CBSE Class 12th Board Mathematics exam was conducted on 8th March 2025 from 10:30 AM to 1:30 PM.The Mathematics theory paper is worth 80 marks, and the internal assessment is worth 20 marks. The CBSE Mathematics question paper includes MCQ (1 mark each), short-answer type questions (2 and 3 marks each), and long-answer type questions (4 and 6 marks each), totaling 80 marks.
CBSE Class 12 2025 Mathematics 65-5-2 question paper with solution PDF is available here for download.
CBSE Class 12 2025 Mathematics 65-5-2 Question Paper with Solution PDF
CBSE Class 12 2025 Mathematics Question Paper with Answer Key | ![]() |
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If \[ f(x) = \begin{cases} \frac{\sin^2(ax)}{x^2}, & x \neq 0
1, & x = 0 \end{cases} \]
is continuous at \( x = 0 \), then the value of \( a \) is:
(A) 1
(B) -1
(C) ±1
(D) 0
View Solution
The principal value of \( \cot^{-1} \left( -\frac{1}{\sqrt{3}} \right) \) is:
A) −π/3
(B) −2π/3
(C) π/3
(D) 2π/3
View Solution
If A and B are two square matrices of the same order, then (A + B)(A - B) is equal to:
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
View Solution
If \[ A = \begin{bmatrix} 5 & 0
0 & 5 \end{bmatrix}, \]
then \( A^3 \) is:
If \( \left| \begin{matrix} 2x & 5
12 & x \end{matrix} \right| = \left| \begin{matrix} 6 & -5
4 & 3 \end{matrix} \right| \), then the value of x is:
View Solution
If \( P(A \cup B) = 0.9 \) and \( P(A \cap B) = 0.4 \), then \( P(A) + P(B) \) is:
View Solution
If a matrix A is both symmetric and skew-symmetric, then A is:
View Solution
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)
View Solution
The area of the region enclosed between the curve \( y = |x| \), x-axis, \( x = -2 \) and \( x = 2 \) is:
(A) 4/3
(B) 16
(C) 8/3
(D) 8
View Solution
\( \int \frac{\cos 2x}{\sin^2 x \cos^2 x} \, dx \) is equal to:
View Solution
If \( \int_0^a \frac{1}{1 + 4x^2} \, dx = \frac{\pi}{8} \), then the value of \(a\) is:
View Solution
If \( f(x) = \lfloor x \rfloor \) 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.
The integrating factor of the differential equation \( \frac{dx}{dy} = \frac{x \log x}{2 \log x - y} \) is:
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)
View Solution
The respective values of \( |\vec{a}| \) and \( |\vec{b}| \), if given \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512 \quad and \quad |\vec{a}| = 3 |\vec{b}|, \]
are:
(A) 48 and 16
(B) 3 and 1
(C) 24 and 8
(D) 6 and 2
View Solution
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
View Solution
The sum of the order and degree of the differential equation \[ \left( 1 + \left( \frac{dy}{dx} \right)^2 \right) \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^3 \]
is:
(A) 2
(B) 5/2
(C) 3
(D) 4
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.
(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.
(D)Assertion (A) is false, but Reason (R) is true.
View Solution
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 \( x = \pm \sqrt{-1} \notin A \).
(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.
(D)Assertion (A) is false, but Reason (R) is true.
View Solution
Find the domain of \( \sec^{-1}(2x + 1) \).
The radius of a cylinder is decreasing at a rate of 2 cm/s and the altitude is increasing at a rate of 3 cm/s. Find the rate of change of volume of this cylinder when its radius is 4 cm and altitude is 6 cm.
(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.
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.
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.
(ii) The probability that she buys a box of colours given she buys the colouring book.
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.
(ii) The expectation of the number of oranges.
Find the particular solution of the differential equation \( \frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0 \); given that \( y = 0 \), when \( x = 1 \).
29. (a) Find: \[ \int \frac{2x}{(x^2 + 3)(x^2 - 5)} \, dx \]
29. (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
\
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 \(\vec{b}\).
Draw a rough sketch of the curve \( y = \sqrt{x} \). Using integration, find the area of the region bounded by the curve \( y = \sqrt{x} \), \( x = 4 \), and the x-axis, in the first quadrant.
.
An amount of ₹ 10,000 is put into three investments at the rate of 10%, 12% and 15% per annum. The combined annual income of all three investments is ₹ 1,310, however, the combined annual income of the first and second investments is ₹ 190 short of the income from the third. Use matrix method and find the investment amount in each at the beginning of the year.
(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 \( (-1, -1, 2) \).
(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.
(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 \( x = \frac{\sqrt{6}}{6} \).
(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.
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