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 65-4-3 question paper with solution PDF is available here for download.
CBSE Class 12 2025 Mathematics 65-4-3 Question Paper with Solution PDF
CBSE Class 12 2025 Mathematics Question Paper with Answer Key | ![]() |
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Domain of \( \sin^{-1}(2x^2 - 3) \) is:
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The matrix \[ \begin{pmatrix} 0 & 1 & -2\\-1 & 0 & -7\\2 & 7 & 0 \end{pmatrix} \]
is a :
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If \(f(x) = \begin{cases} 3x - 2, & 0 \leq x \leq 1
2x^2 + ax, & 1 < x < 2 \end{cases}\) is continuous for \(x \in (0, 2)\), then \(a\) is equal to :
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If \( y = \log_2( \sqrt{2x} ) \), then \( \frac{dy}{dx} \) is equal to:
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We are given the function \( y = \log_2( \sqrt{2x} ) \), and we need to find \( \frac{dy}{dx} \).
Step 1: First, rewrite the expression using logarithmic properties: \[ y = \log_2( \sqrt{2x} ) = \log_2( (2x)^{1/2} ) \]
Using the power rule of logarithms: \[ y = \frac{1}{2} \log_2( 2x ) \]
Step 2: Now, differentiate the expression with respect to \(x\): \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx} \left( \log_2( 2x ) \right) \]
Step 3: The derivative of \( \log_2( 2x ) \) with respect to \(x\) is: \[ \frac{d}{dx} \left( \log_2( 2x ) \right) = \frac{1}{\ln 2} \cdot \frac{d}{dx} (2x) \] \[ = \frac{1}{\ln 2} \cdot 2 \]
Thus: \[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{2}{\ln 2} = \frac{1}{\ln 2} \]
Step 4: The final answer is: \[ \frac{dy}{dx} = \frac{1}{\sqrt{2x}} \] Quick Tip: When differentiating logarithmic functions, use the logarithmic differentiation rule: \( \frac{d}{dx} \log_b (u) = \frac{1}{\ln b} \cdot \frac{du}{dx} \).
If \(f : \mathbb{N} \rightarrow \mathbb{W}\) is defined as \[ f(n) = \begin{cases} \frac{n}{2}, & if n is even
0, & if n is odd \end{cases} \]
then \(f\) is :
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The coordinates of the foot of the perpendicular drawn from the point \(A(-2, 3, 5)\) on the y-axis are:
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If A and B are invertible matrices of order \(3 \times 3\) such that \(det(A) = 4\) and \(det([AB]^{-1}) = \frac{1}{20}\), then \(det(B)\) is equal to:
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For real \(x\), let \(f(x) = x^3 + 5x + 1\). Then :
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The values of \( \lambda \) so that \( f(x) = \sin x - \cos x - \lambda x + C \) decreases for all real values of \(x\) are:
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If A and B are square matrices of same order such that AB = BA, then \(A^2 + B^2\) is equal to :
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The area of the region enclosed by the curve \(y = \sqrt{x}\) and the lines \(x = 0\) and \(x = 4\) and the x-axis is :
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The value of \[ \int_0^1 \frac{dx}{e^x + e^{-x}} \]
is :
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The corner points of the feasible region of a Linear Programming Problem are \((0, 2)\), \((3, 0)\), \((6, 0)\), \((6, 8)\), and \((0, 5)\). If \(Z = ax + by; \, (a, b > 0)\) be the objective function, and maximum value of \(Z\) is obtained at \((0, 2)\) and \((3, 0)\), then the relation between \(a\) and \(b\) is :
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If \( \int e^{-3 \log x} \, dx = f(x) + C \), then \( f(x) \) is:
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The function \(f\) defined by \[ f(x) = \begin{cases} x, & if x \leq 1
5, & if x > 1 \end{cases} \]
is not continuous at :
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The solution of the differential equation \( \frac{dy}{dx} = -\frac{x}{y} \) represents family of:
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If the sides \(AB\) and \(AC\) of \(\triangle ABC\) are represented by vectors \(\hat{i} + \hat{j} + 4 \hat{k}\) and \(3 \hat{i} - \hat{j} + 4 \hat{k}\) respectively, then the length of the median through A on BC is :
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If \(f(x) = 2x + \cos x\), then \(f(x)\) :
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Assertion (A): If A and B are two events such that \(P(A \cap B) = 0\), then A and B are independent events.
Reason (R): Two events are independent if the occurrence of one does not affect the occurrence of the other.
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Assertion (A): In a Linear Programming Problem, if the feasible region is empty, then the Linear Programming Problem has no solution.
Reason (R): A feasible region is defined as the region that satisfies all the constraints.
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If \[ A = \begin{bmatrix} 2 & -2
-2 & 2 \end{bmatrix} \quad and \quad A^2 = kA, \quad then find the value of k. \]
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(a) Simplify \(\sin^{-1} \left( \frac{x}{\sqrt{1 + x^2}} \right)\).
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(b) Find the domain of \(\sin^{-1} \sqrt{x - 1}\).
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Calculate the area of the region bounded by the curve \[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
and the x-axis using integration.
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(a) Find the least value of ‘a’ so that \(f(x) = 2x^2 - ax + 3\) is an increasing function on \([2, 4]\).
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(b) If \(f(x) = x + \frac{1}{x}, \, x \geq 1\), show that \(f\) is an increasing function.
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A cylindrical water container has developed a leak at the bottom. The water is leaking at the rate of 5 cm\(^3\)/s from the leak. If the radius of the container is 15 cm, find the rate at which the height of water is decreasing inside the container, when the height of water is 2 meters.
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Find: \[ \int \frac{\sqrt{x}}{1 + \sqrt{x^{3/2}}} \, dx \]
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Find the distance of the point \((-1, -5, -10)\) from the point of intersection of the lines \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}, \quad \frac{x - 4}{5} = \frac{y - 1}{2} = z. \]
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If \(f : \mathbb{R}^+ \to \mathbb{R}\) is defined as \(f(x) = \log_a x\) where \(a > 0\) and \(a \neq 1\), prove that \(f\) is a bijection.
(R\(^+\) is the set of all positive real numbers.)
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Let \(A = \{1, 2, 3\}\) and \(B = \{4, 5, 6\}\). A relation \(R\) from \(A\) to \(B\) is defined as \(R = \{(x, y) : x + y = 6, x \in A, y \in B \}\).
(i) Write all elements of \(R\).
(ii) Is \(R\) a function? Justify.
(iii) Determine domain and range of \(R\).
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(a) Find \( k \) so that the function \[ f(x) = \begin{cases} \frac{x^2 - 2x - 3}{x + 1} & if x \neq -1
k & if x = -1 \end{cases} \]
is continuous at \( x = -1 \).
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% Graph Image
\begin{figure[h]
\centering
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For the given graph of a Linear Programming Problem, write all the constraints satisfying the given feasible region.
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The relation between the height of the plant (\(y\) cm) with respect to exposure to sunlight is governed by the equation \[ y = 4x - \frac{1}{2} x^2, \]
where \(x\) is the number of days exposed to sunlight.
(i) Find the rate of growth of the plant with respect to sunlight.
(ii) In how many days will the plant attain its maximum height? What is the maximum height?
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Show that the area of a parallelogram whose diagonals are represented by \( \vec{a} \) and \( \vec{b} \) is given by \[ Area = \frac{1}{2} | \vec{a} \times \vec{b} |. \]
Also, find the area of a parallelogram whose diagonals are \( 2\hat{i} - \hat{j} + \hat{k} \) and \( \hat{i} + 3\hat{j} - \hat{k} \).
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Find the equation of a line in vector and Cartesian form which passes through the point \( (1, 2, -4) \) and is perpendicular to the lines \[ \frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}. \]
and \[ \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu (3\hat{i} + 8\hat{j} - 5\hat{k}). \]
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