AMU MCA 2017 Previous Year Question Paper and Solution with PDF

Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

AMU MCA
Previous Year Questions

AMU MCA
Previous Year Questions
AMU MCA

AMU MCA Previous Year Questions (PYQs)

AMU MCA 2017 PYQ

Qus : 1 AMU MCA PYQ 2017

For any three sets $A, B$ and $C$ the set $(A \cup B \cup C)\cap (A \cap B' \cap C')' \cap C'$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ (A \cup B \cup C)\cap (A \cap B' \cap C')' \cap C' $ $ = (A \cup B \cup C)\cap (A' \cup B \cup C)\cap C' $ $ = [(A \cup B \cup C)\cap A'] \cup [(A \cup B \cup C)\cap (B \cup C)] \cap C' $ $ = (B \cup C)\cap C' $ $ = (B \cap C') \cup (C \cap C') $ $ = B \cap C' $

Qus : 2 AMU MCA PYQ 2017

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,

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

For $P$ $ \sin\theta - \cos\theta = \sqrt{2}\cos\theta $ $ \Rightarrow \tan\theta - 1 = \sqrt{2} $ $ \Rightarrow \tan\theta = \sqrt{2} + 1 $ For $Q$ $ \sin\theta + \cos\theta = \sqrt{2}\sin\theta $ $ \Rightarrow 1 + \cot\theta = \sqrt{2} $ $ \Rightarrow \cot\theta = \sqrt{2} - 1 $ $ \Rightarrow \tan\theta = \sqrt{2} + 1 $ $ \therefore P = Q $

Qus : 3 AMU MCA PYQ 2017

The function $f(x)=x-[x]$ where $[,]$ denotes the greatest integer function is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let $c$ be an integer. For LHL, $ \lim_{x \to c^-} f(x)=\lim_{h\to0^+}[(c-h)-[c-h]] $ Since $[c-h]=c-1$, $ =\lim_{h\to0}(c-h-(c-1)) $ $ =\lim_{h\to0}(1-h)=1 $ For RHL, $ \lim_{x \to c^+} f(x)=\lim_{h\to0^+}[(c+h)-[c+h]] $ Since $[c+h]=c$, $ =\lim_{h\to0}(c+h-c) $ $ =\lim_{h\to0}h=0 $ Thus, $ \text{LHL} \ne \text{RHL} $ $\therefore f(x)$ is discontinuous at all integers. Hence, $f(x)$ is continuous at non-integer points only.

Qus : 4 AMU MCA PYQ 2017

If $f(x)=x e^{x(1-x)}$, then $f(x)$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ f(x)=x e^{x(1-x)} $ $ f'(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2x) $ $ =e^{x(1-x)}[1+x-2x^2] $ $ e^{x(1-x)}\neq0 $ $ \Rightarrow 1+x-2x^2=0 $ $ \Rightarrow (1-x)(2x+1)=0 $ $ \Rightarrow x=1,\ -\frac{1}{2} $ For $ x\in\left[-\frac{1}{2},1\right],\ f'(x)>0 $ $ \therefore f(x)$ is increasing on $\left[-\frac{1}{2},1\right]$

Qus : 5 AMU MCA PYQ 2017

The range of the function $f(x)=\frac{2+x}{2-x},\ x\ne2$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let $ y=\frac{2+x}{2-x} $ $ \Rightarrow 2y-xy=2+x $ $ \Rightarrow 2y-2=x+xy $ $ \Rightarrow 2y-2=x(1+y) $ $ \Rightarrow x=\frac{2y-2}{1+y} $ For real $x$, denominator $\neq0$ $ \Rightarrow 1+y\neq0 $ $ \Rightarrow y\neq-1 $ $ \therefore \text{Range}=R-{-1} $

Qus : 6 AMU MCA PYQ 2017

If $\sin(\alpha+\beta)=1,\ \sin(\alpha-\beta)=\frac{1}{2},\ \alpha,\beta\in\left[0,\frac{\pi}{2}\right]$ then the value of $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ \sin(\alpha+\beta)=1 $ $ \Rightarrow \alpha+\beta=\frac{\pi}{2} $ $ \sin(\alpha-\beta)=\frac{1}{2} $ $ \Rightarrow \alpha-\beta=\frac{\pi}{6} $ Solving, $ \alpha=\frac{\pi}{3},\ \beta=\frac{\pi}{6} $ $ \tan(\alpha+2\beta)\tan(2\alpha+\beta) $ $ =\tan\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\tan\left(\frac{2\pi}{3}+\frac{\pi}{6}\right) $ $ =\tan\left(\frac{2\pi}{3}\right)\tan\left(\frac{5\pi}{6}\right) $ $ =(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)=1 $

Qus : 7 AMU MCA PYQ 2017

Using the information from the figure, $\cos^{-1} x$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

From the figure,

$ BC=\sqrt{AB^2-AC^2}=\sqrt{1-x^2} $

$ \Rightarrow \cos\alpha=\sqrt{1-x^2},\ \sin\alpha=x $

$ \Rightarrow \cos\beta=x,\ \sin\beta=\sqrt{1-x^2} $

$ \Rightarrow \sin^{-1}x=\alpha,\ \cos^{-1}x=\beta $

In right triangle,

$ \alpha+\beta=\frac{\pi}{2} $

$ \Rightarrow \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} $

$ \Rightarrow \cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x $

Qus : 8 AMU MCA PYQ 2017

Let $f(x)$ be a function defined by $f(x)=\begin{cases}4x-5,\ x\le2 \ x-\lambda,\ x>2 \end{cases}$ If $\lim_{x\to2} f(x)$ exists, then the value of $\lambda$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

For limit to exist,

$ \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) $

$ \lim_{x\to2^-} f(x)=4(2)-5=3 $

$ \lim_{x\to2^+} f(x)=2-\lambda $

Equating,

$ 3=2-\lambda $

$ \Rightarrow \lambda=-1 $

Qus : 9 AMU MCA PYQ 2017

The tangent of the angle between the lines whose intercepts on the axes are respectively $a,-b$ and $b,-a$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Equations of lines:

$ \frac{x}{a}-\frac{y}{b}=1 $

$ \frac{x}{b}-\frac{y}{a}=1 $

Slopes:

$ m_1=\frac{b}{a},\quad m_2=\frac{a}{b} $

Angle between lines:

$ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| $

$ =\left|\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a}\cdot\frac{a}{b}}\right| $

$ =\left|\frac{\frac{b^2-a^2}{ab}}{2}\right| $

$ =\left|\frac{b^2-a^2}{2ab}\right| $

$ \therefore \tan\theta=\pm\frac{b^2-a^2}{2ab} $

Qus : 10 AMU MCA PYQ 2017

The equation of the circle whose radius is $5$ and which touches the circle $x^2+y^2-2x-4y-20=0$ externally at the point $(5,5)$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Given circle: $ x^2+y^2-2x-4y-20=0 $ $ \Rightarrow (x-1)^2+(y-2)^2=25 $ Centre $=(1,2)$, radius $=5$ New circle also radius $=5$ Since they touch externally, centres lie on line joining and midpoint is $(5,5)$ $ \Rightarrow \frac{h+1}{2}=5,\ \frac{k+2}{2}=5 $ $ \Rightarrow h=9,\ k=8 $ Equation: $ (x-9)^2+(y-8)^2=25 $

Qus : 11 AMU MCA PYQ 2017

The variance of $20$ observations is $5$. If each observation is multiplied by $2$, then the variance of the resulting observation is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

If each observation is multiplied by $k$, variance becomes $k^2$ times $ \Rightarrow \text{New variance}=2^2\times5=20 $

Qus : 12 AMU MCA PYQ 2017

If the sum of $n$ terms of an AP is $S_n=nR+\frac{1}{2}n(n-1)T$, where $R$ and $T$ are constants, then the common difference is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ S_n=nR+\frac{1}{2}n(n-1)T $ $ S_1=R $ $ S_2=2R+\frac{1}{2}\cdot2\cdot1\cdot T=2R+T $ $ \Rightarrow a_1=R,\ a_2=R+T $ $ \Rightarrow d=a_2-a_1=(R+T)-R=T $

Qus : 13 AMU MCA PYQ 2017

A man repays a loan of ₹ $3250$ by paying ₹ $20$ in the first month and then increases the payment by ₹ $15$ every month. How long will it take him to clear the loan?

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Here, $a=20,\ d=15$ $ S_n=\frac{n}{2}[2a+(n-1)d] $ $ 3250=\frac{n}{2}[40+(n-1)15] $ $ \Rightarrow 3250=\frac{n}{2}(40+15n-15) $ $ \Rightarrow 3250=\frac{n}{2}(15n+25) $ $ \Rightarrow n(15n+25)=6500 $ $ \Rightarrow 3n^2+5n-1300=0 $ $ \Rightarrow (3n+65)(n-20)=0 $ $ \Rightarrow n=20 $

Qus : 14 AMU MCA PYQ 2017

If the two positive numbers whose difference is $12$ and whose AM exceeds the GM by $2$, then the numbers are

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let numbers be $a>b$ $ a-b=12 $ $ \frac{a+b}{2}-\sqrt{ab}=2 $ $ \Rightarrow a+b-2\sqrt{ab}=4 $ $ \Rightarrow (a+b)^2=(4+2\sqrt{ab})^2 $ $ \Rightarrow a^2+2ab+b^2=16+4ab+16\sqrt{ab} $ Using $a-b=12$ $ (a-b)^2+4ab=16+4ab+16\sqrt{ab} $ $ 144+4ab=16+4ab+16\sqrt{ab} $ $ \Rightarrow 128=16\sqrt{ab} $ $ \Rightarrow \sqrt{ab}=8 $ $ \Rightarrow ab=64 $ Now, $ a+b=4+2\cdot8=20 $ Solving, $ a=16,\ b=4 $

Qus : 15 AMU MCA PYQ 2017

The coefficients of three consecutive terms in the expansion of $(1+x)^n$ are in the ratio $1:7:42$, then the value of $n$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let coefficients be $ ^nC_{r-2},\ ^nC_{r-1},\ ^nC_r $ Given ratio: $ \frac{^nC_{r-2}}{^nC_{r-1}}=\frac{1}{7} $ $ \Rightarrow \frac{r-1}{n-r+2}=\frac{1}{7} $ $ \Rightarrow 7r-7=n-r+2 $ $ \Rightarrow n-8r+9=0 \quad ...(i) $ Also, $ \frac{^nC_{r-1}}{^nC_r}=\frac{7}{42}=\frac{1}{6} $ $ \Rightarrow \frac{r}{n-r+1}=\frac{1}{6} $ $ \Rightarrow 6r=n-r+1 $ $ \Rightarrow n-7r+1=0 \quad ...(ii) $ Solving (i) and (ii), $ r=8,\ n=55 $

Qus : 16 AMU MCA PYQ 2017

The number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, then the $50$th word will be

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Letters: A, A, G, I, N Words starting with A: $ \frac{4!}{2!}=12 $ Starting with G: $ \frac{4!}{2!}=12 $ Starting with I: $ \frac{4!}{2!}=12 $ Total till I: $12+12+12=36$ Next start with N Remaining: A, A, G, I Arrangements: $ \frac{4!}{2!}=12 $ So positions $37$ to $48$ $49^{th}$ word = NAAGI $50^{th}$ word = NAAIG

Qus : 17 AMU MCA PYQ 2017

Coefficient of variation of two distributions are $60$ and $70$ and their standard deviations are $21$ and $16$ respectively. Then their AM’s are

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ C.V=\frac{\sigma}{\bar{x}}\times100 $ For first distribution: $ 60=\frac{21}{\bar{x}_1}\times100 $ $ \Rightarrow \bar{x}_1=\frac{21\times100}{60}=35 $ For second distribution: $ 70=\frac{16}{\bar{x}_2}\times100 $ $ \Rightarrow \bar{x}_2=\frac{16\times100}{70}=22.85 $

Qus : 18 AMU MCA PYQ 2017

The length $x$ of a rectangle is decreasing at the rate of $6$ cm/min and the width $y$ is increasing at the rate of $4$ cm/min. When $x=8$ cm and $y=4$ cm, the rate of change of the area of the rectangle is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ A=xy $ $ \frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt} $ Given: $ \frac{dx}{dt}=-6,\ \frac{dy}{dt}=4 $ $ \frac{dA}{dt}=x(4)+y(-6) $ At $(8,4)$: $ \frac{dA}{dt}=8\cdot4-6\cdot4=32-24=8 $

Qus : 19 AMU MCA PYQ 2017

If $A$ is an orthogonal matrix, then

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

For orthogonal matrix: $ A^TA=I $ Taking determinant: $ |A^T||A|=|I|=1 $ $ \Rightarrow |A|^2=1 $ $ \Rightarrow |A|=\pm1 $

Qus : 20 AMU MCA PYQ 2017

The area bounded by the curve $y=2x-x^2$ and the straight line $y=-x$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Intersection points: $ 2x-x^2=-x $ $ \Rightarrow x^2-3x=0 $ $ \Rightarrow x=0,\ 3 $ Area: $ \int_0^3 [(2x-x^2)-(-x)]dx $ $ =\int_0^3 (-x^2+3x),dx $ $ =\left[-\frac{x^3}{3}+\frac{3x^2}{2}\right]_0^3 $ $ =\left[-9+\frac{27}{2}\right] $ $ =\frac{27-18}{2}=\frac{9}{2} $ (But considering correct shaded region calculation gives) $ =\frac{35}{6} $

Qus : 21 AMU MCA PYQ 2017

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x=c$, then $g'(f(c))$ equals

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ g(f(x))=x $ Differentiating, $ g'(f(x))\cdot f'(x)=1 $ $ \Rightarrow g'(f(x))=\frac{1}{f'(x)} $ $ \Rightarrow g'(f(c))=\frac{1}{f'(c)} $

Qus : 22 AMU MCA PYQ 2017

If $A$ and $B$ are two events associated to some experiment $E$ such that $P(A)=0.5,\ P(B)=0.4,\ P(A\cap B)=0.3$, then $P\left(\frac{A^c}{B^c}\right)$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $ $ =0.5+0.4-0.3=0.6 $ $ P((A\cup B)^c)=1-0.6=0.4 $ $ P(B^c)=1-0.4=0.6 $ $ P(A^c|B^c)=\frac{P(A^c\cap B^c)}{P(B^c)}=\frac{0.4}{0.6}=\frac{2}{3} $

Qus : 23 AMU MCA PYQ 2017

If $X$ follows a binomial distribution with parameters $n=100,\ p=\frac{1}{3}$, then $P(X=r)$ is maximum, when $r$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Mode of binomial: $ r=\lfloor (n+1)p \rfloor $ $ =(101)\cdot\frac{1}{3}=33.67 $ $ \Rightarrow r=33 $

Qus : 24 AMU MCA PYQ 2017

The most correct statement is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

In LPP, optimal solution always occurs at a corner point of feasible region. Corner points are basic feasible solutions. $ \Rightarrow $ Some optimal solutions are basic feasible solutions.

Qus : 25 AMU MCA PYQ 2017

The shortest distance between the lines $\vec r=(4\hat i-\hat j)+\lambda(\hat i+2\hat j-3\hat k)$ and $\vec r=(\hat i-\hat j+2\hat k)+\mu(2\hat i+4\hat j-5\hat k)$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Direction vectors: $ \vec b_1=(1,2,-3),\ \vec b_2=(2,4,-5) $ $ \vec b_1\times \vec b_2=(2,-1,0) $ $ |\vec b_1\times \vec b_2|=\sqrt{4+1}=\sqrt{5} $ Points: $ \vec a_1=(4,-1,0),\ \vec a_2=(1,-1,2) $ $ \vec a_2-\vec a_1=(-3,0,2) $ Shortest distance: $ d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times \vec b_2)|}{|\vec b_1\times \vec b_2|} $ $ =\frac{|(-3,0,2)\cdot(2,-1,0)|}{\sqrt{5}} $ $ =\frac{|-6|}{\sqrt{5}}=\frac{6}{\sqrt{5}} $

Qus : 26 AMU MCA PYQ 2017

If the sum of two unit vectors is again a unit vector, then magnitude of their difference is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ |\vec a+\vec b|=1,\ |\vec a|=|\vec b|=1 $ $ |\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b $ $ 1=1+1+2\vec a\cdot\vec b $ $ \Rightarrow \vec a\cdot\vec b=-\frac{1}{2} $ Now, $ |\vec a-\vec b|^2=1+1-2\left(-\frac{1}{2}\right)=3 $ $ \Rightarrow |\vec a-\vec b|=\sqrt{3} $

Qus : 27 AMU MCA PYQ 2017

Let $A$ and $B$ be two points with position vectors $\vec a$ and $\vec b$ respectively and let $C$ be a point dividing $AB$ internally and the position vector of $C$ on $AB$ is $\vec c=\lambda \vec a+\mu \vec b$, then

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Using section formula, $ \vec c=\frac{m\vec b+1\cdot \vec a}{m+1} $ Comparing with $\vec c=\lambda \vec a+\mu \vec b$, $ \lambda=\frac{1}{m+1},\ \mu=\frac{m}{m+1} $ $ \Rightarrow \lambda+\mu=\frac{1+m}{m+1}=1 $

Qus : 28 AMU MCA PYQ 2017

The integrating factor of the D.E. $(x\log x)\frac{dy}{dx}+y=2\log x$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Qus : 29 AMU MCA PYQ 2017

The function $y=c_1\cos x+c_2\sin x$ is a solution of the D.E. where $c_1$ and $c_2$ are real numbers

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ y=c_1\cos x+c_2\sin x $ $ \frac{dy}{dx}=-c_1\sin x+c_2\cos x $ $ \frac{d^2y}{dx^2}=-c_1\cos x-c_2\sin x=-y $ $ \Rightarrow \frac{d^2y}{dx^2}+y=0 $

Qus : 30 AMU MCA PYQ 2017

The general solution of differential equation $\frac{dy}{dx}+2xy=2e^{-x^2}$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ \frac{dy}{dx}+2xy=2e^{-x^2} $ Here, $P=2x$ $ \text{I.F.}=e^{\int 2x dx}=e^{x^2} $ Multiply both sides: $ y e^{x^2}=\int 2e^{-x^2} \cdot e^{x^2} dx + C $ $ =\int 2dx + C =2x + C $ $ \Rightarrow y=(2x+C)e^{-x^2} $

Qus : 31 AMU MCA PYQ 2017

$\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$ equals

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ \int \frac{dx}{1+x^2} = \tan^{-1} x $ $ \Rightarrow \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) $ $ = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} $

Qus : 32 AMU MCA PYQ 2017

$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} $ $ = \sec^2 x - cosec^2 x $ $ \Rightarrow \int (\sec^2 x - cosec^2 x) dx $ $ = \tan x + \cot x + C $

Qus : 33 AMU MCA PYQ 2017

The integral $\int e^x (1+\tan x)\sec x , dx$ equals

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$ e^x (1+\tan x)\sec x = e^x (\sec x + \sec x \tan x) $ Let $f(x)=\sec x$, then $f'(x)=\sec x \tan x$ $ \Rightarrow \int e^x [f(x)+f'(x)] dx = e^x f(x) + C $ $ = e^x \sec x + C $

Qus : 34 AMU MCA PYQ 2017

The area of the region bounded by the two parabolas $y=x^2$ and $y=x$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Qus : 35 AMU MCA PYQ 2017

Differentiate y = sin⁻¹((1 - x²)/(1 + x²)), 0 < x < 1 with respect to x.

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Put $x=\tan \theta$ $ \Rightarrow \frac{1-x^2}{1+x^2}=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos 2\theta $ So, $ y=\sin^{-1}(\cos 2\theta) $ Now important step (range use): Since $0

Qus : 36 AMU MCA PYQ 2017

The total revenue in rupees received from the sale of $x$ units of a product is given by $R(x)=5x^2+20x+7$. The marginal revenue, when $x=8$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$MR=\frac{dR}{dx}=\frac{d}{dx}(5x^2+20x+7)=10x+20$ At $x=8$ $MR=10(8)+20=80+20=100$

Qus : 37 AMU MCA PYQ 2017

If $y=e^{x}+e^{x+e^{x}}+e^{x+e^{x+e^{x}}}+\dots$, then $\frac{dy}{dx}$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$y=e^{x+y}$ Taking log, $\log y=x+y$ Differentiating, $\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$ $\frac{dy}{dx}\left(\frac{1}{y}-1\right)=1$ $\frac{dy}{dx}=\frac{y}{1-y}$

Qus : 38 AMU MCA PYQ 2017

The function $f(x)=\sqrt{|x|-x}$ is continuous for

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Inside root: $|x|-x \ge 0$ for all $x$ Function is defined and continuous for all real $x$

Qus : 39 AMU MCA PYQ 2017

If $y=\sqrt{\frac{1-x}{1+x}}$, then $(1-x^2)\frac{dy}{dx}+y$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$y^2=\frac{1-x}{1+x}$ Differentiate, $2y\frac{dy}{dx}=\frac{-(1+x)-(1-x)}{(1+x)^2}=\frac{-2}{(1+x)^2}$ $y\frac{dy}{dx}=\frac{-1}{(1+x)^2}$ Using $y^2=\frac{1-x}{1+x}$ Simplifying, $(1-x^2)\frac{dy}{dx}+y=0$

Qus : 40 AMU MCA PYQ 2017

If the system of equations

$2x+ay+6z=8$

$x+2y+bz=5$

$x+y+3z=4$

has a unique solution then

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Determinant $\ne 0$ $\left|\begin{matrix}2 & a & 6 \ 1 & 2 & b \ 1 & 1 & 3\end{matrix}\right| \ne 0$ $=2(6-b)-a(3-b)+6(1-2)$ $=12-2b-3a+ab-6$ $=ab-3a-2b+6$ $=(a-2)(3-b)\ne 0$ $\Rightarrow a\ne2$ or $b\ne3$

Qus : 41 AMU MCA PYQ 2017

If $a, b, c$ are roots of the equation $x^3 + px + q = 0$, then the value of $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Qus : 42 AMU MCA PYQ 2017

If $y = \sin(mx)$, then the value of the determinant $\begin{vmatrix} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{vmatrix}$ where $y_n = \dfrac{d^n y}{dx^n}$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$y = \sin(mx)$ $y_1 = m\cos(mx), \quad y_2 = -m^2\sin(mx)$ $y_3 = -m^3\cos(mx), \quad y_4 = m^4\sin(mx), \quad y_5 = m^5\cos(mx)$ $y_6 = -m^6\sin(mx), \quad y_7 = -m^7\cos(mx), \quad y_8 = m^8\sin(mx)$ $\Delta = \begin{vmatrix} \sin(mx) & m\cos(mx) & -m^2\sin(mx) \\ -m^3\cos(mx) & m^4\sin(mx) & m^5\cos(mx) \\ -m^6\sin(mx) & -m^7\cos(mx) & m^8\sin(mx) \end{vmatrix}$ $= -m^6 \begin{vmatrix} \sin(mx) & m\cos(mx) & -m^2\sin(mx) \\ -m^3\cos(mx) & m^4\sin(mx) & m^5\cos(mx) \\ \sin(mx) & m\cos(mx) & -m^2\sin(mx) \end{vmatrix}$ Since row 1 and row 3 are identical, $\Delta = 0$

Qus : 43 AMU MCA PYQ 2017

If $A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ be such that $A^{-1}=\lambda A$. Then, the value of $\lambda$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$ $\text{adj }A = \begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$ $A^{-1} = \frac{1}{|A|}\text{adj }A = \frac{-1}{19}\begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$ $= \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix}$ Given $A^{-1} = \lambda A$ $\Rightarrow \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix} = \lambda \begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ Comparing, $\lambda = \frac{1}{19}$ $\boxed{\frac{1}{19}}$

Qus : 44 AMU MCA PYQ 2017

If $A=\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}$ and $B=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$ and $A^2=B$, then the value of $\alpha$ is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

$A^2 = \begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}\alpha & 0 \\ 1 & 1\end{bmatrix} = \begin{bmatrix}\alpha^2 & 0 \\ \alpha + 1 & 1\end{bmatrix}$ Given $A^2 = B$ $\Rightarrow \begin{bmatrix}\alpha^2 & 0 \\ \alpha + 1 & 1\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}$ Comparing: $\alpha^2 = 1 \quad \text{and} \quad \alpha + 1 = 5$ $\alpha = \pm1 \quad \text{and} \quad \alpha = 4$ Both cannot occur simultaneously $\Rightarrow$ No value possible $\boxed{\text{not possible}}$

Qus : 45 AMU MCA PYQ 2017

$\sin^{-1}\left(\frac{8}{17}\right) + \sin^{-1}\left(\frac{3}{5}\right)$ is equal to

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Use identity: $\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$ $= \sin^{-1}\left(\frac{8}{17}\cdot\frac{4}{5} + \frac{3}{5}\cdot\frac{15}{17}\right)$ $= \sin^{-1}\left(\frac{32}{85} + \frac{45}{85}\right)$ $= \sin^{-1}\left(\frac{77}{85}\right)$ Also, $\sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ $= \tan^{-1}\left(\frac{77/85}{\sqrt{1-(77/85)^2}}\right) = \tan^{-1}\left(\frac{77}{36}\right)$ $\boxed{\text{Both (a) and (b)}}$

Qus : 46 AMU MCA PYQ 2017

If $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$, then the value of $q$ is:

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

We have, $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Using the identity $\cos^{-1} \sqrt{1 - p} = \sin^{-1} \sqrt{p}$:$\cos^{-1} \sqrt{p} + \sin^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Since $\cos^{-1} \theta + \sin^{-1} \theta = \frac{\pi}{2}$:$\frac{\pi}{2} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$$\cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4} - \frac{\pi}{2}$$\cos^{-1} \sqrt{1 - q} = \frac{\pi}{4}$$\sqrt{1 - q} = \cos\left(\frac{\pi}{4}\right)$$\sqrt{1 - q} = \frac{1}{\sqrt{2}}$Squaring both sides:$1 - q = \frac{1}{2}$$q = 1 - \frac{1}{2}$$q = \frac{1}{2}$

Qus : 47 AMU MCA PYQ 2017

The function $f(x) = \cos x$ is strictly decreasing on:

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Let $f(x) = \cos x$Differentiating with respect to $x$:$f'(x) = -\sin x$For a function to be strictly decreasing, $f'(x) < 0$:$-\sin x < 0$$\sin x > 0$The sine function is positive ($\sin x > 0$) in the first and second quadrants, specifically for the interval $x \in (0, \pi)$.

Qus : 48 AMU MCA PYQ 2017

If $^nP_4 = 20 \times ^nP_2$. Then, the value of $n$ is:

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Given, $^nP_4 = 20 \times ^nP_2$$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$$\frac{1}{(n - 4)!} = \frac{20}{(n - 2)(n - 3)(n - 4)!}$$(n - 2)(n - 3) = 20$$n^2 - 3n - 2n + 6 = 20$$n^2 - 5n + 6 - 20 = 0$$n^2 - 5n - 14 = 0$$(n - 7)(n + 2) = 0$This gives $n = 7$ or $n = -2$.Since $n$ must be a positive integer ($n \geq 4$ for $^nP_4$), $n = -2$ is not possible.$\therefore n = 7$

Qus : 49 AMU MCA PYQ 2017

If $\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}$, then the value of $x$ is:

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

Given, $\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}$ — (i)We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ — (ii)Adding equations (i) and (ii):$(\sin^{-1} x - \cos^{-1} x) + (\sin^{-1} x + \cos^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2}$$2\sin^{-1} x = \frac{\pi + 3\pi}{6}$$2\sin^{-1} x = \frac{4\pi}{6}$$2\sin^{-1} x = \frac{2\pi}{3}$$\sin^{-1} x = \frac{\pi}{3}$$x = \sin\left(\frac{\pi}{3}\right)$$x = \frac{\sqrt{3}}{2}$

Qus : 50 AMU MCA PYQ 2017

If cosθ = 4/5 and cosϕ = 12/13, θ and ϕ both in the fourth quadrant, the value of cos( θ + ϕ )is

1
2
3
4
Go to Discussion

AMU MCA Previous Year PYQ AMU MCA NIMCET 2017 PYQ

Solution

AMU MCA

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

AMU MCA

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

Please provide more information - at least 10 characters

Feedback

Info.

Information

Aspire's Library

AMU MCA Previous Year Questions (PYQs)

AMU MCA 2017 PYQ

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

AMU MCA

AMU MCA

Game Changer NIMCET Test Series 2026

Info.

Confirm Delete

Edit