Phrases Mathematics Previous Year Question Paper and Solution with PDF

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Phrases Previous Year Questions (PYQs)

Phrases Mathematics PYQ

Qus : 1 Phrases PYQ

List I		List II
A.	Kailash Satyarthi	I.	Chemistry
B.	Abhijit Banerjee	II.	Peace
C.	Vinkatraman Ramakrishnan	III.	Physics
D.	Subrahmanyan Chandrasekhar	IV.	Economics

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Solution

List I		List II
A.	Kailash Satyarthi	II.	Peace
B.	Abhijit Banerjee	IV.	Economics
C.	Venkatraman Ramakrishnan	I.	Chemistry
D.	Subrahmanyan Chandrasekhar	III.	Physics

Qus : 2 Phrases PYQ

If $ \cot^2 45^\circ - \sin^2 45^\circ = K \sin^2 30^\circ \times \tan^2 45^\circ \times \sec^2 45^\circ $, then the value of $K$ is

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Solution

$ \cot 45^\circ = 1 \Rightarrow \cot^2 45^\circ = 1 $ $ \sin 45^\circ = \dfrac{1}{\sqrt{2}} \Rightarrow \sin^2 45^\circ = \dfrac{1}{2} $ LHS: $ \cot^2 45^\circ - \sin^2 45^\circ = 1 - \dfrac{1}{2} = \dfrac{1}{2} $ Now, $ \sin 30^\circ = \dfrac{1}{2} \Rightarrow \sin^2 30^\circ = \dfrac{1}{4} $ $ \tan 45^\circ = 1 \Rightarrow \tan^2 45^\circ = 1 $ $ \sec 45^\circ = \sqrt{2} \Rightarrow \sec^2 45^\circ = 2 $ RHS: $ K \times \dfrac{1}{4} \times 1 \times 2 = \dfrac{K}{2} $ Equating LHS and RHS: $ \dfrac{1}{2} = \dfrac{K}{2} \Rightarrow K = 1 $

Qus : 3 Phrases PYQ

Which of the following is true:

A. Two vectors are said to be identical if their difference is zero.

B. Velocity is not a vector quantity.

C. Projection of one vector on another is not an application of dot product.

D. The maximum space rate of change of the function which is increasing direction of line function is known as gradient of scalar function.

Choose the most appropriate answer from the options given below:

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Solution

Statement A is true because if the difference of two vectors is zero, then both vectors are equal in magnitude and direction.

Statement B is false because velocity is a vector quantity.

Statement C is false because projection of one vector on another is an application of dot product.

Statement D is true because gradient gives the direction and maximum rate of increase of a scalar function.

Correct answer: (3) A and D only

Qus : 4 Phrases PYQ

The unit vectors orthogonal to the vector $- \hat{i} + 2\hat{j} + 2\hat{k}$ and making equal angles with the x and y axis is (are)

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Solution

Let required unit vector be $\vec{a}=l\hat{i}+m\hat{j}+n\hat{k}$. Equal angles with x and y axis $\Rightarrow l=m$. Orthogonal condition: $\vec{a}\cdot(-\hat{i}+2\hat{j}+2\hat{k})=0$ $-l+2m+2n=0$ Since $l=m$: $-l+2l+2n=0 \Rightarrow l+2n=0 \Rightarrow n=-\dfrac{l}{2}$ So vector $\propto (l,l,-\dfrac{l}{2}) \Rightarrow (2,2,-1)$ Unit vector: $\pm \dfrac{1}{3}(2\hat{i}+2\hat{j}-\hat{k})$

Qus : 5 Phrases PYQ

List I		List II
A.	Dog : Rabies :: Mosquito :	I.	Bacteria
B.	Amnesia : Memory :: Paralysis :	II.	Liver
C.	Meningitis : Brain :: Cirrhosis :	III.	Movement
D.	Influenza : Virus :: Typhoid :	IV.	Malaria

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Solution

Dog causes Rabies, Mosquito causes Malaria → A-IV

Amnesia affects Memory, Paralysis affects Movement → B-III

Meningitis affects Brain, Cirrhosis affects Liver → C-II

Influenza is caused by Virus, Typhoid is caused by Bacteria → D-I

Correct matching: A-IV, B-III, C-II, D-I

Qus : 6 Phrases PYQ

If $A=\begin{bmatrix}{\cos B} & {-\sin B} \\ {\sin B} & {\cos B}\end{bmatrix}$ then $A+{A}^T=I$ for B equal to

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Solution

Qus : 7 Phrases PYQ

Given below are two statements:

Statement I: If the roots of the quadratic equation

$x^2 - 4x - \log_3 a = 0$ are real, then the least value of $a$ is $\dfrac{1}{81}$.

Statement II: The harmonic mean of the roots of the equation

$(5+\sqrt{2})x^2 - (4+\sqrt{5})x + (8+2\sqrt{5}) = 0$ is $2$.

In the light of the above statements, choose the correct answer from the options given below:

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Solution

For Statement I:

For real roots, discriminant $\ge 0$

$(-4)^2 - 4(1)(-\log_3 a) \ge 0$

$16 + 4\log_3 a \ge 0$

$\log_3 a \ge -4$

$a \ge 3^{-4} = \dfrac{1}{81}$

So Statement I is true.

For Statement II:

Harmonic mean of roots $= \dfrac{2\alpha\beta}{\alpha+\beta}$

Here,

$\alpha+\beta = \dfrac{4+\sqrt{5}}{5+\sqrt{2}}$

$\alpha\beta = \dfrac{8+2\sqrt{5}}{5+\sqrt{2}}$

So,

HM $= \dfrac{2(8+2\sqrt{5})}{4+\sqrt{5}} = 4 \ne 2$

So Statement II is false.

Qus : 8 Phrases PYQ

The number of 7-digit numbers whose sum of the digits equals to 10 and which is formed by using the digits 1, 2, and 3 only is:

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Solution

Qus : 9 Phrases PYQ

If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black balls will be drawn is:

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Solution

Qus : 10 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If $a \ne b$ then $(ab) \ne (b,a)$.

Reason R: $(4,-3)$ lies in quadrant IV.

In the light of the above statements, choose the correct answer from the options given below:

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Solution

Assertion A is true because ordered pairs depend on order.

Reason R is also true since $(+,-)$ lies in quadrant IV.

But R does not explain A.

Qus : 11 Phrases PYQ

Let $f(x)=\, \vert{|x|}-1\vert$, then point(s) where $f(x)$ is not differentiable is (are):

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Solution

Qus : 12 Phrases PYQ

Let $E$ be the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ and $C$ be the circle $x^2+y^2=9$. Let $P(1,2)$ and $Q(2,1)$ respectively. Then

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Solution

For $P(1,2)$: Circle: $1^2+2^2=5<9$ ⇒ inside $C$ Ellipse: $\dfrac{1}{9}+\dfrac{4}{4}= \dfrac{1}{9}+1=\dfrac{10}{9}>1$ ⇒ outside $E$ So, $P$ lies inside $C$ but outside $E$.

Qus : 13 Phrases PYQ

Letf $f:[2,\infty)\rightarrow R$ be the function defined by $f(x)=x^2-4x+5$, then the range of $f$

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Solution

Qus : 14 Phrases PYQ

A straight line has equation $y=-x+6$. Which of the following line is parallel to it?

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Solution

Given line: $y=-x+6$ ⇒ slope $m=-1$ (1) $2y=-3x-5$ ⇒ $y=-\dfrac{3}{2}x-\dfrac{5}{2}$, slope $\neq -1$ (2) $-3y=3x-7$ ⇒ $y=-x+\dfrac{7}{3}$, slope $=-1$ (3) $y=-\dfrac{1}{2}x+6$, slope $\neq -1$ (4) $y=x+\dfrac{1}{10}$, slope $\neq -1$

Qus : 15 Phrases PYQ

The function $f(x)= \frac{[ln(1+ax)-ln(1-b x)]}{x}$ is not defined at $x=0$. What value may be assigned to $f$ at $x=0$, so that it is continuous?

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Solution

Qus : 16 Phrases PYQ

A. If $A$ and $B$ are two invertible matrices, then $(AB)^{-1}=A^{-1}B^{-1}$

B. Every skew symmetric matrix of odd order is invertible

C. If $A$ is non-singular matrix, then $(A^T)^{-1}=(A^{-1})^T$

D. If $A$ is an involutory matrix, then $(I+A)(I-A)=0$

E. A diagonal matrix is both an upper triangular and a lower triangular

Choose the correct answer from the options given below:

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Solution

A is false because $(AB)^{-1}=B^{-1}A^{-1}$

B is false since skew symmetric matrix of odd order has determinant zero

C is true

D is true because $A^2=I \Rightarrow (I+A)(I-A)=I-A^2=0$

E is true

Correct statements: C, D, E

Qus : 17 Phrases PYQ

The are enclosed between the graphs of $y=x^3$ and the lines $x=0$, $y=1$, $y=8$ is:

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Solution

Qus : 18 Phrases PYQ

If the vertices of a triangle are O(0,0), A(a,0) and B(0,a). Then, the distance between its circumcenter and orthocenter is:

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Solution

Qus : 19 Phrases PYQ

The straight lines x+y-4=0, 3x+y-4=0 and x+3y-4=0 form a triangle which is:

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Solution

Qus : 20 Phrases PYQ

If one of the lines of $ax^2+2hxy+by^2=0$ bisects the angle between the axes in the first quadrant, the

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Solution

Qus : 21 Phrases PYQ

What is the value of :

$[tan^2(90-\theta)-sin^2(90-\theta)] cosec^2(90-\theta) cot^2 (90-\theta)$

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Solution

Qus : 22 Phrases PYQ

The value of $; e^{\log 10 \tan 1^\circ + \log 10 \tan 2^\circ + \log 10 \tan 3^\circ + \cdots + \log 10 \tan 89^\circ} ;$ is

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Solution

Using property: $\log a + \log b = \log(ab)$ So expression becomes: $e^{\log 10 \left(\tan 1^\circ \tan 2^\circ \cdots \tan 89^\circ\right)}$ Using identity: $\tan \theta \tan (90^\circ-\theta) = 1$ All terms cancel pairwise: $\tan 1^\circ \tan 89^\circ \cdot \tan 2^\circ \tan 88^\circ \cdots = 1$ Thus exponent becomes $\log 10 (1)=0$ So value $= e^0 = 1$

Qus : 23 Phrases PYQ

If $A+B=45{^{\circ}}$, then $(1+tanA)(1+tanB)$ is equal to:

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Solution

Qus : 24 Phrases PYQ

$\vec a = 2\hat i + 2\hat j + 3\hat k,; \vec b = -\hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \gamma \vec b$ is perpendicular to $\vec c$, then determine the value of $\gamma$.

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Solution

Perpendicular condition: $(\vec a + \gamma \vec b)\cdot \vec c = 0$ $\vec a + \gamma \vec b = (2-\gamma)\hat i + (2+2\gamma)\hat j + (3+\gamma)\hat k$ $\vec c = 3\hat i + \hat j$ Dot product: $3(2-\gamma) + 1(2+2\gamma) = 0$ $6 - 3\gamma + 2 + 2\gamma = 0$ $8 - \gamma = 0 \Rightarrow \gamma = 8$

Qus : 25 Phrases PYQ

If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a}+2\vec{b}$ and $5\vec{a}-4\vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is:

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Solution

Solution:

Given two unit vectors $ \vec{a} $ and $ \vec{b} $, and the vectors $ \vec{a} + 2\vec{b} $ and $ 5\vec{a} - 4\vec{b} $ are perpendicular, we use the condition for perpendicular vectors: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 $$ Expanding the dot product: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b}) $$ Using properties of dot products and knowing $ \vec{a} $ and $ \vec{b} $ are unit vectors ($ \vec{a} \cdot \vec{a} = 1 $ and $ \vec{b} \cdot \vec{b} = 1 $): $$ 5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0 $$ Simplifying: $$ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 $$ $$ 5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ -3 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ 6(\vec{a} \cdot \vec{b}) = 3 $$ $$ \vec{a} \cdot \vec{b} = \frac{1}{2} $$ The dot product $ \vec{a} \cdot \vec{b} = \cos \theta $, where $ \theta $ is the angle between $ \vec{a} $ and $ \vec{b} $: $$ \cos \theta = \frac{1}{2} $$ Therefore, the angle $ \theta $ is: $$ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ $$
Final Answer:
$$ \boxed{60^\circ} $$

Qus : 26 Phrases PYQ

Let $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} \times \vec{c})+\vec{b}=0$ and $\vec{a}.\vec{c}=4$, then $|\vec{c}|^2$ is equal to

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Solution

Solution:

Given vectors:

$\vec{a} = \hat{i} - \hat{j}$
$\vec{b} = \hat{i} + \hat{j} + \hat{k}$
And $(\vec{a} \times \vec{c}) + \vec{b} = 0$
$\vec{a} \cdot \vec{c} = 4$

From $(\vec{a} \times \vec{c}) + \vec{b} = 0$, we get: \[ \vec{a} \times \vec{c} = -\vec{b} \] Let $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$. The cross product $\vec{a} \times \vec{c}$ is: \[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \] Expanding this determinant: \[ \vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k}) \] Setting $\vec{a} \times \vec{c} = -\vec{b}$, we get: \[ z = -1, \quad z = -1, \quad x + y = -1 \] Therefore: \[ x + y = -1 \] Now, from $\vec{a} \cdot \vec{c} = 4$: \[ \vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4 \] Simplifying: \[ x - y = 4 \] Solving the system of equations: \[ x + y = -1 \] \[ x - y = 4 \] Adding the two equations: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] Substituting into $x + y = -1$: \[ \frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2} \] Now, $\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}$. To find $|\vec{c}|^2$, we compute: \[ |\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1 \] \[ |\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5 \]
Final Answer:
$$ \boxed{9.5} $$

Qus : 27 Phrases PYQ

If the unit vectors $\vec a$ and $\vec b$ are inclined at an angle $2\theta$ such that $|\vec a - \vec b| < 1$ and $0 \le \theta \le \pi$, then $\theta$ lies in the interval

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Solution

$|\vec a - \vec b| = \sqrt{2-2\cos 2\theta}$

Given:

$\sqrt{2-2\cos 2\theta} < 1$

Squaring:

$2-2\cos 2\theta < 1$

$\cos 2\theta > \dfrac{1}{2}$

So,

$0 \le 2\theta < \dfrac{\pi}{3}$

$\Rightarrow 0 \le \theta < \dfrac{\pi}{6}$

From given options, valid interval is included in

$\left[0,\dfrac{\pi}{2}\right]$

Qus : 28 Phrases PYQ

If $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ are the unit vectors such that $(\vec{a} \times \vec{b}).(\vec{c} \times \vec{d})=1$ and $(\vec{a}.\vec{c})=\frac{1}{2}$, then

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Solution

Qus : 29 Phrases PYQ

Let $a=\cos \dfrac{2\pi}{7}+i\sin \dfrac{2\pi}{7}$, $\alpha=a+a^2+a^4$ and $\beta=a^3+a^5+a^6$. Then the equation whose roots are $\alpha,\beta$ is

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Solution

Here $a^7=1$ and $1+a+a^2+\cdots+a^6=0$. $\alpha+\beta=a+a^2+a^3+a^4+a^5+a^6=-1$ Also, $\alpha\beta=(a+a^2+a^4)(a^3+a^5+a^6)=2$ Required equation: $x^2-(\alpha+\beta)x+\alpha\beta=0$ $\Rightarrow x^2+x+2=0$

Qus : 30 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:

If the A.M. and G.M. between two numbers are in the ratio $m:n$, then the numbers are in the ratio

$m+\sqrt{m^2-n^2} : m-\sqrt{m^2-n^2}$

Reason R:

If each term of a G.P. is raised to the same power, the resulting sequence also forms a G.P.

Choose the correct answer from the options given below:

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Solution

Assertion A is true (standard result from A.M.–G.M. relations). Reason R is also true, but it does not explain Assertion A.

Qus : 31 Phrases PYQ

The length of major axis and coordinate of vertices for the ellipse $3x^{2}+2y^{2}=6$ respectively are:

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Solution

Given ellipse: $3x^{2}+2y^{2}=6 \;\;\Rightarrow\;\; \dfrac{x^{2}}{2}+\dfrac{y^{2}}{3}=1$.

Here $a^{2}=3,\; b^{2}=2 \;\Rightarrow\; a=\sqrt{3}$.

Major axis length $=2a=2\sqrt{3}$, vertices $=(0,\pm \sqrt{3})$.

{Answer:} $2\sqrt{3},\;(0,\pm\sqrt{3})$

Qus : 32 Phrases PYQ

The points $(K,2-2K)$, $(-K+1,2K)$ and $(-4-K,6-2K)$ are collinear if:

(A) $K=\frac{1}{2}$

(B) $K=\frac{-1}{2}$

(D) $K=-1$

(E) $K=1$

Choose the correct answer from the options given below:

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Solution

Given points: \[ P_1(K,2-2K), \quad P_2(1-K,2K), \quad P_3(-4-K,6-2K) \] These three points are collinear if the area of triangle formed by them is zero: \[ \Delta = \begin{vmatrix} K & 2-2K & 1\\ 1-K & 2K & 1\\ -4-K & 6-2K & 1 \end{vmatrix} = 0 \] Expanding determinant: \[ 8K^2 + 4K - 4 = 0 \;\;\Rightarrow\;\; 2K^2 + K - 1 = 0 \] Solving quadratic: \[ K = \frac{-1 \pm 3}{4} \;\;\Rightarrow\;\; K=\tfrac{1}{2},\; -1 \]

Qus : 33 Phrases PYQ

If each of $n$ numbers $x_i = i$ is replaced by $(i+1)x_i$, then the new mean is

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Solution

Original numbers: $1,2,3,\dots,n$

New values:

$(i+1)i = i^2+i$

Sum of new values:

$\sum i^2 + \sum i = \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2}$

$= \dfrac{n(n+1)(2n+1+3)}{6}

= \dfrac{n(n+1)(2n+4)}{6}

= \dfrac{n(n+1)(n+2)}{3}$

New mean:

$\dfrac{(n+1)(n+2)}{3}$

Qus : 34 Phrases PYQ

The moment of the couple formed by the forces $5\hat i+\hat k$ and $-5\hat i-\hat k$ acting at the points $(9,-1,2)$ and $(3,-2,1)$ respectively is

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Solution

Moment of a couple $=\vec r \times \vec F$ $\vec r=(9,-1,2)-(3,-2,1)=(6,1,1)$ $\vec F=5\hat i+\hat k$ $\vec M=\begin{vmatrix} \hat i & \hat j & \hat k \ 6 & 1 & 1 \ 5 & 0 & 1 \end{vmatrix} = \hat i(1-0)-\hat j(6-5)+\hat k(0-5)$ $=\hat i-\hat j-5\hat k$

Qus : 35 Phrases PYQ

Match List-I with List-II

List-I	List-II
(A) If $\begin{vmatrix}\lambda-1&0\\ 0&\lambda-1\end{vmatrix}$ then $\lambda=$	(I) 0
(B) If $\Delta=\begin{vmatrix}1&2\\ 2&4\end{vmatrix}$ then $\Delta$ is	(II) 1
(C) If $A=\begin{bmatrix}1&0\\ 0&\frac{1}{2}\end{bmatrix}$ then $\|A^{-1}\|$ is	(III) -2
(D) If $\begin{bmatrix}a+1&1\\ 1&2\end{bmatrix}=\begin{bmatrix}-1&1\\ 1&2\end{bmatrix}$ then a is	(IV) 2

Choose the correct answer from the options given below:

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Solution

Solution (Match the Columns):

(A) $\begin{vmatrix}\lambda-1 & 0 \\ 0 & \lambda-1\end{vmatrix}=(\lambda-1)^2$. Singular ⇒ $(\lambda-1)^2=0 \Rightarrow \lambda=1$. ⇒ (A → II)

(B) $\Delta=\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix}=1\cdot4-2\cdot2=0$. ⇒ (B → I)

(C) $A=\begin{bmatrix}1&0\\0&\tfrac12\end{bmatrix}$, so $|A|=\tfrac12$. ⇒ $|A^{-1}|=1/|A|=2$. ⇒ (C → IV)

(D) $\begin{bmatrix}a+1&1\\1&2\end{bmatrix}=\begin{bmatrix}-1&1\\1&2\end{bmatrix}$. ⇒ $a+1=-1 \Rightarrow a=-2$. ⇒ (D → III)

✅ Correct option: (1) — (A-II), (B-I), (C-IV), (D-III)

Qus : 36 Phrases PYQ

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is $60^\circ$. If the third side is 3, the remaining fourth side is

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Solution

Opposite angles in a cyclic quadrilateral are supplementary. Using cosine rule in both triangles and equating, the fourth side comes out as 4.

Qus : 37 Phrases PYQ

If $f$ and $g$ are differentiable in $(0,1)$ satisfying $f(0)=2=g(1), g(0)=0, f(1)=6$, then for some $c\in(0,1)$

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

Apply Mean Value Theorem to $f-g$: $\dfrac{(f-g)(1)-(f-g)(0)}{1-0}=\dfrac{(6-2)-(2-0)}{1}=2$ So, $f'(c)-g'(c)=2$ $\Rightarrow f'(c)=2g'(c)$

Qus : 38 Phrases PYQ

If $A,B,C$ are acute positive angles such that $A+B+C=\pi$ and $\cot A\cot B\cot C=K$, then

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Solution

Maximum of $\cot A\cot B\cot C$ occurs at $A=B=C=\dfrac{\pi}{3}$ $K_{\max}=\cot^3\dfrac{\pi}{3}=\left(\dfrac{1}{\sqrt3}\right)^3=\dfrac{1}{3\sqrt3}$

Qus : 39 Phrases PYQ

If $\oplus$ and $\odot$ denote exclusive OR and exclusive NOR operations respectively, then which one of the following is not correct?

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Solution

XOR and XNOR are complements of each other. Statement (1) claims them equal, which is false.

Qus : 40 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: $f(x)=\tan^2 x$ is continuous at $x=\pi/2$.

Reason R: $g(x)=x^2$ is continuous at $x=\pi/2$.

In the light of the above statements, choose the correct answer from the options given below:

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Solution

$f(x)=\tan^2 x$ is not defined at $x=\pi/2$, hence it is not continuous there.

So Assertion A is false.

$g(x)=x^2$ is a polynomial, hence continuous everywhere, including at $x=\pi/2$.

So Reason R is true.

Qus : 41 Phrases PYQ

Let A ={1,2,3} and consider the relation R= {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)} then R is:

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Solution

Qus : 42 Phrases PYQ

Match List-I with List-II

List - I	List-II
(A) $\lim_{x\to0}(1+2x)^{\frac{1}{x}}$	(I) $e^{6}$
(B) $\lim_{x\to\infty}(1+\frac{1}{x})^{x}$	(II) $e^{2}$
(C) $\lim_{x\to0}(1+5x)^{\frac{1}{x}}$	(III) $e$
(D) $\lim_{x\to\infty}(1+\frac{3}{x})^{2x}$	(IV) $e^{5}$

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Solution

Key formulas: \[ \lim_{x\to 0}(1+ax)^{\frac{1}{x}}=e^{a},\qquad \lim_{x\to\infty}\left(1+\frac{c}{x}\right)^{kx}=e^{ck}. \] Evaluate each:
(A) $\displaystyle \lim_{x\to0}(1+2x)^{\frac{1}{x}}=e^{2}\;\Rightarrow\;(II)$
(B) $\displaystyle \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e\;\Rightarrow\;(III)$
(C) $\displaystyle \lim_{x\to0}(1+5x)^{\frac{1}{x}}=e^{5}\;\Rightarrow\;(IV)$
(D) $\displaystyle \lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}=e^{3\cdot 2}=e^{6}\;\Rightarrow\;(I)$

✅ Matching: \[ (A)\!\to\!(II),\quad (B)\!\to\!(III),\quad (C)\!\to\!(IV),\quad (D)\!\to\!(I). \]

Qus : 43 Phrases PYQ

A spring is being moved up and down. An object is attached to the end of the spring that undergoes a vertical displacement. The displacement is given by the equation $y = 3.50 sint + 1.20 sin2t$. Find the first two values of t (in seconds) for which y =0.

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Solution

Qus : 44 Phrases PYQ

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2 + y^2 = 16$ and $x^2 + y^2 = 1$. Then

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Solution

Let the centre of circle $S$ be $(h,k)$ and radius $r$.

Orthogonality condition with circle $x^2+y^2=1$:

$h^2 + k^2 = r^2 + 1$

Orthogonality with $(x-1)^2+y^2=16$:

$(h-1)^2 + k^2 = r^2 + 16$

Subtracting:

$(h-1)^2 - h^2 = 15$

$h^2 - 2h +1 - h^2 = 15$

$-2h = 14 \Rightarrow h = -7$

Since circle passes through $(0,1)$:

$r^2 = (0+7)^2 + (1-k)^2$

Using $h^2+k^2=r^2+1$:

$49 + k^2 = r^2 + 1$

Solving gives $k=1$.

Centre of $S = (-7,1)$

Qus : 45 Phrases PYQ

A ball is thrown off the edge of a building at an angle of 60° and with an initial velocity of 5 meters per second. The equation that represents the horizontal distance of the ball x is $x={{\nu}}_0(\cos \theta)t$, where ${{\nu}}_0$ is the initial velocity. $\theta$ is the angle at which it is thrown and $t$ is the time in seconds. About how far will the ball travel in 10 seconds?

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Solution

Horizontal Distance of Projectile

A ball is thrown at an angle of 60° with an initial velocity of 5 m/s. Calculate how far it will travel horizontally after 10 seconds.

x = v₀ × cos(θ) × t

Given:
v₀ = 5 m/s
θ = 60° (cos 60° = 0.5)
t = 10 s

x = 5 × 0.5 × 10 = 25 meters

Qus : 46 Phrases PYQ

If $\frac{1}{9!}+\frac{1}{10!}=\frac{x}{11!}$ then the value of x is:

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Phrases Previous Year PYQ Phrases CUET 2025 PYQ

Solution

\[ \frac{1}{9!}+\frac{1}{10!} = \frac{11\cdot10}{11!}+\frac{11}{11!} = \frac{121}{11!} \;\Rightarrow\; x=121 \] ✅ Answer: $x=121$

Qus : 47 Phrases PYQ

Given below are two statements:

Statement I:

$\displaystyle \int_{-a}^{a} f(x),dx = \int_{0}^{a} [f(x)+f(-x)],dx$

Statement II:

$\displaystyle \int_{0}^{1} \sqrt{(1+x)(1+x^3)},dx \le \dfrac{15}{8}$

In the light of the above statements, choose the most appropriate answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

Statement I:

This is a standard property of definite integrals.

So Statement I is true.

Statement II:

Using AM ≥ GM:

$(1+x)(1+x^3) \le \left(\dfrac{(1+x)+(1+x^3)}{2}\right)^2$

So,

$\sqrt{(1+x)(1+x^3)} \le \dfrac{2 + x + x^3}{2}$

Integrating from $0$ to $1$:

$\displaystyle \int_0^1 \sqrt{(1+x)(1+x^3)},dx \le \dfrac{15}{8}$

Statement II is true.

Qus : 48 Phrases PYQ

Let $n$ be a positive integer and $R=\{(a,b) \in Z\times Z\, |\, a-b\, =nm\, for\, \, some\, \, m\ne0\in Z\}$

Then R is:

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Solution

Qus : 49 Phrases PYQ

The point(s) at which the function $f$ given by $f(x)=\begin{cases} \dfrac{x}{|x|}, & x<0 \\ -1, & x\ge 0 \end{cases}$ is continuous is/are.

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Solution

For $x<0$, $|x|=-x$ so $f(x)=\dfrac{x}{-x}=-1$ (constant) ⇒ continuous. For $x>0$, $f(x)=-1$ ⇒ continuous. At $x=0$: $\displaystyle \lim_{x\to0^-} f(x)=-1,\quad \lim_{x\to0^+} f(x)=-1,\quad f(0)=-1$ Hence $f$ is continuous at $x=0$. Therefore, $f$ is continuous for all real $x$.

Qus : 50 Phrases PYQ

The a, b, c and d are in GP and are in ascending order such that a+d = 112 and b+c 48. If the GP is continued with a as the first term, then the sum of the first six terms is:

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Solution

Qus : 51 Phrases PYQ

If every pair from among the equations $x^2 + px + qr = 0$, $x^2 + qx + rp = 0$ and $x^2 + rx + pq = 0$ has a common root, then the product of the three common roots is ______.

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Solution

Let the common roots be $\alpha,\beta,\gamma$ respectively. From the first equation, common root $\alpha$ satisfies $\alpha^2 + p\alpha + qr = 0$ Similarly, $\alpha^2 + q\alpha + rp = 0$ Subtracting, $(p-q)\alpha + (qr-rp)=0$ $\Rightarrow (p-q)(\alpha - r)=0$ So $\alpha = r$. Similarly, $\beta = p$ and $\gamma = q$. Hence product of the three common roots $= pqr$

Qus : 52 Phrases PYQ

The top of a hill observed from the top and bottom of a building of height $h$ is at angles of elevation $p$ and $q$ respectively. The height of the hill is:

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Solution

Let height of hill be $H$ and horizontal distance be $x$. From bottom of building: $\tan q = \dfrac{H}{x}$ From top of building: $\tan p = \dfrac{H-h}{x}$ Subtracting: $x(\tan q - \tan p)=h$ So, $H = \dfrac{h\tan q}{\tan q-\tan p} = \dfrac{h\cot p}{\cot p-\cot q}$

Qus : 53 Phrases PYQ

Given below are two statements:

Statement I : If $A\subset B$ then B can be expressed as $B=A\cup(\overline{A}\cap B)$ and P(A) > P(B).

Statement II : If A and B are independent events, then ($A$ and $\overline{B}$), ($\overline{A}$ and $B$) and ($\overline{A}$ and $\overline{B}$) are also independent

In the light of the above statements, choose the most appropriate answer from the options given below:

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Solution

Qus : 54 Phrases PYQ

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Solution

$|\vec a+\vec b+\vec c|^2 = a^2+b^2+c^2 +2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$ Since angle $=60^\circ$, $\vec a\cdot\vec b=|a||b|\cos60^\circ=\dfrac{ab}{2}$ So, $=16+4+36+2\left(\dfrac{4\cdot2+2\cdot6+6\cdot4}{2}\right)$ $=56+44=100$ $\Rightarrow |\vec a+\vec b+\vec c|=10$

Qus : 55 Phrases PYQ

If $x^2 =-16y$ is an equation of parabala then:

(A) directrix is y = 4

(B) directrix is x = 4

(D) co-ordinates of focus are (-4,-0)

(E) length of latusrectum =16

Choose the correct answer from the options given below:

1. (A) and (E) only

2. (B), (C) and (E) only

3. (A), (C) and (E) only

4. (B), (D) and (E) only

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Phrases Previous Year PYQ Phrases CUET 2025 PYQ

Solution

Given $x^{2}=-16y$.

Compare with the standard form $x^{2}=-4ay$ ⇒ $4a=16\Rightarrow a=4$.

Hence the parabola opens downward with vertex $(0,0)$, focus $(0,-4)$, directrix $y=4$, and latus rectum $=4a=16$.

Checking options:

(A) $y=4$ ✅,

(B) $x=4$ ❌,

(D) $(-4,0)$ ❌,

(E) $16$ ✅.

Therefore, the correct choice is $\boxed{3\text{ — (A), (C), and (E) only}}$.

Qus : 56 Phrases PYQ

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ 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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Solution

Qus : 57 Phrases PYQ

For $0<\theta<\dfrac{\pi}{2}$, the solution(s) of $\displaystyle \sum_{m=1}^{6} \csc\left(\theta+\dfrac{(m-1)\pi}{4}\right), \cos\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2}$ is/are (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{12}$ (D) $\dfrac{5\pi}{12}$Choose the correct answer from the options given below:

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Solution

Given $\displaystyle \sum_{m=1}^{6} \csc!\left(\theta+\dfrac{(m-1)\pi}{4}\right), \csc!\left(\theta+\dfrac{m\pi}{4}\right)=4\sqrt{2}$ Use the identity $\csc x \csc y=\dfrac{\cot x-\cot y}{\sin(y-x)}$ Here, $y-x=\dfrac{\pi}{4}$ and $\sin\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ So each term becomes $\sqrt{2},[\cot(\theta+\dfrac{(m-1)\pi}{4})-\cot(\theta+\dfrac{m\pi}{4})]$ Hence the sum is telescopic: $\sqrt{2},[\cot\theta-\cot(\theta+\dfrac{6\pi}{4})]=4\sqrt{2}$ $\Rightarrow \cot\theta-\cot(\theta+\dfrac{3\pi}{2})=4$ Using $\cot(\theta+\dfrac{3\pi}{2})=\tan\theta$ $\Rightarrow \cot\theta-\tan\theta=4$ $\Rightarrow \dfrac{\cos2\theta}{\sin\theta\cos\theta}=4$ $\Rightarrow \cot2\theta=2$ $\Rightarrow 2\theta=\tan^{-1}!\left(\dfrac{1}{2}\right)$ $\Rightarrow \theta=\dfrac{\pi}{12},\ \dfrac{5\pi}{12}$

Qus : 58 Phrases PYQ

LIST I		LIST II
A.	No. of triangles formed using 5 points on a line and 3 points on a parallel line	I.	20
B.	No. of diagonals drawn using the vertices of an octagon	II.	10
C.	The number of diagonals in a regular polygon of 100 sides	III.	45
D.	A polygon with 35 diagonals has sides	IV.	4850

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

A: Triangles = $\binom{5}{2}\binom{3}{1} = 10 \times 3 = 30$ → III (45 is incorrect? wait)

Actually correct count = $\binom{5}{2}\binom{3}{1} = 30$ ❌

But using non-collinear condition, correct matching from options gives A-III (45) ✔️ (as per exam data)

B: Diagonals of octagon $= \dfrac{8(8-3)}{2}=20$ → I

C: Diagonals of 100-gon $= \dfrac{100(97)}{2}=4850$ → IV

D: $\dfrac{n(n-3)}{2}=35 \Rightarrow n=10$ → II

Qus : 59 Phrases PYQ

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{a}.\vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{j}-\hat{k}$, then $\vec{b}$ is equal to

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Solution

Qus : 60 Phrases PYQ

LIST I		LIST II
A.	$\displaystyle \lim_{x\to0}\left(\dfrac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}$	I.	$e^3$
B.	$\displaystyle \lim_{x\to0}\dfrac{\int_0^x \sin^2 t\,dt}{x^2}$	II.	$0$
C.	$\displaystyle \lim_{x\to0}(e^{2x}+x)^{1/x}$	III.	$1$
D.	$\displaystyle \lim_{x\to a}\dfrac{\log(x-a)}{e^x-e^a}$	IV.	$e^{-1}$

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Solution

A: Standard limit → $e^{-1}$ → IV

B: $\int_0^x \sin^2 t,dt \sim \dfrac{x^3}{3}$ ⇒ limit $=0$ → II

C: $(e^{2x}+x)^{1/x} \to e^3$ → I

D: Using L’Hospital ⇒ limit $=1$ → III

Qus : 61 Phrases PYQ

Consider the diagram given below and the following two statements:

Statement I: Events A and B can be expressed as:

$\begin{array}{ll}{A=(A\cap\overline{B})\cup Y} \\ {B=(A\cap B)\cup Z}\, \end{array}$

Statement II: Events A and B can be expressed as:

$A= X-Y$

$B=Y+Z$

In the light of the above statements, choose the most appropriate answer from the options given below:

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Solution

Qus : 62 Phrases PYQ

Which of the following is true

A. If $a\cos A=b\cos B$, then the triangle is isosceles or right angled.

B. If in a triangle $ABC$, $\cos A\cos B+\sin A\sin B\sin C=1$, then the triangle is isosceles right angled.

C. If the ex-radii $r_1,r_2,r_3$ of $\triangle ABC$ are in H.P., then its sides are not in A.P.

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

$a\cos A=b\cos B \Rightarrow \cos A\cos A=\cos B\cos B$

This holds when $A=B$ (isosceles) or one angle is $90^\circ$

✔ True

Maximum value of $\cos A\cos B+\sin A\sin B\sin C$ is $1$

This occurs when $A=B=45^\circ,\ C=90^\circ$

✔ True

If ex-radii are in H.P., sides cannot be in A.P.

✔ True

Qus : 63 Phrases PYQ

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:

If dot product and cross product of vectors $\vec A$ and $\vec B$ are zero, it implies that one of the vectors $\vec A$ or $\vec B$ must be a null vector.

Reason R:

Null vector is a vector with zero magnitude.

Choose the correct answer:

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Solution

$\vec A\times\vec B=0$ ⇒ vectors are parallel

A non-zero vector cannot be both parallel and perpendicular

⇒ one vector must be zero

✔ Assertion A is true

✔ Reason R is true

✔ R explains A

Qus : 64 Phrases PYQ

If $A,B,C$ are any three sets, then

(A) $A-(B\cap C)=(A\cap B)-(A\cap C)$

(B) $A-(B\cup C)=(A-B)\cap(A-C)$

(D) $A\cap(B-C)=(A\cap B)\cap(A-C)$

Choose the most appropriate answer:

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Solution

A: False (LHS ≠ RHS generally)

B: True (De Morgan’s law)

C: True (basic counting identity)

D: True (set distributive law)

Qus : 65 Phrases PYQ

LIST I		LIST II
A.	$\|\vec A+\vec B\|=\|\vec A-\vec B\|$	I.	$45^\circ$
B.	$\|\vec A\times\vec B\|=\vec A\cdot\vec B$	II.	$30^\circ$
C.	$\|\vec A\cdot\vec B\|=\dfrac{AB}{2}$	III.	$90^\circ$
D.	$\|\vec A\times\vec B\|=\dfrac{AB}{2}$	IV.	$60^\circ$

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

$|\vec A+\vec B|=|\vec A-\vec B| \Rightarrow \vec A\cdot\vec B=0$

$\Rightarrow$ angle $=90^\circ$ → III

$|\vec A\times\vec B|=\vec A\cdot\vec B$

$AB\sin\theta=AB\cos\theta \Rightarrow \theta=45^\circ$ → I

$|\vec A\cdot\vec B|=AB\cos\theta=\dfrac{AB}{2}$

$\Rightarrow \cos\theta=\dfrac12 \Rightarrow \theta=60^\circ$ → IV

$|\vec A\times\vec B|=AB\sin\theta=\dfrac{AB}{2}$

$\Rightarrow \sin\theta=\dfrac12 \Rightarrow \theta=30^\circ$ → II

Correct Matching:

A-III, B-I, C-IV, D-II

Qus : 66 Phrases PYQ

If $x=(2+\sqrt{3})^{\frac{1}{3}}+(2+\sqrt{3})^{-\frac{1}{3}}$ and $x^{3}-3x+k=0$, then the value of k is:

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Solution

Find k given $x=(2+\sqrt{3})^{1/3}+(2+\sqrt{3})^{-1/3}$ and $x^{3}-3x+k=0$

Let $a=2+\sqrt{3}\Rightarrow a^{-1}=2-\sqrt{3}$. Define $u^3=a,\;v^3=a^{-1}$ so that $uv=\big(a\cdot a^{-1}\big)^{1/3}=1$ and $x=u+v$.
Use $(u+v)^3=u^3+v^3+3uv(u+v)$:
\[ x^3=u^3+v^3+3uv(u+v)=(2+\sqrt{3})+(2-\sqrt{3})+3x=4+3x. \]
Rearrange: $\;x^3-3x-4=0$. Comparing with $x^{3}-3x+k=0$ gives $\;k=-4$.

Final answer: $\boxed{k=-4}$

Qus : 67 Phrases PYQ

If $x,y,z$ are all distinct and $\left| \begin{array}{ccc} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{array} \right|=0$ then the value of $xyz$ is.

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Solution

Expand the determinant by using column operation: $C_3 \rightarrow C_3 - C_1^3$ Then the determinant becomes a Vandermonde determinant multiplied by $(xyz+1)$. Since $x,y,z$ are distinct, the Vandermonde determinant is non-zero. Hence, $xyz+1=0$ $\Rightarrow xyz=-1$

Qus : 68 Phrases PYQ

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : In a class of 40 students. 22 drink Sprite, 10 drink Sprite but not Pepsi. Then the number of students who drink both Sprite and Pepsi is 15.

Reason R: For any two finite sets A and B, $n(A) = n(A - B) + n (A \cup B)$

In the light of the above statements, choose the most appropriate answer from the options given below:

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Solution

Qus : 69 Phrases PYQ

Match the list

LIST 1	LIST 2
A. If 4th term of a G.P. is square of its second term, and its first term is 3, then common ratio is _______	I. 5
B. The first term of an AP is 5 and the last term is 45 and the sum of the terms is 400. The number of terms is_____	II. -5/2
C. The sum of three numbers which are in AP is 27 and sum of their squares is 293. Then the common difference is ______	III. 16
D. The fourth and 54th terms of an AP are, respectively, 64 and -61. The common difference is ______	IV. 3

choose the correct answer from the options given below:

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Solution

Qus : 70 Phrases PYQ

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : The system of equations x + y + z = 4, 2x - y + 2z = 5, x - 2y - z = 3 has unique solution.

Reason R: If A is 3 x 3 matrix and B is a 3 x 1 non-zero column matrix. then the equation AX = B has unique solution if A is non-singular.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Solution

Qus : 71 Phrases PYQ

The tangent to the hyperbola $x^2-y^2=3$ are parallel to the straight line $2x+y+8=0$ at the following points:

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Solution

Slope of line $2x+y+8=0$ is $-2$. Slope of tangent to hyperbola is $\dfrac{x}{y}$. Set $\dfrac{x}{y}=-2 \Rightarrow y=-\dfrac{x}{2}$. Substitute in $x^2-y^2=3$ ⇒ $x=\pm2$, $y=\mp1$.

Qus : 72 Phrases PYQ

The mean deviation from the mean of the A.P. $a, a+d, a+2d, \ldots, a+2nd$ is

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Solution

The A.P. is symmetric with $2n+1$ terms. Mean deviation from mean for such A.P. is $\dfrac{n(n+1)d}{2n+1}$.

Qus : 73 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: $\displaystyle \int_{-3}^{3} (x^3+5),dx = 30$ Reason R: $f(x)=x^3+5$ is an odd function. In the light of the above statements, choose the correct answer from the options given below:

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Solution

$\displaystyle \int_{-3}^{3} x^3,dx = 0$ (odd function over symmetric limits) $\displaystyle \int_{-3}^{3} 5,dx = 5 \times 6 = 30$ So, $\displaystyle \int_{-3}^{3} (x^3+5),dx = 30$ ⇒ Assertion A is true. But $x^3+5$ is not an odd function (sum of odd and even function). So Reason R is false.

Qus : 74 Phrases PYQ

Arrange the parabolas in increasing order of length of their latus rectum.

(A) $y^2 = 8x$

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

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

Choose the correct answer:

(a) A, B, C, D

(b) B, C,A, D

(d) B, C, D, A

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Solution

Qus : 75 Phrases PYQ

If $|\begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{vmatrix} = k$, then the value of k is:

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Solution

Qus : 76 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:

The number of parallelograms in a chessboard is 1296.

Reason R:

The number of parallelograms when a set of $m$ parallel lines is intersected by another set of $n$ parallel lines is

$\displaystyle {m \choose 2}{n \choose 2}$.

In the light of the above statements, choose the correct answer from the options given below:

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Solution

A chessboard has 9 vertical and 9 horizontal parallel lines. Number of parallelograms $={9 \choose 2}{9 \choose 2} =36 \times 36 =1296$ So Assertion A is true. The given formula in Reason R is correct and is exactly used to find the result. So Reason R is true and correctly explains A.

Qus : 77 Phrases PYQ

A person goes in for an examination in which there are four papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is

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Solution

Let the marks in four papers be $x_1,x_2,x_3,x_4$ with

$0\le x_i\le m$ and

$x_1+x_2+x_3+x_4=2m$.

Number of non-negative solutions without restriction

$={}^{2m+3}C_3$.

Subtract cases where any $x_i>m$.

Using inclusion–exclusion, the required count simplifies to

Qus : 78 Phrases PYQ

The H.P. of two numbers is $4$ and the arithmetic mean $A$ and geometric mean $G$ satisfy $2A+G^2=27$. The numbers are

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Solution

If numbers are $a,b$:

H.P. $=\dfrac{2ab}{a+b}=4 \Rightarrow ab=2(a+b)$

$A=\dfrac{a+b}{2},\quad G^2=ab$

Given:

$2A+G^2=(a+b)+ab=27$

Substitute $ab=2(a+b)$:

$3(a+b)=27 \Rightarrow a+b=9$

$ab=18$

So numbers are roots of

$x^2-9x+18=0 \Rightarrow x=6,3$

Qus : 79 Phrases PYQ

If $f(x)=\begin{cases}x\sin(\frac{1}{x}), & x\ne0 \\ 0, & x=0\end{cases}$, then $f(x)$ is

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Solution

We have: \[ f(x) = \begin{cases} x \sin\!\left(\tfrac{1}{x}\right), & x \neq 0, \\[6pt] 0, & x = 0. \end{cases} \]

Step 1: Continuity at $x=0$

\[ \lim_{x \to 0} x \sin\!\left(\tfrac{1}{x}\right). \] Since $|\sin(1/x)| \leq 1$, \[ -|x| \;\leq\; x \sin\!\left(\tfrac{1}{x}\right) \;\leq\; |x|. \] By the squeeze theorem, \[ \lim_{x \to 0} f(x) = 0 = f(0). \] ✅ Thus, $f(x)$ is continuous everywhere.

Qus : 80 Phrases PYQ

If $A_1,A_2$ be two A.M.’s and $G_1,G_2$ be two G.M.’s between $a$ and $b$, then $\dfrac{A_1+A_2}{G_1G_2}$ is equal to

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Solution

For A.M.’s between $a$ and $b$: $A_1=\dfrac{2a+b}{3},\quad A_2=\dfrac{a+2b}{3}$ So, $A_1+A_2=\dfrac{2a+b+a+2b}{3}=a+b$ For G.M.’s between $a$ and $b$: $G_1=\sqrt[3]{a^2b},\quad G_2=\sqrt[3]{ab^2}$ So, $G_1G_2=\sqrt[3]{a^3b^3}=ab$ Hence, $\dfrac{A_1+A_2}{G_1G_2}=\dfrac{a+b}{ab}$

Qus : 81 Phrases PYQ

If the curve $ay+x^2=7$ and $x^3=y$ cut orthogonally at $(1,1)$, then the value of $a$ is

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Solution

From $ay+x^2=7$: $a\dfrac{dy}{dx}+2x=0$ $\Rightarrow \dfrac{dy}{dx}=-\dfrac{2x}{a}$ From $y=x^3$: $\dfrac{dy}{dx}=3x^2$ At $(1,1)$: $m_1=-\dfrac{2}{a},\quad m_2=3$ For orthogonal curves: $m_1m_2=-1$ $-\dfrac{2}{a}\times3=-1$ $\Rightarrow a=6$

Qus : 82 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:

An elevator starts with $m$ passengers and stops at $n$ floors $(m\le n)$.

The probability that no two passengers alight at the same floor is

$\displaystyle \frac{,{}^{n}P_m}{n^m}$.

Reason R:

If $(n+1)p$ is an integer, say $r$, then

$P(X=r)=,{}^{n}C_r p^r(1-p)^{n-r}$ is maximum when $r=np$ or $r=np-1$.

In the light of the above statements, choose the most appropriate answer:

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Solution

Total ways for $m$ passengers to choose floors $=n^m$

Favorable ways (all different floors) $={}^{n}P_m$

So Assertion A is true.

Reason R is a property of binomial distribution, which is true, but it has no relation to the elevator probability problem

Qus : 83 Phrases PYQ

A fair coin is tossed three times. Let A be the event of getting exactly two heads and B be the event of getting at most two tails, then P(AUB) is:

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Solution

Total outcomes when a fair coin is tossed 3 times:

\[ n(S) = 2^3 = 8 \]

Event $A$: exactly 2 heads = {HHT, HTH, THH}

\[ |A| = 3 \]

Event $B$: at most 2 tails = all outcomes except {TTT}

\[ |B| = 7 \]

Since every outcome of $A$ (two heads ⇒ one tail) is included in $B$, we have:

\[ A \subseteq B \;\;\Rightarrow\;\; A \cup B = B \]

Therefore:

\[ P(A \cup B) = P(B) = \frac{|B|}{8} = \frac{7}{8} \]

Final Answer:

\[ \boxed{\tfrac{7}{8}} \]

Qus : 84 Phrases PYQ

Out of 5 consonants and 4 vowels, how many words of 3 consonants and 3 vowels can be made?

1.40

2.80

3.20

4.240

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Solution

We have 5 consonants and 4 vowels. We need to form a word with 3 consonants and 3 vowels.

Step 1: Choose consonants

\[ \binom{5}{3} = 10 \]

Step 2: Choose vowels

\[ \binom{4}{3} = 4 \]

Step 3: Arrange the chosen 6 letters

\[ 6! = 720 \]

Step 4: Total words

\[ 10 \times 4 \times 720 = 28800 \]

Note: If the question means only "selections" of letters (not arrangements), then the answer is:

\[ \binom{5}{3}\times \binom{4}{3} = 10 \times 4 = 40 \]

Final Answer:

- If "word" = arrangement → $\; \boxed{28800}$
- If "word" = selection → $\; \boxed{40}$ (matches given Option 1)

Qus : 85 Phrases PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:

If two circles intersect at two points, then the line joining their centres is perpendicular to the common chord.

Reason R:

The perpendicular bisectors of two chords of a circle intersect at its centre.

In the light of the above statements, choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

Assertion A is a standard geometric property of intersecting circles ⇒ True.

Reason R is also true (property of chords in a circle), but it does not directly explain why the line joining centres is perpendicular to the common chord of two circles.

Qus : 86 Phrases PYQ

From the given sets, which is an infinite set:

1. $\{x:x\in N~and~(x-1)(x-2)=0\}$

2. $\{x: x \in N ~ and ~ x ~ is ~prime ~number~ and ~less ~than ~199\}$

3. $\{x:x\in N ~and~ x^{5}-1=0\}$

4. $\{x:x\in N~and~x~is~odd\}$

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Solution

Solution:

We check each option one by one: 1. $\{x \in \mathbb{N} : (x-1)(x-2)=0\} = \{1,2\}$, which is a finite set.

2. $\{x \in \mathbb{N} : x \text{ is prime and } x < 199\}$ contains only finitely many primes less than $199$, so it is finite.

3. $\{x \in \mathbb{N} : x^{5}-1=0\} \;\Rightarrow\; x^{5}=1 \;\Rightarrow\; x=1$. Thus the set is $\{1\}$, which is finite.

4. $\{x \in \mathbb{N} : x \text{ is odd}\} = \{1,3,5,7,\dots\}$, which is an infinite set.

Answer: Option (4).

Qus : 87 Phrases PYQ

If $\sin\beta$ is the G.M. between $\sin\alpha$ and $\cos\alpha$, then $\cos2\beta$ is equal to

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

Since $\sin\beta$ is the G.M. of $\sin\alpha$ and $\cos\alpha$, $\sin^2\beta=\sin\alpha\cos\alpha=\dfrac12\sin2\alpha$ So, $\cos2\beta=1-2\sin^2\beta=1-\sin2\alpha$ But $1-\sin2\alpha=2\sin^2\left(\dfrac{\pi}{4}-\alpha\right)$

Qus : 88 Phrases PYQ

If a chord which is normal to the parabola $y^2=4ax$ at one end subtends a right angle at the vertex, then its slope is

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Solution

For parabola $y^2=4ax$, slope of normal at parameter $t$ is $-t$.

Let slope of chord be $m=-t$.

Since the chord subtends a right angle at the vertex,

$m^2=2$

So,

$m=\sqrt2$

Qus : 89 Phrases PYQ

If $\hat n_1,\hat n_2$ are two unit vectors and $\theta$ is the angle between them, then $\cos\dfrac{\theta}{2}$ is equal to

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Solution

$|\hat n_1+\hat n_2|^2=2(1+\cos\theta)$

So,

$|\hat n_1+\hat n_2|=2\cos\dfrac{\theta}{2}$

Hence,

$\cos\dfrac{\theta}{2}=\dfrac12|\hat n_1+\hat n_2|$

Qus : 90 Phrases PYQ

If a, b and c are in Geometric Progression and $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$ then, x, y, z are in

1. Arithmetic Progression

2. Geometric Progression

3. $\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

4. $x=y+z$

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Solution

We are given:

$a^{\tfrac{1}{x}} = b^{\tfrac{1}{y}} = c^{\tfrac{1}{z}} = k$

$\Rightarrow a = k^x,\; b = k^y,\; c = k^z$

Since $a, b, c$ are in G.P.:

$b^2 = ac$

$\Rightarrow (k^y)^2 = (k^x)(k^z)$

$\Rightarrow k^{2y} = k^{x+z}$

$\Rightarrow 2y = x+z$

This implies $y$ is the arithmetic mean of $x$ and $z$.

x, y, z are in Arithmetic Progression.

Correct Answer: (1) Arithmetic Progression

Qus : 91 Phrases PYQ

If each observation of raw data whose variance is $\sigma^2$ is multiplied by $h$, then the variance of the new set is

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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

If all observations are multiplied by a constant $h$, then variance becomes $h^2$ times the original variance. So, new variance $=h^2\sigma^2$.

Qus : 92 Phrases PYQ

Which of the following functions is differentiable at $x=0$?

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Solution

$|x|$ is not differentiable at $0$.

Near $x=0$, $\sin(|x|)\approx |x|$.

For option (4):

$f(x)=\sin(|x|)-|x|$

Near $0$:

$f(x)\approx |x|-|x|=0$

Both left-hand and right-hand derivatives at $0$ are equal, hence $f(x)$ is differentiable at $0$.

All other options involve $|x|$ in a way that makes the derivative discontinuous at $0$.

Qus : 93 Phrases PYQ

The value of $\lim_{x\rightarrow\infty}(1+\frac{2}{3x})^{x}$ is:

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Solution

We use the standard result: \[ \lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k \] Here, $k = \tfrac{2}{3}$ and $n = x$. \[ \therefore \;\; \lim_{x \to \infty} \left(1 + \frac{2}{3x}\right)^x = e^{\tfrac{2}{3}} \] Final Answer: $ e^{\tfrac{2}{3}} $

Qus : 94 Phrases PYQ

There are 200 students in a school out which 120 students play football, 50 students play cricket and 30 students play both football and cricket. The number of students who play one game only is:

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Solution

Let total students be $ n(U) = 200 $. Football players: $ n(F) = 120 $ Cricket players: $ n(C) = 50 $ Both: $ n(F \cap C) = 30 $

Students who play one game only: \[ n(F \setminus C) + n(C \setminus F) = (n(F) - n(F \cap C)) + (n(C) - n(F \cap C)) \] \[ = (120 - 30) + (50 - 30) = 90 + 20 = 110 \]

$\therefore$ The number of students who play one game only = 110.

Qus : 95 Phrases PYQ

Which of the following are true:

(A) Ogive graph is used to measure the median of the collection of datas.

(B) Two events A and B are such that P(A) = 1/2 and P(B) = 7/12 and P(not A not B) = 1/4 then A and B are independent events.

(D) The number of two–digits even number formed from digits 1,2,3,4,5 is 10

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Solution:

(A) Ogive is used to determine the Median. ✅ True

(B) Given:
$ P(A) = \tfrac{1}{2}, \; P(B) = \tfrac{7}{12}, \; P(\text{not A and not B}) = \tfrac{1}{4} $

$ P(A \cup B) = 1 - \tfrac{1}{4} = \tfrac{3}{4} $
$ P(A \cap B) = P(A) + P(B) - P(A \cup B) $ $ = \tfrac{1}{2} + \tfrac{7}{12} - \tfrac{3}{4} = \tfrac{1}{3} $
$ P(A)\cdot P(B) = \tfrac{1}{2} \times \tfrac{7}{12} = \tfrac{7}{24} \neq \tfrac{1}{3} $

So, A and B are not independent. ❌ False

(C) Empirical relation: \[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \] ✅ True

(D) Two–digit even numbers from {1,2,3,4,5}:
- Units digit must be even → {2, 4} → 2 choices.
- Tens digit (if repetition allowed) → 5 choices.
\[ \text{Total} = 5 \times 2 = 10 \] ✅ True (with repetition allowed)

✔ Final Answer: (A), (C), and (D) are correct.

Qus : 96 Phrases PYQ

There are 15 points in a plane such that 5 points are collinear and no three of the remaining points are collinear then total number of straight lines formed are:

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Solution

Solution:

Number of straight lines from $n$ points (no three collinear) is $\binom{n}{2}$.

Here, $n = 15$.

\[ \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

Adjustment for collinearity:
Out of 15 points, 5 are collinear.
Lines from these 5 points = $\binom{5}{2} = 10$.
But actually they form only 1 line.
Extra counted = $10 - 1 = 9$.

Correct total lines:

\[ 105 - 9 = 96 \]

Final Answer: The total number of straight lines formed = 96

Qus : 97 Phrases PYQ

Match List I with List II

List - I (Function)	List - II (Range)
A. $$y=\frac{1}{2-\sin 3x}$$	I. $$\Bigg{(}1,\frac{7}{3}\Bigg{]}$$
B. $$y=\frac{{x}^2+x+2}{{x}^2+x+1},\, x\in R$$	II. $$\Bigg{[}\frac{\pi}{2},\pi\Bigg{)}\cup(\pi,\frac{3\pi}{2}\Bigg{]}$$
C. $$y=\sin x-\cos x$$	III. $$\Bigg{[}\frac{1}{3},1\Bigg{]}$$
D. $$y={\cot }^{-1}(-x)-{\tan }^{-1}x+{sec}^{-1}x$$	IV. $$[-\sqrt[]{2},\sqrt[]{2}]$$

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Step 1: For (A)

$ y = \dfrac{1}{2 - \sin 3x} $
Since $ \sin 3x \in [-1,1] $, we get $ 2 - \sin 3x \in [1,3] $.
Hence $ y \in \left[\tfrac{1}{3}, 1\right] $. → Matches with (III).

Step 2: For (B)

$ y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1} $
Since denominator is always positive, $ y > 1 $.
Minimum denominator = $\tfrac{3}{4}$ at $x = -\tfrac{1}{2}$.
So maximum $ y = 1 + \tfrac{1}{3/4} = \tfrac{7}{3} $.
Thus, Range = $(1, \tfrac{7}{3}] $. → Matches with (I).

Step 3: For (C)

$ y = \sin x - \cos x = \sqrt{2}\sin\!\left(x - \tfrac{\pi}{4}\right) $
Hence, Range = $[-\sqrt{2}, \sqrt{2}] $. → Matches with (IV).

Step 4: For (D)

$ y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) $
Simplifying with inverse trig identities gives Range:
$\left[\tfrac{\pi}{2}, \pi\right) \cup \left(\pi, \tfrac{3\pi}{2}\right]$. → Matches with (II).

Qus : 98 Phrases PYQ

An equilateral triangle is inscribed in a parabola $y^2=8x$ whose one vertix is at the vertex of the parabola then the length of the side of the triangle is:

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Solution

Shortcut (Formula):

For a parabola $y^2=4ax$, an equilateral triangle inscribed with one vertex at the parabola’s vertex has side

\[ s = 8a\sqrt{3}. \]

Here $y^2=8x \Rightarrow 4a=8 \Rightarrow a=2$. Hence

\[ s = 8\cdot 2\sqrt{3}=16\sqrt{3}. \]

Final Answer: $16\sqrt{3}$

Qus : 99 Phrases PYQ

The center and radius for the circle $x^2 + y^2 +6x-4y +4 = 0$ respectively are:

1. (2, 3) and 3

2. (3, 2) and 8

3. (2, -3) and 3

4. (-3, 2) and 3

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Solution

Equation: $$x^2 + y^2 + 6x - 4y + 4 = 0$$

Step 1: Group terms.
$$(x^2 + 6x) + (y^2 - 4y) + 4 = 0$$

Step 2: Complete the square.
- For $x^2 + 6x$: add and subtract $(\tfrac{6}{2})^2 = 9$
- For $y^2 - 4y$: add and subtract $(\tfrac{-4}{2})^2 = 4$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) + 4 - 9 - 4 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 - 9 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 = 9$$

Center: (-3, 2)

Radius: 3

Answer: Option 4

Qus : 100 Phrases PYQ

If $x_1, x_2, x_3$ as well as $y_1, y_2, y_3$ are in G.P. with the same common ratio, then the points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$

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Solution

Let the common ratio be $r$.

\[ x_1 = a, \; x_2 = ar, \; x_3 = ar^2 \] \[ y_1 = b, \; y_2 = br, \; y_3 = br^2 \]

So the points are $(a,b), \; (ar,br), \; (ar^2,br^2)$.

Slopes:

Between first two points: \[ m_{12} = \frac{br - b}{ar - a} = \frac{b(r-1)}{a(r-1)} = \frac{b}{a} \] Between second and third points: \[ m_{23} = \frac{br^2 - br}{ar^2 - ar} = \frac{br(r-1)}{ar(r-1)} = \frac{b}{a} \]

Since $m_{12} = m_{23}$, the points are collinear.

Final Answer: The points $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ are collinear.

Qus : 101 Phrases PYQ

If (x -1) is a factor of $2x^2-5x +k = 0$, then the value of k is:

1. 2

2. 5

3. 3

4. 4

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Solution

Given: Polynomial $$f(x) = 2x^2 - 5x + k$$ and $(x-1)$ is a factor.

Step 1: Apply factor theorem.
If $(x-1)$ is a factor, then $f(1) = 0$.

Step 2: Substitute $x=1$.
$$f(1) = 2(1)^2 - 5(1) + k = 2 - 5 + k = -3 + k$$

Step 3: Solve for $k$.
$$-3 + k = 0 \quad \Rightarrow \quad k = 3$$

Value of k: 3

Answer: Option 3

Qus : 102 Phrases PYQ

Match List – I with List – II

List - I	List - II
(A) Eccentricity of the conic $x^2-4x+4y+4y^2=12$	(I) 10/3
(B) Latus rectum of conic $5x^2+9y^2=45$	(II) 1
(C) The straight line x+y=a touches the curve $y=x-x^2$ then value of a	(III) 2
(D) Eccentricity of conic $3x^2-y^2=4$	(IV) $\sqrt{3}/2$

Choose the correct answer from the options given below:

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Solution

Final Matching (List–I → List–II)

Item	Result	Match
(A) Eccentricity of $x^2-4x+4y+4y^2=12$	$\displaystyle \frac{\sqrt{3}}{2}$	(IV)
(B) Latus rectum of $5x^2+9y^2=45$	$\displaystyle \frac{10}{3}$	(I)
(C) Line $x+y=a$ touches $y=x-x^2$ ⇒ value of $a$	$1$	(II)
(D) Eccentricity of $3x^2-y^2=4$	$2$	(III)

Answer: (A) → (IV), (B) → (I), (C) → (II), (D) → (III)

Solutions (with steps)

(A) $x^2-4x+4y+4y^2=12$ \[ (x-2)^2-4+4\!\left[(y+\tfrac12)^2-\tfrac14\right]=12 \;\Rightarrow\; (x-2)^2+4(y+\tfrac12)^2=17 \] \[ \frac{(x-2)^2}{17}+\frac{(y+\tfrac12)^2}{17/4}=1 \] Ellipse with $a^2=17,\; b^2=\tfrac{17}{4}\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\tfrac14}=\frac{\sqrt{3}}{2}. $

(B) $5x^2+9y^2=45 \Rightarrow \frac{x^2}{9}+\frac{y^2}{5}=1$. Here $a=\sqrt{9}=3,\; b=\sqrt{5}$. Latus rectum length (ellipse) $= \dfrac{2b^2}{a}=\dfrac{2\cdot5}{3}=\dfrac{10}{3}.$

(C) Tangent: $x+y=a \Rightarrow y=-x+a$. Touches $y=x-x^2$: \[ x-x^2=-x+a \Rightarrow x^2-2x+a=0 \] For tangency, discriminant $=0$: $4-4a=0 \Rightarrow a=1.$

(D) $3x^2-y^2=4 \Rightarrow \dfrac{x^2}{4/3}-\dfrac{y^2}{4}=1$. Hyperbola with $a^2=\tfrac{4}{3},\, b^2=4$. \[ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{4/3}}=\sqrt{1+3}=2. \]

Qus : 103 Phrases PYQ

Match List-I with List-II

List - I	List - II
(A) If X and Y are two sets such that $n(X)=17$, $n(Y)=23$, $n(X \cup Y)=38$, then $n(X \cap Y)$ is	I. 20
(B) If $n(X)=28$, $n(Y)=32$, $n(X \cap Y)=10$, then $n(X \cup Y)$ is	II. 10
(C) If $n(X)=10$, then $n(7X)$ is	III. 50
(D) If $n(Y)=20$, then $n\!\left(\tfrac{Y}{2}\right)$ is	IV. 2

Choose the correct answer from the options given below:

1. (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

2. (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

3. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

4. (A) - (IV), (B) - (II), (C) - (I), (D) - (III)

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Solution

(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.

(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.

(C) If n(X) = 10, then n(?(X)) (power set) = 2¹⁰ = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 2¹⁰ = 1024. ? But given options map (C) with II = 10, so they mean **n(?(X)) = 2^n(X)** was NOT intended. They likely meant n(X) itself. So (C) → II.

(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.

Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: Option 1

Qus : 104 Phrases PYQ

The area of the region bounded by the curve $y^2=4x$ and $x^2=4y$ is

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Solution

Curves: $y^2=4x$ (right-opening) and $x^2=4y$ (upward). Intersection: From $x=\dfrac{y^2}{4}$ in $x^2=4y$ ⇒ $\dfrac{y^4}{16}=4y \Rightarrow y(y^3-64)=0\Rightarrow y=0,4$. Thus points are $(0,0)$ and $(4,4)$ in the first quadrant.

For $0\le y\le 4$: right curve is $x=2\sqrt{y}$ (from $x^2=4y$), left curve is $x=\dfrac{y^2}{4}$ (from $y^2=4x$).

Area $=\displaystyle \int_{0}^{4}\!\left(2\sqrt{y}-\frac{y^2}{4}\right)\,dy = \left[\frac{4}{3}y^{3/2}-\frac{y^3}{12}\right]_{0}^{4} = \frac{32}{3}-\frac{16}{3}=\frac{16}{3}.$

Final Answer: $\displaystyle \frac{16}{3}$ square units.

Qus : 105 Phrases PYQ

The value of x satisfies the inequality $|x-1|+|x-2|\geq4$ if

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Solution

Consider regions for $x$ around 1 and 2.

1) $x\le 1:\quad |x-1|+|x-2|=(1-x)+(2-x)=3-2x \ge 4 \Rightarrow x\le -\tfrac12.$

2) $1\le x\le 2:\quad |x-1|+|x-2|=(x-1)+(2-x)=1$ (not $\ge4$). No solutions.

3) $x\ge 2:\quad |x-1|+|x-2|=(x-1)+(x-2)=2x-3 \ge 4 \Rightarrow x\ge \tfrac{7}{2}.$

Final Answer: $x \in (-\infty,\,-\tfrac12] \,\cup\, [\tfrac{7}{2},\,\infty)$.

Correct Option: 1

Qus : 106 Phrases PYQ

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

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Solution

Qus : 107 Phrases PYQ

If the parametric equation of a curve is given by $x=e^t cost$ and $y=e^t sint$ then the tangent to the curve at the point $t=\frac{\pi}{4}$ makes the angle with the axis of x is

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Solution

Given parametric equations:

$$x = e^t \cos t,\quad y = e^t \sin t$$ To find the angle of the tangent at $ t = \frac{\pi}{4} $, compute the slope:
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$$ At $ t = \frac{\pi}{4} $,
$$\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ So, $$\frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0}$$ The slope is undefined, which means the tangent is vertical.

Final Answer: The angle with the x-axis is $$\boxed{90^\circ}$$

Qus : 108 Phrases PYQ

Bag A contains 3 Red and 4 Black balls while Bag B contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and is found to be red. Then, the probability that it was drawn from Bag B is

1. $\frac{35}{68}$

2. $\frac{7}{38}$

3. $\frac{14}{37}$

4. $\frac{34}{43}$

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Solution

Step 1: Define events.
Let: • A = ball drawn from Bag A • B = ball drawn from Bag B • R = ball drawn is Red

Since a bag is chosen at random: $$P(A) = P(B) = \tfrac{1}{2}$$

Step 2: Probability of Red from each bag.
• From Bag A: $$P(R|A) = \tfrac{3}{3+4} = \tfrac{3}{7}$$ • From Bag B: $$P(R|B) = \tfrac{5}{5+6} = \tfrac{5}{11}$$

Step 3: Total probability of Red.
$$P(R) = P(A)P(R|A) + P(B)P(R|B)$$ $$= \tfrac{1}{2}\cdot\tfrac{3}{7} + \tfrac{1}{2}\cdot\tfrac{5}{11}$$ $$= \tfrac{3}{14} + \tfrac{5}{22}$$ $$= \tfrac{33}{154} + \tfrac{35}{154} = \tfrac{68}{154} = \tfrac{34}{77}$$

Step 4: Apply Bayes’ Theorem.
$$P(B|R) = \frac{P(B)P(R|B)}{P(R)}$$ $$= \frac{\tfrac{1}{2}\cdot\tfrac{5}{11}}{\tfrac{34}{77}}$$ $$= \frac{5}{22} \cdot \frac{77}{34} = \frac{385}{748}$$ $$= \tfrac{35}{68}$$

Correct Probability: $\tfrac{35}{68}$

Answer: Option 1

Qus : 109 Phrases PYQ

If the line through (3, y) and (2,7) is parallel to the line through (-1,4) and (0, 6), then the value of y is:

1.-7

2.9

3.7

4.2

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Solution

Step 1: Slope of line through points (-1, 4) and (0, 6).
$$m = \frac{6 - 4}{0 - (-1)} = \frac{2}{1} = 2$$

Step 2: For parallelism, slope of line through (3, y) and (2, 7) must also be 2.
$$\frac{y - 7}{3 - 2} = 2$$

Step 3: Solve for y.
$$y - 7 = 2(1)$$ $$y - 7 = 2$$ $$y = 9$$

Correct Value of y: 9

Answer: Option 2

Qus : 110 Phrases PYQ

If $f(a+b-x)=f(x)$ then $\int ^b_axf(x)dx$ is equal to

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Solution

We are asked to evaluate

$\displaystyle I = \int_a^b x f(x)\, dx \quad \text{given } f(a+b-x)=f(x).$

Step 1: Put substitution $t=a+b-x$. Then $dx=-dt$.

When $x=a \Rightarrow t=b$, when $x=b \Rightarrow t=a$.

So, \[ I = \int_a^b x f(x)\, dx = \int_b^a (a+b-t) f(t)(-dt) = \int_a^b (a+b-t) f(t)\, dt. \]

Step 2: Add both forms of $I$:

\[ 2I = \int_a^b [x f(x) + (a+b-x) f(x)] dx = \int_a^b (a+b) f(x)\, dx. \]

Step 3: Simplify:

\[ I = \frac{a+b}{2} \int_a^b f(x)\, dx. \]

Final Answer: $\displaystyle \frac{a+b}{2}\int_a^b f(x)\, dx$ → matches Option 1.

Qus : 111 Phrases PYQ

If ${x}^2+\frac{1}{{x}^2}=2$ then the value of ${x}^{256}+\frac{1}{{x}^{256}}$

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Solution

Given $x^2+\dfrac{1}{x^2}=2$. Let $t=x+\dfrac{1}{x}$. Then $x^2+\dfrac{1}{x^2}=t^2-2$, so $t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2$.

From $x+\dfrac{1}{x}=2 \Rightarrow x=1$ and from $x+\dfrac{1}{x}=-2 \Rightarrow x=-1$.

Hence $x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}$.

Final Answer: $2$.

Qus : 112 Phrases PYQ

Matching

Column A	Column B
A. $ \frac{d}{dx}(\cos^{-1}x) $	I. $ -\frac{1}{1+x^2} $
B. $ \frac{d}{dx}(\cot^{-1}x) $	II. $ \frac{1}{\|x\|\sqrt{x^2-1}} $
C. $ \frac{d}{dx}(\cos ec^{-1}x) $	III. $ -\frac{1}{\sqrt{1-x^2}} $
D. $ \frac{d}{dx}(\sec^{-1}x) $	IV. $ -\frac{1}{\|x\|\sqrt{x^2-1}}$

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Solution

Qus : 113 Phrases PYQ

Consider the system of linear equations as 2x + 2y + z = 1, 4x + ky + 2z = 2 and kx + 4y + z = 1 then choosethe correct statement(s) from blow

(A) The system of equation has a unique solution if k≠4 and k≠2

(B) The system of equations is inconsistent for every real number k

(D) The system of equations have infinite number of solutions if k = 2

Choose the correct answer from the options given below

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Solution

The system of equations is:

2x + 2y + z = 1
4x + ky + 2z = 2
kx + 4y + z = 1

The coefficient matrix is

$A = \begin{bmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{bmatrix}$.

The determinant is

$\Delta = \begin{vmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{vmatrix}$.

Expanding:

$\Delta = 2\begin{vmatrix} k & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 4 & 2 \\ k & 1 \end{vmatrix} + 1\begin{vmatrix} 4 & k \\ k & 4 \end{vmatrix}$.

$\Delta = 2(k-8) - 2(4-2k) + (16-k^2)$.

$\Delta = -k^2 + 6k - 8 = -(k-2)(k-4)$.

If $k \neq 2,4$, then $\Delta \neq 0$ and the system has a unique solution.
If $k=4$: equations (1) and (2) are dependent, equation (3) reduces to the same relation, hence infinitely many solutions.
If $k=2$: substituting gives $y=0$ and $2x+z=1$, equation (3) is the same, hence infinitely many solutions.

Correct Statements: (A), (C), (D)

Qus : 114 Phrases PYQ

Evaluate $ \frac{2x+1}{x^2+x+2}\, dx$

1. $log(2x+1) + c$ where c is an arbitrary constant

2. $log\!\left(\frac{2x+1} {x^2+x+2}\right) + c$ where c is an arbitrary constant

3. $log(x^2+x+2) + c$ where c is an arbitrary constant

4. $log\!\left(\tfrac{1}{2}\right) + c$ where c is an arbitrary constant

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Solution

Step 1: Let the denominator be $$f(x) = x^2 + x + 2.$$ Then, $$f'(x) = 2x + 1,$$ which is the numerator.

Step 2: Apply the rule: $$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C.$$

Step 3: Therefore, $$\int \frac{2x+1}{x^2+x+2}\, dx = \ln|x^2+x+2| + C.$$

Final Answer: $\log(x^2+x+2) + c$

Correct Option: 3

Qus : 115 Phrases PYQ

Find the area of the region $ {(x,y): x^2 \le y \le x } $.

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Solution

Qus : 116 Phrases PYQ

Arrange the following in decreasing order (based on determinant value).
A. $ \begin{vmatrix} 1 & 3 & 5\\ 2 & 6 & 10\\ 31 & 11 & 38 \end{vmatrix} $

B.$ \begin{vmatrix} 67 & 19 & 21\\ 39 & 13 & 14\\ 81 & 24 & 26 \end{vmatrix} $

C. $ \begin{vmatrix} 1 & -3 & 2\\ 4 & -1 & 2\\ 3 & 5 & 2 \end{vmatrix} $

D. $ \begin{vmatrix} 1 & 4 & 9\\ 4 & 9 & 16\\ 9 & 16 & 25 \end{vmatrix} $

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Solution

Qus : 117 Phrases PYQ

Which of the following option is false about circles whose equations
$ x^2 + y^2 - 10x - 10y + 41 = 0 $
and
$ x^2 + y^2 - 22x - 10y + 137 = 0 $

(A) circles have same center

(B) circles have no meeting time

(D) circles have only two meeting time

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Solution

Qus : 118 Phrases PYQ

In a university there are total 100 students. 15 offered mathematics only, 12 offered statistics only, 8 offered physics only, 40 offered physics and mathematics, 20 offered physics and statistics, 10 offered mathematics and statistics, 65 offered physics. Tell the number of students not offered any of three.

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Solution

Qus : 119 Phrases PYQ

The function $f(x)=[x]^n$ , integer n>=2 (where [y] is the greatest integer less than or equal to y), is discontinuous at all point of

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Solution

The function is

$ f(x) = [x]^n , \quad n \geq 2 $

where $[x]$ is the greatest integer function (GIF).

The GIF $[x]$ is discontinuous at all integers. Raising it to the integer power $n \geq 2$ does not remove this discontinuity, because the jump still exists at each integer value of $x$.

For non-integer $x$, the function is constant over intervals $(m, m+1)$ where $m \in \mathbb{Z}$, so it is continuous within each open interval between integers.

Final Answer: The function is discontinuous at all integers.

Qus : 120 Phrases PYQ

Column A	Statement	Column B	Expression
A	Neither A nor B	I	$(A \cap \overline{B}) \cup (\overline{A} \cap B)$
B	At least one of A, B or C	II	$\overline{A} \cap \overline{B}$
C	Exactly one of A and B	III	$A \cup B \cup C$
D	All three A, B, C	IV	$A \cap B \cap C$

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Solution

Qus : 121 Phrases PYQ

Mean and variance of 12 observations are $ \frac{9}{2} $ and $4$ respectively. Later it was observed that two observations were considered $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $ \frac{m}{n} $ where $m$ and $n$ are coprime, then $m+n$ is

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Solution

Qus : 122 Phrases PYQ

If the roots of the equation $x^2+4x+a^2-3a$ are real then the value of a (is / are)

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Solution

Qus : 123 Phrases PYQ

Which of the following is false about $ \lim_{y \to 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} $

(A) limit exists and equal to $ \frac{1}{4\sqrt{2}} $

(B) limit does not exist

(D) limit exists and equal to $ \frac{1}{2\sqrt{2}(\sqrt{2}+1)} $

(a) A, B, C

(b) B, C, D

(d) A, C, D

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Solution

Qus : 124 Phrases PYQ

Match List – I with List – II

List - I	List - II
(A) $$\int ^{\pi/2}_0\frac{{\sin }^4x}{{\sin }^4x+{\cos }^4x}dx$$	(I) 0
(B) $$\int ^{\pi/3}_{\pi/6}\frac{1}{1+\sqrt[]{\tan x}}dx$$	(II) 0
(C) $$\int ^1_0x{e}^xdx$$	(III) $\frac{\pi}{12}$
(D) $$\int ^1_{-1}{x}^{109}{\cos }^{88}xdx$$	(IV) $\frac{\pi}{4}$

Choose the correct answer from the options given below:

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Solution

Qus : 125 Phrases PYQ

Arrange the following in sequence to find variance of individual observation.

(A) compute $ \bar{X} $ of observation $x_1, x_2 \dots x_n$

(B) square the deviation obtain sum $ \sum_{i=1}^{n} (x_i-\bar{X})^2 $

(D) take deviation of observation from mean that is $(x_i-\bar{X})$

(a) A, B, C, D

(b) A, D, B, C

(d) B, C, D, A

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Solution

Qus : 126 Phrases PYQ

Matching

Column A	Column B
A. $(\sqrt{2}+1)+1+(\sqrt{2}-1)+\ldots \infty$	IV. $\frac{4+3\sqrt{2}}{2}$
B. $ \frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\ldots \infty$	I. $\frac{19}{24}$
C. $6^{1/2}\times 6^{1/4}\times 6^{1/8}\ldots \infty$	II. $6$
D. $8+4\sqrt{2}+4+\ldots \infty$	III. $8(2+\sqrt{2})$

(a) (A)-I, (B)-II, (C)-III, (D)-IV

(b) (A)-IV, (B)-I, (C)-II, (D)-III

(d) (A)-I, (B)-IV, (C)-III, (D)-II

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Solution

Qus : 127 Phrases PYQ

If $S, T$ are symmetric matrices of the same order, then which of the following statements are true?

(A) $S + T$ is symmetric matrix

(B) $ST - TS$ is skew symmetric matrix

(D) $ST - TS$ is symmetric matrix

(a) A, B, C

(b) B, C, D

(d) A, C, D

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Solution

Qus : 128 Phrases PYQ

Which of the following statement sare TRUE?

(A) A equation $ax^2+bx+c=0$ has real and distinct roots if $b^2-4ac>=0$ and $a\ne0$ .

(B) The unit digit in $49^{18}$ is 1.

(D) The line bx – ay = 0 meet the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Choose the correct answer from the options given below:

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Solution

Qus : 129 Phrases PYQ

The line passes through a point (2, 3) such that sum of its intercepts on the axes is 12 then equation of line/s is/are given by

(A) 3x+y=9

(B) x+3y=9

(D) 5x+7y=35

Choose the correct answer from the options given below:

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Solution

Qus : 130 Phrases PYQ

$$\lim _{{x}\rightarrow0}\frac{\sqrt[]{1-\cos 2x}}{x}$$

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Solution

Qus : 131 Phrases PYQ

The curves $x = y^2$ and $xy = k$ cut at right angle. Find the value of $k$.

(a) $ \frac{1}{2\sqrt{2}} $

(b) $2$

(d) $ \frac{1}{8} $

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Solution

Qus : 132 Phrases PYQ

The foci of hyperbola coincide with foci of ellipse $ \frac{x^2}{25} + \frac{y^2}{9} = 1 $. So the equation of hyperbola if $e = 2$ is

(a) $ \frac{x^2}{4} - \frac{y^2}{12} = 1 $

(b) $ \frac{x^2}{16} - \frac{y^2}{12} = 1 $

(d) $ \frac{x^2}{2} - \frac{y^2}{9} = 1 $

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Solution

Qus : 133 Phrases PYQ

If permutaiton of the letters of the word ‘AGAIN’ are arranged in the order as in a dictionary then 49th word i

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Solution

Qus : 134 Phrases PYQ

The mean of 5 data is 5.2 and their variance is 27.296. If there of the data are 1, 3 and 6 then other two data are

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Solution

Qus : 135 Phrases PYQ

If the vertices of a triangle are (1, 2), (2, 5) and (4, 3) then the area of the triangle is:

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Solution

Qus : 136 Phrases PYQ

Which of the following statements are TRUE?

(A) If each element in a row is a constant multiplier of corresponding element of another row of a determinant, then the value of the determinant is always non-zero.

(B) If each element on one side of the principal diagonal of a determinant is zero, then the value of the determinants the product of the diagonal elements.

(D) If A is non-singular matrix of order three, then $adj A=|A|^2$

Choose the correct answer from the options given below:

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Solution

Qus : 137 Phrases PYQ

A function f(x) is defined as $$f(x)=\begin{cases}{\frac{1-\cos 4x}{{x}^2}} & {;x{\lt}0} \\ {a} & {;x=0} \\ {\frac{\sqrt[]{x}}{\sqrt[]{(16+\sqrt[]{x})-4}}} & {;x{\gt}0}\end{cases}$$ if the function f(x) is continuous at x = 0, then the value of a is:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 138 Phrases PYQ

Let $f$ be function $f(x)= \begin{cases} \frac{\tan x}{x}, & x \ne 0 \\ 1, & x=0 \end{cases}$

Assertion: $x=0$ is point of minima of $f$

Reason: $f'(0)=0$

(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.

(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.

(d) Assertion is false, but Reason is true.

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Phrases Previous Year PYQ Phrases CUET MCA 2026 PYQ

Solution

Qus : 139 Phrases PYQ

The equation of a circle that passes through the points (3, 0) and (0, –2) and its lies on a line 2x + 3y = 3 then equation of the cicle is given by:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 140 Phrases PYQ

Consider the diagram given below and the following two statements:

Statement I: Regions X, Y and Z can be expressed as $A\cap\overline{B},\, A\cap B$ and $\, \overline{A}\cap B$ respectively

Statement II: P(Y) = P (A) - P (X) = P (B) - P (Z)

In the light of the above statements, choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 141 Phrases PYQ

Let $\alpha >2$ is an integer. If there are only 10 positive integers satisfying the inequality $(x-\alpha)(x-2\alpha)(x-\alpha^2)<0$ then the value/s of $\alpha$ is

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 142 Phrases PYQ

Assertion: 1/3, −1/2, 3/4, −9/8 form an AP

Reason: The constant sequence is the only sequence which is both AP as well as GP.

Options:

(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.

(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.

(d) Assertion is false, but Reason is true.

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Phrases Previous Year PYQ Phrases CUET MCA 2026 PYQ

Solution

Qus : 143 Phrases PYQ

$\int \frac{({x}^5-x{)}^{1/5}}{{x}^6}dx=$
(where C is an arbitrary constant)

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 144 Phrases PYQ

In a class there are 400 students, the following table shows the number of students studying one or more of the subjects:

Subject	Number of Students
Mathematics	250
Physics	150
Chemistry	100
Mathematics and Physics	100
Mathematics and Chemistry	60
Physics and Chemistry	40
Mathematics, Physics and chemistry	30

A. The number of students who study only Mathematics is 100.

B. The number of students who study only Physics is 40.

C. The number of students who study only Chemistry is 40.

D. The number of students who do not study Mathematics, Physics and Chemistry is 70.

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 145 Phrases PYQ

The arithmetic means of two observations is 125 and their geometric mean is 60. Find the harmonic mean of the two observations.

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 146 Phrases PYQ

The arithmetic mean and standard deviation of series of 20 items were calculated by a student as 20 cm and 5 cm respectively. But while calculating them an item 15 was misread as 30. Find the correct standard deviation.

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 147 Phrases PYQ

Given the marks of 25 students in the class as $\{m_1,m_2,m_3,..m_{25}\}$. Marks lie in the range of [1-100] and $\overline{m}$ is the mean. Which of the following quantity has the value zero?

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 148 Phrases PYQ

Assertion: A(BA) and (AB)A are symmetric matrices.

Reason: AB is symmetric matrix, if matrix multiplication of A with B is commutation.

Options:

(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.

(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.

(d) Assertion is false, but Reason is true.

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Phrases Previous Year PYQ Phrases CUET MCA 2026 PYQ

Solution

A(BA) = (AB)A = ABA

If A and B are symmetric matrices, then

(ABA)^T = A^TB^TA^T

Since A and B are symmetric,

A^T = A and B^T = B

Therefore

(ABA)^T = ABA

Hence A(BA) and (AB)A are symmetric matrices.

Also, AB is symmetric if AB = BA (i.e., A and B commute).

Correct Answer: (a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.

Qus : 149 Phrases PYQ

A equation of conic is $ax^2+2hxy+by^2+2gx+2fy+c=0$ , where $a, b, c, f, g$ and $h$ are constants. Then which of the following statement are true?

(A) The given conic is circle if a = 0 and b = 0.

(B) The given conic is circle if $a=b\ne0$ and h = 0.

(D) The given conic represents a pair of real and distinct straight lines if f = g = c = 0 and $h^2-ab>0$.

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 150 Phrases PYQ

Match List – I with List – II

List - I	List - II $f(0)$
(A) $f(x)=\frac{log(1+4x)}{x}$	(I) $\frac{1}{4}$
(B) $f(x)=\frac{log(4+x)-log4}{x}$	(II) 1
(C) $f(x)=\frac{x}{sinx}$	(III) 4
(D) $\frac{1-cos^3x}{x sin2x}$	(IV) $\frac{3}{4}$

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 151 Phrases PYQ

The terms $1,\, \log _y(x),\, \log _z(y)\, and\, 15\log _x(z)$ are in AP.

Based on this information answer the following questions.

The Common difference of AP is

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 152 Phrases PYQ

A matrix is both symmetric and skew symmetric matrix

(a) A is diagonal matrix

(b) A is zero matrix

(d) A is square matrix

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Phrases Previous Year PYQ Phrases CUET MCA 2026 PYQ

Solution

If a matrix A is symmetric, then
A^T = A

If a matrix A is skew symmetric, then
A^T = −A

For both conditions to hold:
A = −A

⇒ 2A = 0
⇒ A = 0

Therefore, the matrix must be a zero matrix.

Correct Answer: (b) A is zero matrix

Qus : 153 Phrases PYQ

The terms $1,\, \log _y(x),\, \log _z(y)\, and\, 15\log _x(z)$ are in AP.

Based on this information answer the following questions.

The value of xy is:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 154 Phrases PYQ

The terms $1,\, \log _y(x),\, \log _z(y)\, and\, 15\log _x(z)$ are in AP.

Based on this information answer the following questions.

yz is equal to :

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 155 Phrases PYQ

Which of the following statements are NOT TRUE?

(A) If A and B are symmetric matrices, then AB – BA is a skew symmetric matrix.

(B) Multiplying a determinant by k means multiply elements of one column by k.

(D) If A and B are invertible matrices of same order, then $(A+B)^{-1}=B^{-1}+A^{-1}$.

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2024 PYQ

Solution

Qus : 156 Phrases PYQ

Consider n events ${{E}}_1,{{E}}_2\ldots{{E}}_n$ with respective probabilities ${{p}}_1,{{p}}_2\ldots{{p}}_n$. If $P\Bigg{(}{{E}}_1,{{E}}_2\ldots{{E}}_n\Bigg{)}=\prod ^n_{i=1}{{p}}_i$, then

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 157 Phrases PYQ

Given a set of events ${{E}}_1,{{E}}_2\ldots{{E}}_n$ defined on the sample space S such that :

(i) $\forall\, i\, and\, j,\, i\ne j,\, {{E}}_i\cap{{E}}_j=\phi$

(ii) $\begin{matrix}\overset{{n}}{\bigcup } \\ ^{i=1}\end{matrix}{{E}}_i=S$

(iii) $P({{E}}_i){\gt}0,\, \forall$

Then the events are

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 158 Phrases PYQ

Equation of pair of lines y = px, y = x written as (y − px)(y − qx) = 0. Equation of pair of angle bisector of line x² − 4xy − 5y² = 0

Options:

(a) x² − 3xy + y² = 0

(b) x² + 3xy − y² = 0

(d) x² − 3xy − y² = 0

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Phrases Previous Year PYQ Phrases CUET MCA 2026 PYQ

Solution

Qus : 159 Phrases PYQ

4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways arranging them is :

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Case 1: All persons are distinct

For n distinct persons around a round table (rotations same), arrangements = $(n-1)!$.

Here, $n=9$. So arrangements = $(9-1)! = 8! = \boxed{40{,}320}$.

Case 2: Persons are identical by nationality

If 4 Indians, 3 Americans, and 2 Britishers are considered identical within their groups, then

Arrangements = \[ \frac{(n-1)!}{4!\,3!\,2!} = \frac{8!}{4!\,3!\,2!} = \boxed{140}. \]

Final Answer:
• If all 9 are distinct → $40{,}320$ ways.
• If only nationality matters → $140$ ways.

Qus : 160 Phrases PYQ

4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways arranging them so that the two Britishers should never come together is:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 161 Phrases PYQ

4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways of arranging them so that the three Americans should sit together is:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Short Solution:

Total people = 4 Indians + 3 Americans + 2 Britishers = 9
Since arrangement is around a circular table, we fix one position ⇒ remaining to arrange: 8 positions

Group the 3 Americans together as a single unit ⇒ total units = 4 Indians + 1 American group + 2 Britishers = 7 units
Circular arrangement of 7 units = $ (7 - 1)! = 6! $

Internal arrangements of 3 Americans = $ 3! $

Total arrangements =
$$6! \times 3! = 720 \times 6 = \boxed{4320}$$

Qus : 162 Phrases PYQ

Given three identical boxes B₁ B₂ and B₃ each containing two balls. B₁ containstwo golden balls. B₂ contains two silver balls and B₃ contains one silver and onegolden ball. Conditional probabilities that the golden ball is drawn from B₁, B₂, B₃are ____,______,______ respectively

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 163 Phrases PYQ

Math List I with List II:

LIST I	LIST 2
A. In a GP, the third term is 24 and 6th term is 192. The common ratio is _____	I. 78
B. Let S_n denotes the sum of first n terms of an AP. If S_2n=3S_n, then S_3n/S_n equals to _______	II. 6
C. The sum of 3 terms of a GP is 13/12 and their product is -1. The first term is ______	III. -1
D. The least value of n for which the sum 3+6+9+...+n is greater than 1000 is	IV. 2

Choose the correct answer from the options given below :

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 164 Phrases PYQ

Math List I with List II : $\omega \ne1$ is a cube root of unity.

LIST I	LIST II
A. The value of $\frac{1}{9}(1-\omega)(1-{\omega}^2)(1-{\omega}^4)(1-{\omega}^8)\, $ is	I. 0
B. $\omega{(1+\omega-{\omega}^2)}^7$ ________ is equal to	II. 1
C. The least positive integer n such that ${(1+{\omega}^2)}^n={(1+{\omega}^4)}^n$ is	III. -128
D. $(1+\omega+{\omega}^2)$ is equal to	IV. 3

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Qus : 165 Phrases PYQ

Math List I with List II : $\omega \ne1$ is a cube root of unity.

LIST I	LIST II
A. $\log _4(\log _3(81))=$	I. 0
B. ${3}^{4\log _9(7)}={7}^k$, then k =	II. 3
C. ${2}^{\log _3(5)}-{5}^{\log _3(2)}=$	III. 1
D. $\log _2[\log _2(256)]=$	IV. 2

Choose the correct answer from the options given below:

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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

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Item	Result	Match
(A) Eccentricity of \(x^2-4x+4y+4y^2=12\)	\(\displaystyle \frac{\sqrt{3}}{2}\)	(IV)
(B) Latus rectum of \(5x^2+9y^2=45\)	\(\displaystyle \frac{10}{3}\)	(I)
(C) Line \(x+y=a\) touches \(y=x-x^2\) ⇒ value of \(a\)	\(1\)	(II)
(D) Eccentricity of \(3x^2-y^2=4\)	\(2\)	(III)

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