CBSE Class 12 2025 Mathematics 65-5-3 Question Paper Set-3: Download Solutions with Answer Key

The principal value of \( \cot^{-1} \left( -\frac{1}{\sqrt{3}} \right) \) is:

• (A) \( -\frac{\pi}{3} \)
• (B) \( -\frac{2\pi}{3} \)
• (C) \( \frac{\pi}{3} \)
• (D) \( \frac{2\pi}{3} \)

If \[ A = \begin{bmatrix} 1 & 0 & 0
0 & 5 & 0
0 & 0 & -2 \end{bmatrix}, \]
then \( |A| \) is:

• (A) 0
• (B) -10
• (C) 10
• (D) 1

If \( A = kB \), where \( A \) and \( B \) are two square matrices of order \( n \) and \( k \) is a scalar, then:

• (A) \( |A| = k|B| \)
• (B) \( |A| = k^n|B| \)
• (C) \( |A| = k + |B| \)
• (D) \( |A| = |B|^k \)

If \( f(x) = \begin{cases} \dfrac{\sin^2(ax)}{x^2}, & x \ne 0
1, & x = 0 \end{cases} \) is continuous at \( x = 0 \), then the value of \( a \) is:

• (A) \( 1 \)
• (B) \( -1 \)
• (C) \( \pm 1 \)
• (D) \( 0 \)

If \( \left| \begin{array}{cc} 2x & 5
12 & x \end{array} \right| = \left| \begin{array}{cc} 6 & -5
4 & 3 \end{array} \right| \), then the value of \( x \) is:

• (A) 3
• (B) 7
• (C) \( \pm 7 \)
• (D) \( \pm 3 \)

If \( P(A \cup B) = 0.9 \) and \( P(A \cap B) = 0.4 \), then \( P(\bar{A}) + P(\bar{B}) \) is:

• (A) 0.3
• (B) 1
• (C) 1.3
• (D) 0.7

If \( A = \left[ \begin{array}{ccc} 5 & 0 & 0
3 & 0 & 5
0 & 0 & 5 \end{array} \right] \), then \( A^3 \) is:

• (A) \( \left[ \begin{array}{ccc} 125 & 0 & 0
  0 & 125 & 0
  0 & 0 & 125 \end{array} \right] \)
• (B) \( \left[ \begin{array}{ccc} 125 & 0 & 0
  0 & 0 & 125
  0 & 0 & 125 \end{array} \right] \)
• (C) \( \left[ \begin{array}{ccc} 15 & 0 & 0
  0 & 45 & 0
  0 & 0 & 15 \end{array} \right] \)
• (D) \( \left[ \begin{array}{ccc} 5^3 & 0 & 0
  0 & 5 & 0
  0 & 0 & 5 \end{array} \right] \)

Let \( A \) and \( B \) be two matrices of suitable orders. Then, which of the following is not correct?

• (A) \( (A')' = A \)
• (B) \( (kA)' = kA' \), \( k \) is scalar
• (C) \( (A' + B')' = A + B \)
• (D) \( (AB)' = A'B' \)

The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \) and \( x = 2 \) is:

• (A) 8
• (B) 16
• (C) 0
• (D) \( \dfrac{16}{3} \)

If \( f(x) = [x] \), \( x \in \mathbb{R} \) is the greatest integer function, then the correct statement is:

• (A) \( f \) is continuous but not differentiable at \( x = 2 \)
• (B) \( f \) is neither continuous nor differentiable at \( x = 2 \)
• (C) \( f \) is continuous as well as differentiable at \( x = 2 \)
• (D) \( f \) is not continuous but differentiable at \( x = 2 \)

\( \int \frac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} dx \) is equal to:

• (A) \( 2(\sin x + x \cos \theta) + C \)
• (B) \( 2(\sin x - x \cos \theta) + C \)
• (C) \( 2(\sin x + \sin \theta) + C \)
• (D) \( 2(\sin x - x \sin \theta) + C \)

Evaluate: \( \int_0^1 \frac{2x}{5x^2 + 1} dx \)

• (A) \( \frac{1}{5} \log 6 \)
• (B) \( \frac{1}{5} \log 5 \)
• (C) \( \frac{1}{2} \log 6 \)
• (D) \( \frac{1}{2} \log 5 \)

The slope of the curve \( y = -x^3 + 3x^2 + 8x - 20 \) is maximum at:

• (A) \( (1, -10) \)
• (B) \( (1, 10) \)
• (C) \( (10, 1) \)
• (D) \( (-10, 1) \)

The integrating factor of the differential equation \( \frac{dx}{dy} = \frac{-(1 + \sin x)}{x + y \cos x} \) is:

• (A) \( \log \cos x \)
• (B) \( 1 + \sin x \)
• (C) \( e^{1 + \sin x} \)
• (D) \( e^{\log \cos x} \)

For a Linear Programming Problem (LPP), the given objective function \( Z = 3x + 2y \) is subject to constraints:
\[ x + 2y \leq 10, \] \[ 3x + y \leq 15, \] \[ x, y \geq 0. \]





The correct feasible region is:

• (A) ABC
• (B) AOEC
• (C) CED
• (D) Open unbounded region BCD

The sum of the order and degree of the differential equation \[ \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}} = \frac{d^2 y}{dx^2} \]
is:

• (A) 2
• (B) \( \dfrac{5}{2} \)
• (C) 3
• (D) 4

The respective values of \( |\vec{a}| \) and \( |\vec{b}| \), if given: \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512, \quad |\vec{a}| = 3|\vec{b}| \]
are:

• (A) 48 and 16
• (B) 3 and 1
• (C) 24 and 8
• (D) 6 and 2

Let \( \vec{a} \) be a position vector whose tip is the point \( (2, -3) \). If \( \overrightarrow{AB} = \vec{a} \), where coordinates of \( A \) are \( (-4, 5) \), then the coordinates of \( B \) are:

• (A) \( (-2, -2) \)
• (B) \( (2, -2) \)
• (C) \( (-2, 2) \)
• (D) \( (2, 2) \)
• (A) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
• (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• (C) Assertion (A) is true, but Reason (R) is false.

Assertion (A): The shaded portion of the graph represents the feasible region for the given Linear Programming Problem (LPP).

Reason (R): The region representing \( Z = 50x + 70y \) such that \( Z < 380 \) does not have any point common with the feasible region.


 

Assertion (A): Let \( A = \{ x \in \mathbb{R} : -1 \leq x \leq 1 \} \). If \( f : A \to A \) be defined as \( f(x) = x^2 \), then \( f \) is not an onto function.
Reason (R): If \( y = -1 \in A \), then \( x = \pm \sqrt{-1} \notin A \).

Find the domain of \( \sin^{-1}(x^2 - 3) \).

Let the volume of a metallic hollow sphere be constant. If the inner radius increases at the rate of 2 cm/s, find the rate of increase of the outer radius when the radii are 2 cm and 4 cm respectively.

A man needs to hang two lanterns on a straight wire whose end points have coordinates A (4, 1, -2) and B (6, 2, -3). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.

(a) Differentiate \(\frac{\sin x}{\sqrt{\cos x}}\) \text{ with respect to x.

(b) If y = 5 \cos x - 3 \sin x, \text{ prove that \frac{d^2y{dx^2 + y = 0.

(a) Find a vector of magnitude 5 which is perpendicular to both the vectors \( 3\hat{i} - 2\hat{j} + \hat{k} and 4\hat{i} + 3\hat{j} - 2\hat{k} \).

(b) Let \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c\) \text{ be three vectors such that \mathbf{a \times \mathbf{b = \mathbf{a \times \mathbf{c \text{ and \mathbf{a \times \mathbf{b \neq 0. \text{ Show that \mathbf{b = \mathbf{c.

Find the interval/intervals in which the function \( f(x) = \sin 3x - \cos 3x \), \( 0 < x < \frac{\pi}{2} \) is strictly increasing.

If \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\) such that \(|\vec{a}| = 3\), \(|\vec{b}| = 5\), \(|\vec{c}| = 7\), then find the angle between \(\vec{a}\) and \(\vec{b}\).

• (3) ^2 + (5)^2 + 2(3)(5)\cos\theta = 49 \Rightarrow 9 + 25 + 30\cos\theta = 49 \Rightarrow 34 + 30\cos\theta = 49 \Rightarrow 30\cos\theta = 15 \Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = 60^\circ \] % Quick Tip % Quick Tip \begin{quicktipbox} When three vectors add to zero, the triangle rule applies: the sum of any two vectors is the negative of the third. You can use the cosine law to find angles between vectors. \end{quicktipbox} % Topic - Vector Algebra \hrule \textbf{OR}

(b)
If \(\vec{a}\) and \(\vec{b}\) are unit vectors inclined at an angle \(\theta\), prove that: \[ \frac{1}{2} |\vec{a} - \vec{b}| = \sin\left(\frac{\theta}{2}\right) \]

Solve the differential equation: \[ x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos\left(\frac{y}{x}\right) + x \]

The probability that a student buys a colouring book is 0.7, and a box of colours is 0.2. The probability that she buys a colouring book, given that she buys a box of colours, is 0.3. Find:

• (i) The probability that she buys both the colouring book and the box of colours.
• (i) By definition of conditional probability: \[ P(C|B) = \frac{P(C \cap B)}{P(B)} \Rightarrow P(C \cap B) = P(C|B) \cdot P(B) = 0.3 \cdot 0.2 = 0.06 \]

A fruit box contains 6 apples and 4 oranges. A person picks out a fruit three times with replacement. Find:

• (i) The probability distribution of the number of oranges he draws.
• (i) Probability distribution: \[ \begin{aligned} P(X = 0) &= \binom{3}{0} (0.4)^0 (0.6)^3 = 1 \cdot 1 \cdot 0.216 = 0.216
  P(X = 1) &= \binom{3}{1} (0.4)^1 (0.6)^2 = 3 \cdot 0.4 \cdot 0.36 = 0.432
  P(X = 2) &= \binom{3}{2} (0.4)^2 (0.6)^1 = 3 \cdot 0.16 \cdot 0.6 = 0.288
  P(X = 3) &= \binom{3}{3} (0.4)^3 (0.6)^0 = 1 \cdot 0.064 \cdot 1 = 0.064
  \end{aligned} \]

30. (a) Find: \[ \int \frac{2x}{(x^2 + 3)(x^2 - 5)} \, dx \]

30. (b) Evaluate: \[ \int_1^5 \left( |x-2| + |x-4| \right) \, dx \]

In the Linear Programming Problem (LPP), find the point/points giving the maximum value for \( Z = 5x + 10y \text{ subject to the constraints:

x + 2y \leq 120

x + y \geq 60

x - 2y \geq 0

x \geq 0, y \geq 0
 

In a rough sketch, mark the region bounded by \( y = 1 + |x + 1| \), \( x = -2 \), \( x = 2 \), and \( y = 0 \). Using integration, find the area of the marked region.

Three students run on a racing track such that their speeds add up to 6 km/h. However, double the speed of the third runner added to the speed of the first results in 7 km/h. If thrice the speed of the first runner is added to the original speeds of the other two, the result is 12 km/h. Using the matrix method, find the original speed of each runner.

(a) For a positive constant \( a \), differentiate \( \left( t + \frac{1}{t} \right)^a \) with respect to \( t \), where \( t \) is a non-zero real number.

(b) Find \( \frac{dy}{dx} \) if \( x^3 + y^3 + x^2 = a^b \), where \( a \) and \( b \) are constants.

(a) Find the foot of the perpendicular drawn from the point \( (1, 1, 4) \) on the line \( \frac{x+2}{5} = \frac{y+1}{2} = \frac{z-4}{-3} \).

(b) Find the point on the line \( \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-4}{3} \) at a distance of \( \sqrt{2} \) units from the point \( (-1, -1, 2) \).

(i) Is \( f \) a bijective function?

(ii) Give reasons to support your answer to (i).

iii)(a) Let \( R \) be a relation defined by the teacher to plan the seating arrangement of students in pairs, where \( R = \{(x, y) : x, y are Roll Numbers of students such that y = 3x \} \).
List the elements of \( R \). Is the relation \( R \) reflexive, symmetric, and transitive? Justify your answer.

iii)(b) Let \( R \) be a relation defined by \( R = \{(x, y) : x, y are Roll Numbers of students such that y = x^3 \} \).
List the elements of \( R \). Is \( R \) a function? Justify your answer.

(i) Calculate the probability of a randomly chosen seed to germinate.

(ii) What is the probability that it is a cabbage seed, given that the chosen seed germinates?

(i) Taking length = breadth = \( x \) m and height = \( y \) m, express the surface area \( S \) of the box in terms of \( x \) and its volume \( V \), which is constant.

(ii) Find \( \frac{dS}{dx} \).

(iii) (a) Find a relation between \( x \) and \( y \) such that the surface area \( S \) is minimum.

(iii) (b) If surface area \( S \) is constant, the volume \( V = \frac{1}{4}(Sx - 2x^3) \), \( x \) being the edge of the base. Show that the volume \( V \) is maximum for \( x = \frac{\sqrt{6}}{6} \).