touch each other then whichof the following is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 820 MCA NIMCET PYQ

If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that OPQ is an equilateral triangle, where O is the centre of the hyperbola, then which of the following is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 821 MCA NIMCET PYQ

Consider the following frequency distribution table.

Class interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	180	$f_1$	34	180	136	$f_2$	50

If the total frequency is 685 & median is 42.6 then the values of $f_1$ and $f_2$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Median & Frequency Table

Given: Median = 42.6, Total Frequency = 685

Using Median Formula:

$ \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h $

Median Class: 40–50
Lower boundary $ L = 40 $
Frequency $ f = 180 $
Class width $ h = 10 $
Cumulative freq before median class $ F = 214 + f_1 $

Substituting values:

$ 42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82} $

Using total frequency:

$ 662 + f_2 = 685 \Rightarrow f_2 = \boxed{23} $

✅ Final Answer: $ f_1 = 82,\quad f_2 = 23 $

Qus : 822 MCA NIMCET PYQ

Let $f(x) = x^2 - bx + c$, b is an odd positive integer. If f(x)=0 has two prime numbers as roots and b + c = 35, then the global minimum value of f(x) is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: $f(x) = x^2 - bx + c$
where $b$ is an odd positive integer and $b + c = 35$
Let $\alpha, \beta$ be two prime roots of $f(x) = 0$
By Vieta's formulae:
$\alpha + \beta = b$
$\alpha \beta = c$
Now, $b + c = 35$
$\Rightarrow (\alpha + \beta) + \alpha\beta = 35$
Since $b$ is odd and $\alpha + \beta = b$ is odd
$\Rightarrow$ one of $\alpha, \beta$ must be even prime
$\therefore \alpha = 2$ (only even prime)
$\Rightarrow 2 + \beta + 2\beta = 35$
$\Rightarrow 3\beta = 33$
$\Rightarrow \beta = 11$
$\therefore b = \alpha + \beta = 2 + 11 = 13$
$\therefore c = \alpha\beta = 2 \times 11 = 22$
Verification: $b + c = 13 + 22 = 35$ ✓
So $f(x) = x^2 - 13x + 22$
Global minimum occurs at $x = \dfrac{b}{2} = \dfrac{13}{2}$
Minimum value $= f\left(\dfrac{13}{2}\right) = \left(\dfrac{13}{2}\right)^2 - 13\left(\dfrac{13}{2}\right) + 22$
$= \dfrac{169}{4} - \dfrac{169}{2} + 22$
$= \dfrac{169 - 338 + 88}{4}$
$= \dfrac{-81}{4}$
$\therefore \boxed{\text{Global Minimum} = -\dfrac{81}{4}}$

Qus : 823 MCA NIMCET PYQ

are four vectors such that

is collinear with

and

is collinear with

then

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 824 MCA NIMCET PYQ

$\int_0^\pi [cotx]dx$ where [.] denotes the greatest integer function, is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 825 MCA NIMCET PYQ

In ΔABC, if a = 2, b = 4 and ∠C = 60°, then A and B are respectively equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 826 MCA NIMCET PYQ

The sum of infinite terms of decreasing GP is equal to the greatest value of the function $f(x) = x^3 + 3x – 9$ in the interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

? GP and Function Relation

Given: $ f(x) = x^3 + 3x - 9 $

The sum of infinite GP = max value of $ f(x) $ on [−2, 3]

The difference between first two terms = $ f'(0) $

Step 1: $ f(x) $ is increasing ⇒ Max at $ x = 3 $

$ f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27 $

Step 2: $ f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3 $

⇒ $ a(1 - r) = 3 $

Step 3: Solve:
$ a = 27(1 - r) $
$ \Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3} $

✅ Final Answer: $ r = \frac{2}{3} $

Qus : 827 MCA NIMCET PYQ

The vertex of the parabola whose focus is (-1,1) and directrix is 4x + 3y - 24 = 0 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given: Focus $S(-1, 1)$ and directrix $4x + 3y - 24 = 0$
The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 834 MCA NIMCET PYQ

A is targeting B, B and C are targeting A. Probability of hitting the target by A, B and C are $\frac{2}{3}, \frac{1}{2}$ and $\frac{1}{3}$ respectively. If A is hit then the probability that B hits the target and C does not, is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 835 MCA NIMCET PYQ

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Given averages: Boys = 52, Girls = 42, Combined = 50.

Let the proportion of boys be $p$ (so girls = $1-p$). Weighted mean gives: $$52p + 42(1-p) = 50$$ $$52p + 42 - 42p = 50 \Rightarrow 10p = 8 \Rightarrow p = 0.8.$$

Percentage of boys $= 0.8 \times 100\% = \boxed{80\%}$.

Qus : 836 MCA NIMCET PYQ

The value of $\int ^{\pi/3}_{-\pi/3}\frac{x\sin x}{{\cos }^2x}dx$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 837 MCA NIMCET PYQ

There are $4$ books on fairy tales, $5$ novels and $3$ plays. In how many ways can they be arranged in the order books on fairy tales, novels and then plays so that books of same category are put together

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given:
$4$ books on fairy tales, $5$ novels, $3$ plays
The order of arrangement is fixed as:
Fairy Tales $\rightarrow$ Novels $\rightarrow$ Plays
Step 1: Arrange books within each category
Fairy tales books can be arranged among themselves $= 4!= 24$ ways
Novels can be arranged among themselves $= 5! = 120$ ways
Plays can be arranged among themselves $= 3! = 6$ ways
Step 2: Total number of arrangements
$= 4! \times 5! \times 3!$
$= 24 \times 120 \times 6$
$\therefore \boxed{= 17280 \text{ ways}}$

Qus : 838 MCA NIMCET PYQ

If a, b, c are three non-zero vectors with no two of which are collinear, a + 2b is collinear with c and b + 3c is collinear with a , then | a + 2b + 6c | will be equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 839 MCA NIMCET PYQ

If the angles of a triangle are in the ratio 2 : 3 : 7, then the ratio of the sides opposite to these angles is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 840 MCA NIMCET PYQ

How many even integers between 4000 and 7000 have four different digits?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

1: Thousands digit

The number is 4-digit and must be between 4000 and 7000.
So the thousands digit can be: 4, 5, or 6.

Step 2: Units digit

The number must be even, so the units digit must be one of {0, 2, 4, 6, 8}.

3: Count cases

Case 1: Thousands digit = 4
Units digit ≠ 4 → possible units = {0, 2, 6, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224.

Case 2: Thousands digit = 5
Units digit can be {0, 2, 4, 6, 8} → 5 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 5 × 8 × 7 = 280.

Case 3: Thousands digit = 6
Units digit ≠ 6 → possible units = {0, 2, 4, 8} → 4 choices.
Hundreds digit = 8 choices, Tens digit = 7 choices.
Total = 4 × 8 × 7 = 224.

4: Add results

Total = 224 + 280 + 224 = 728.

Final Answer:

There are 728 such even integers.

Qus : 841 MCA NIMCET PYQ

The equation of the tangent at any point of curve $x=a cos2t, y=2\sqrt{2} a sint$ with $m$ as its slope is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 842 MCA NIMCET PYQ

Suppose a population $A$ has $100$ observations $101,102,\ldots,200$ and another population $B$ has $100$ observations $151,152,\ldots,250$. If $V_A$ and $V_B$ represent variance of the two populations respectively, then $\dfrac{V_A}{V_B}$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Population $A$: $101, 102, \ldots, 200$
Population $B$: $151, 152, \ldots, 250$
Using Property:
Variance is independent of change of origin
i.e., if each observation is increased or decreased by a constant, variance remains unchanged
Here, each observation of $B$ = each observation of $A$ $+ 50$
$\Rightarrow$ Population $B$ is obtained by adding $50$ to each observation of population $A$
$\Rightarrow V_B = V_A$
$\therefore \dfrac{V_A}{V_B} = \boxed{1}$

Qus : 843 MCA NIMCET PYQ

Vertices of the vectors i - 2j + 2k , 2i + j - k and 3i - j + 2k form a triangle. This triangle is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 844 MCA NIMCET PYQ

Suppose that A and B are two events with probabilities $P(A) =\frac{1}{2} \, P(B)=\frac{1}{3}$ Then which of the following is true?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 845 MCA NIMCET PYQ

If $\prod ^n_{i=1}\tan ({{\alpha}}_i)=1\, \forall{{\alpha}}_i\, \in\Bigg{[}0,\, \frac{\pi}{2}\Bigg{]}$ where i=1,2,3,...,n. Then maximum value of $\prod ^n_{i=1}\sin ({{\alpha}}_i)$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 846 MCA NIMCET PYQ

If $\vec{a}, \vec{b}$ are vector such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a}\times (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k})\times \vec{b}$, then a possible value of $(\vec{a}+\vec{b})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 847 MCA NIMCET PYQ

If the volume of a parallelepiped whose adjacent edges are

a = 2i + 3j + 4k,

b = i + αj + 2k

Solution

Qus : 852 MCA NIMCET PYQ

A harbour lies in a direction 60° South of West from a fort and at a distance 30 km from it, a ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 853 MCA NIMCET PYQ

If $\overrightarrow{{a}}$ and $\overrightarrow{{b}}$ are vectors in space, given by $\overrightarrow{{a}}=\frac{\hat{i}-2\hat{j}}{\sqrt[]{5}}$ and $\overrightarrow{{b}}=\frac{2\hat{i}+\hat{j}+3\hat{k}}{\sqrt[]{14}}$, then the value of$(2\vec{a} + \vec{b}).[(\vec{a} × \vec{b}) × (\vec{a} – 2\vec{b})]$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 854 MCA NIMCET PYQ

Area of the greatest rectangle that can be inscribed in the ellipse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$

Qus : 855 MCA NIMCET PYQ

If $\frac{tanx}{2}=\frac{tanx}{3}=\frac{tanx}{5}$ and x + y + z = π, then the value of tan²x + tan²y + tan²z is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 856 MCA NIMCET PYQ

If $x, y, z$ are three consecutive positive integers, then $log (1 + xz)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 857 MCA NIMCET PYQ

Let $\vec{A} = 2\hat{i} + \hat{j} – 2\hat{k}$ and $\vec{B} = \hat{i} + \hat{j}$, If $\vec{C}$ is a vector such that $|\vec{C} – \vec{A}| = 3$ and the angle between A × B and C is ${30^{\circ}}$, then $|(\vec{A} × \vec{B}) × \vec{C}|$ = 3 then the value of $\vec{A}.\vec{C}$ is equal to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Qus : 858 MCA NIMCET PYQ

. Two common tangents to the circles $x^2 + y^2 = 2a^2$ and parabola $y^2 = 8ax$ are

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given circle: $x^2 + y^2 = 2a^2$
Given parabola: $y^2 = 8ax$
**Equation of tangent to parabola:**
For parabola $y^2 = 8ax$, tangent is:
$y = mx + \dfrac{2a}{m}$ ...(i)
**Condition for tangent to circle:**
For line $y = mx + \dfrac{2a}{m}$ to be tangent to circle $x^2 + y^2 = 2a^2$,
distance from centre $(0,0)$ = radius $= \sqrt{2}\ a$
$\Rightarrow \dfrac{\left|\dfrac{2a}{m}\right|}{\sqrt{1 + m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2a}{|m|\sqrt{1+m^2}} = \sqrt{2}\ a$
$\Rightarrow \dfrac{2}{|m|\sqrt{1+m^2}} = \sqrt{2}$
$\Rightarrow |m|\sqrt{1+m^2} = \sqrt{2}$
Squaring both sides:
$\Rightarrow m^2(1+m^2) = 2$
$\Rightarrow m^4 + m^2 - 2 = 0$
$\Rightarrow (m^2 + 2)(m^2 - 1) = 0$
$\Rightarrow m^2 = 1 \quad (\because m^2 = -2 \text{ is not possible})$
$\Rightarrow m = \pm 1$
**Substituting in (i):**
When $m = 1$: $y = x + 2a$
When $m = -1$: $y = -x - 2a$
$\therefore$ The two common tangents are:
$\therefore \boxed{y = x + 2a \quad \text{and} \quad y = -(x + 2a)}$

Qus : 859 MCA NIMCET PYQ

Qus : 863 MCA NIMCET PYQ

If cos x = tan y , cot y = tan z and cot z = tan x, then sinx =

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 864 MCA NIMCET PYQ

The value of $tan(\frac{7\pi}{8})$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 865 MCA NIMCET PYQ

If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

$a, b, c, d$ are in H.P.

$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in A.P.

$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$

Given A.M. of $ab, bc, cd$ is $9$:

$\frac{ab + bc + cd}{3} = 9$

$\Rightarrow ab + bc + cd = 27$

Now multiply: $\frac{2}{b} = \frac{1}{a} + \frac{1}{c} $

$\Rightarrow 2ac = b(a + c)$

$\frac{2}{c} = \frac{1}{b} + \frac{1}{d} \Rightarrow 2bd = c(b + d)$

Multiply both: $4abcd = bc(a + c)(b + d)$

Cancel $bc$:

$4ad = (a + c)(b + d)$

Expand RHS:

$4ad = ab + ad + bc + cd$

$\Rightarrow 3ad = ab + bc + cd$

But $ab + bc + cd = 27$:

$\Rightarrow 3ad = 27$

$\Rightarrow ad = 9$

$\boxed{9}$

Qus : 866 MCA NIMCET PYQ

The value of $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

We need to find: $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$

Qus : 867 MCA NIMCET PYQ

The value of

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 868 MCA NIMCET PYQ

If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}|=13$, $|\vec{b}|=5$ and $\vec{a} . \vec{b} =60$then the value of $|\vec{a} \times \vec{b}|$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 869 MCA NIMCET PYQ

Find foci of the equation $x^2 + 2x – 4y^2 + 8y – 7 = 0$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Finding Foci of a Conic

Given Equation: $ x^2 + 2x - 4y^2 + 8y - 7 = 0 $

Step 1: Complete the square

⇒ $ (x + 1)^2 - 4(y - 1)^2 = 4 $

Rewriting: $ \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 $

This is a horizontal hyperbola with:

Center: $ (-1, 1) $
$ a^2 = 4 $, $ b^2 = 1 $
$ c = \sqrt{a^2 + b^2} = \sqrt{5} $

✅ Foci: $ (-1 \pm \sqrt{5},\ 1) $

Qus : 870 MCA NIMCET PYQ

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° degree and 45° repectively. If the lighthouse is 100 m high, the distance between the two ships is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:

$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$, $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.

Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.

Qus : 871 MCA NIMCET PYQ

The permutations of ${a,b,c,d,e,f,g}$ are listed in lexicographic order. Which of the following permutations are just before and just after the permutation $bacdefg$ ?

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 872 MCA NIMCET PYQ

The value of sin 10°sin 50°sin 70° is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

sin10° sin50° sin70°

= sin10° sin(60°−10°) sin(60°+10°)

= 1/4 sin3x10°

=1/4x1/2=1/8

Qus : 873 MCA NIMCET PYQ

Two towers face each other separated by a distance of 25 meters. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. The height (in meters) of the second tower is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Two Towers – Height of the Second Tower

Distance between towers’ bases = 25 m

Let the first tower be $AB$ (top $A$, base $B$) and the second tower be $CD$ (top $C$, base $D$). The bases $B$ and $D$ are 25 m apart.

From the top $A$: angle of depression to base $D$ is $60^\circ$ and to top $C$ is $30^\circ$.

From right $\triangle ABD$: \[ \tan 60^\circ=\frac{AB}{BD}=\frac{AB}{25}\;\Rightarrow\; AB=25\sqrt{3}. \]
From right $\triangle ACD$: vertical difference $=AB-CD$ and horizontal $=25$. \[ \tan 30^\circ=\frac{AB-CD}{25}=\frac{1}{\sqrt{3}} \;\Rightarrow\; AB-CD=\frac{25}{\sqrt{3}}. \]
Substitute $AB=25\sqrt{3}$: \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]

Answer: $ \displaystyle CD=\frac{50}{\sqrt{3}} \approx 28.87\ \text{m} $

Qus : 874 MCA NIMCET PYQ

The locus of the mid-point of all chords of the parabola $y^2 = 4x$ which are drawn through its vertex is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2023 PYQ

Solution

Locus of Midpoint of Chords

Given Parabola: $ y^2 = 4x $

Condition: Chords pass through the vertex $ (0, 0) $

Let the other end of the chord be $ (x_1, y_1) $, so the midpoint is:

$ M = \left( \frac{x_1}{2}, \frac{y_1}{2} \right) = (h, k) $

Since the point lies on the parabola: $ y_1^2 = 4x_1 $

⇒ $ (2k)^2 = 4(2h) $

⇒ $ 4k^2 = 8h $

⇒ $ \boxed{k^2 = 2h} $

✅ Locus of midpoints: $ y^2 = 2x $

Qus : 875 MCA NIMCET PYQ

The foci of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ and the hyperbola $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ coincide. Then the value of $b^2$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given ellipse: $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ ...(i)
Given hyperbola: $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ ...(ii)
**For Hyperbola:**
Rewriting (ii) in standard form:
$\dfrac{x^2}{\dfrac{144}{25}} - \dfrac{y^2}{\dfrac{81}{25}} = 1$
Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$
For hyperbola, $c^2 = a^2 + b^2$
$\Rightarrow c^2 = \dfrac{144}{25} + \dfrac{81}{25} = \dfrac{225}{25} = 9$
$\therefore c = 3$
So, foci of hyperbola $= (\pm 3,\ 0)$
**For Ellipse:**
Here, $a^2 = 16$
For ellipse, $c^2 = a^2 - b^2$
$\Rightarrow c^2 = 16 - b^2$
Since foci of ellipse and hyperbola coincide:
$\Rightarrow 16 - b^2 = 9$
$\Rightarrow b^2 = 16 - 9$
$\therefore \boxed{b^2 = 7}$

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