CBSE Class 12 Mathematics Question Paper 2025 with Solutions

The CBSE Class 12 Mathematics question paper 2025 is the benchmark resource for students aiming for a 95+ score in board examinations. Reviewing the CBSE Class 12 Mathematics question paper 2025 with solutions allows candidates to understand the latest paper pattern, step-marking allocation, and chapter-wise weightage prescribed by the Central Board of Secondary Education. Practicing authentic previous year question paper CBSE sets bridges the gap between conceptual understanding in NCERT textbooks and the analytical application demanded in board exams.

CBSE Class 12 Mathematics Question Paper Structure and Blueprint

The Class 12 Mathematics theory examination carries a total of 80 marks with a time duration of 3 hours (180 minutes). The qualifying passing mark is 26 out of 80 (33%) in theory, alongside the mandatory 33% in internal assessment (20 marks). The question paper is divided into five mandatory sections comprising 38 questions in total.

Complete Questions with Detailed Step-by-Step Solutions

Below is a curated selection of representative questions from all five sections of the CBSE Class 12 Mathematics examination, solved using official CBSE marking guidelines.

Section A: 1-Mark Questions (MCQ & Assertion-Reason)

Question 1: If A is a non-singular square matrix of order 3 × 3 such that |A| = 5, find the value of |adj(A)|.
(a) 5     (b) 25     (c) 125     (d) 1/5

Solution:

1. For any square matrix A of order n × n, the standard property of adjugate matrices states: |adj(A)| = |A|^{n − 1}.
2. Here, the matrix order is n = 3 and |A| = 5.
3. Substitute the given values: |adj(A)| = 5^{3 − 1} = 5² = 25.

Correct Option: (b) 25

Question 2: Determine the order and degree (if defined) of the differential equation: (d²y/dx²)³ + (dy/dx)⁴ + sin(dy/dx) = 0.
(a) Order = 2, Degree = 3     (b) Order = 2, Degree = 4     (c) Order = 2, Degree = Not Defined     (d) Order = 1, Degree = Not Defined

Solution:

1. The highest order derivative present in the equation is d²y/dx², which means the Order = 2.
2. The degree of a differential equation is the highest exponent of the highest-order derivative when the equation is expressed as a polynomial in derivatives.
3. Because the term sin(dy/dx) is present, this equation cannot be expressed as a polynomial in its derivatives. Therefore, the Degree is Not Defined.

Correct Option: (c) Order = 2, Degree = Not Defined

Question 3: Assertion (A): The function f(x) = |x − 2| is continuous at x = 2 but not differentiable at x = 2.
Reason (R): If the left-hand derivative and right-hand derivative of a continuous function are not equal at a point x = a, the function is not differentiable at that point.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.

Solution:

1. Compute Left-Hand Derivative (LHD) at x = 2: lim_h→0⁻ (|2 − h − 2| − 0)/(−h) = lim_h→0 h/(−h) = −1.
2. Compute Right-Hand Derivative (RHD) at x = 2: lim_h→0⁺ (|2 + h − 2| − 0)/(h) = lim_h→0 h/h = 1.
3. Since LHD ≠ RHD (−1 ≠ 1), f(x) is not differentiable at x = 2 despite being continuous. Reason (R) defines the exact criteria for non-differentiability.

Correct Option: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).

Section B: 2-Mark Questions (Short Answer Type I)

Question 4: Find the principal value of: tan⁻¹(−√3) + cos⁻¹(−1/2) + sin⁻¹(1/2).

Solution:

1. Principal value branch for tan⁻¹(x) is (−π/2, π/2):
   tan⁻¹(−√3) = −tan⁻¹(√3) = −π/3. [0.5 Mark]
2. Principal value branch for cos⁻¹(x) is [0, π]:
   cos⁻¹(−1/2) = π − cos⁻¹(1/2) = π − π/3 = 2π/3. [0.5 Mark]
3. Principal value branch for sin⁻¹(x) is [−π/2, π/2]:
   sin⁻¹(1/2) = π/6. [0.5 Mark]
4. Sum the evaluated components:
   Value = (−π/3) + (2π/3) + (π/6) = π/3 + π/6 = 3π/6 = π/2. [0.5 Mark]

Section C: 3-Mark Questions (Short Answer Type II)

Question 5: Evaluate the definite integral: ∫₀^π/2 [ √(sin x) / ( √(sin x) + √(cos x) ) ] dx.

Solution:

1. Let I = ∫₀^π/2 [ √(sin x) / ( √(sin x) + √(cos x) ) ] dx     … (Equation 1) [0.5 Mark]
2. Apply the definite integral property: ∫₀^a f(x) dx = ∫₀^a f(a − x) dx.
   I = ∫₀^π/2 [ √(sin(π/2 − x)) / ( √(sin(π/2 − x)) + √(cos(π/2 − x)) ) ] dx
   Since sin(π/2 − x) = cos x and cos(π/2 − x) = sin x:
   I = ∫₀^π/2 [ √(cos x) / ( √(cos x) + √(sin x) ) ] dx     … (Equation 2) [1 Mark]
3. Add Equation 1 and Equation 2:
   2I = ∫₀^π/2 [ ( √(sin x) + √(cos x) ) / ( √(sin x) + √(cos x) ) ] dx
   2I = ∫₀^π/2 1 dx [0.5 Mark]
4. Integrate and evaluate limits:
   2I = [ x ]₀^π/2 = π/2 − 0 = π/2 [0.5 Mark]
   I = π/4. [0.5 Mark]

Question 6: Solve the linear differential equation: dy/dx + y cot x = 2x + x² cot x, given that y = 0 when x = π/2.

Solution:

1. This is a first-order linear differential equation of the standard form: dy/dx + P(x)y = Q(x), where P(x) = cot x and Q(x) = 2x + x² cot x. [0.5 Mark]
2. Find the Integrating Factor (I.F.):
   I.F. = e^{∫ cot x dx} = e^{ln |sin x|} = sin x. [0.5 Mark]
3. The general solution formula is: y × (I.F.) = ∫ [ Q(x) × (I.F.) ] dx + C.
   y × sin x = ∫ (2x + x² cot x) sin x dx + C
   y sin x = ∫ (2x sin x + x² cos x) dx + C. [0.5 Mark]
4. Recognize the derivative product rule: d/dx (x² sin x) = 2x sin x + x² cos x.
   Therefore, ∫ (2x sin x + x² cos x) dx = x² sin x.
   General solution: y sin x = x² sin x + C. [0.5 Mark]
5. Apply the initial condition y = 0 at x = π/2:
   0 × sin(π/2) = (π/2)² sin(π/2) + C
   0 = (π²/4)(1) + C ⇒ C = −π²/4. [0.5 Mark]
6. Substitute C into the general equation:
   y sin x = x² sin x − π²/4 ⇒ y = x² − (π²/4) cosec x. [0.5 Mark]

Section D: 5-Mark Questions (Long Answer Type)

Question 7: Solve the following system of linear equations using the matrix method (inverse matrix method):
2x − 3y + 5z = 11
3x + 2y − 4z = −5
x + y − 2z = −3

Solution:

1. Express the system in matrix form AX = B:
   A = &begin;bmatrix& 2 & −3 & 5 \\ 3 & 2 & −4 \\ 1 & 1 & −2 &end;bmatrix&,    X = &begin;bmatrix& x \\ y \\ z &end;bmatrix&,    B = &begin;bmatrix& 11 \\ −5 \\ −3 &end;bmatrix&. [1 Mark]
2. Calculate the determinant of matrix A (|A|):
   |A| = 2[(2)(−2) − (−4)(1)] − (−3)[(3)(−2) − (−4)(1)] + 5[(3)(1) − (2)(1)]
   |A| = 2[−4 + 4] + 3[−6 + 4] + 5[3 − 2]
   |A| = 2(0) + 3(−2) + 5(1) = 0 − 6 + 5 = −1.
   Since |A| ≠ 0, A⁻¹ exists and the system possesses a unique solution given by X = A⁻¹B. [1 Mark]
3. Find the cofactors (C_ij) of matrix A:
   C₁₁ = +(−4 + 4) = 0;   C₁₂ = −(−6 + 4) = 2;   C₁₃ = +(3 − 2) = 1
   C₂₁ = −(6 − 5) = −1;   C₂₂ = +(−4 − 5) = −9;   C₂₃ = −(2 − (−3)) = −5
   C₃₁ = +(12 − 10) = 2;   C₃₂ = −(−8 − 15) = 23;   C₃₃ = +(4 − (−9)) = 13. [1 Mark]
4. Compute the adjoint matrix and inverse matrix A⁻¹:
   adj(A) = [C_ij]^T = &begin;bmatrix& 0 & −1 & 2 \\ 2 & −9 & 23 \\ 1 & −5 & 13 &end;bmatrix&
   A⁻¹ = (1/|A|) × adj(A) = (1/−1) × adj(A) = &begin;bmatrix& 0 & 1 & −2 \\ −2 & 9 & −23 \\ −1 & 5 & −13 &end;bmatrix&. [1 Mark]
5. Calculate X = A⁻¹B:
   x = (0)(11) + (1)(−5) + (−2)(−3) = 0 − 5 + 6 = 1
   y = (−2)(11) + (9)(−5) + (−23)(−3) = −22 − 45 + 69 = 2
   z = (−1)(11) + (5)(−5) + (−13)(−3) = −11 − 25 + 39 = 3.
   Final Answer: x = 1, y = 2, z = 3. [1 Mark]

Section E: 4-Mark Case-Based Integrated Problem

Question 8 (Real-World Case Study on Bayes' Theorem): A national pathology diagnostic chain tests individuals for a specific hereditary blood disorder. Epidemiological data indicates that 1% of the population suffers from this disorder. The diagnostic lab's laboratory screening test has an accuracy rate of 98% in correctly detecting the disease when present (true positive rate). However, the test also produces a false positive result in 2% of healthy individuals.

Based on the clinical data provided, answer the following questions:

1. Find the total probability that a randomly chosen individual tests positive for the disorder. [2 Marks]
2. If an individual tests positive, what is the conditional probability that the person actually suffers from the disorder? [2 Marks]

Solution:

1. Define Events and Prior Probabilities:
   Let E₁ be the event that the chosen person has the disorder: P(E₁) = 1% = 0.01.
   Let E₂ be the event that the person does not have the disorder: P(E₂) = 99% = 0.99.
   Let A be the event that the screening test result is positive.
   Conditional probabilities: P(A|E₁) = 0.98 (Test positive given disease) and P(A|E₂) = 0.02 (Test positive given healthy).
2. (i) Total Probability P(A):
   By the Theorem on Total Probability:
   P(A) = P(E₁) × P(A|E₁) + P(E₂) × P(A|E₂)
   P(A) = (0.01 × 0.98) + (0.99 × 0.02)
   P(A) = 0.0098 + 0.0198 = 0.0296 (or 296/10000 = 37/1250). [2 Marks]
3. (ii) Conditional Probability using Bayes' Theorem P(E₁|A):
   P(E₁|A) = [ P(E₁) × P(A|E₁) ] / P(A)
   P(E₁|A) = 0.0098 / 0.0296 = 98 / 296 = 49 / 148 ≈ 0.3311 (33.11%).
   Thus, despite testing positive, the actual probability of having the disease is approximately 33.11% due to the low base rate in the general population. [2 Marks]

Difficulty Analysis of CBSE Class 12 Mathematics 2025 Paper

The 2025 Mathematics question paper maintained a well-balanced distribution of difficulty levels, adhering to the National Curriculum Framework (NCF) standards that test rote memory, core procedural knowledge, and high-order thinking skills (HOTS).

Unit-Wise Weightage Summary

• Calculus (35 Marks): The highest-scoring unit. Comprises Continuity, Differentiability, Application of Derivatives, Integrals, and Differential Equations.
• Vectors & Three-Dimensional Geometry (14 Marks): Critical for securing full marks in Section C and Section D via standard vector and Cartesian line/plane formulas.
• Algebra (10 Marks): Matrices and Determinants questions are formulaic and high-scoring if calculation errors are avoided.
• Probability (8 Marks): Dominated by Bayes' Theorem and Probability Distribution functions.
• Relations & Functions (8 Marks): Covers Equivalence Relations and Inverse Trigonometric Functions.
• Linear Programming (5 Marks): Single optimization formulation and feasible bounded/unbounded region graphing.

Official Marking Scheme and Step-Marking Guidelines

CBSE strictly enforces a step-by-step marking scheme. Even if a student arrives at an incorrect final numerical answer due to a late-stage computational slip, up to 80% of the total question marks can still be awarded if preceding steps, formulas, and working are mathematically sound.

How to Use This Question Paper for Exam Prep

Maximizing exam performance using the 2025 question paper requires structured practice rather than casual reading. Implement these evidence-based board exam preparation techniques:

1. Simulate Strict Exam Conditions: Sit in an isolated environment for a continuous 3-hour window with an analog clock. Restrict textbook access and complete the entire paper in one sitting.
2. Master 15-Minute Reading Time: Use the initial 15 minutes to select internal choice questions in Section C, D, and E. Identify calculation-heavy problems and plan your solving sequence.
3. Cross-Check with the Official Answer Key: Evaluate your written answers against the marking rubric above. Identify specific deduction patterns such as missing constants, improper determinant signs, or incomplete probability statements.
4. Target Weakness Cycles: If you lost marks in Calculus differential equations or 3D Geometry lines, immediately revise NCERT Exercise problems and exemplar questions for those exact sub-topics.

More Practice Papers and Solutions

Consistent paper-solving is the single most effective way to eliminate exam anxiety and build speed for the CBSE Class 12 board examination. Access complete chapter-wise practice tests, previous 10-year question papers, and AI-powered step-by-step evaluation tools at qptool.theorify.in to strengthen your preparation and secure top marks in Mathematics.