Class 12 Mathematics CBSE Format

CBSE Class 12 Sample Paper 2025 Maths Set 1 with Solutions

Updated for 2025–2026 Board Pattern · 9 Views

CBSE Class 12 Mathematics: CBSE Class 12 Sample Paper 2025 Maths Set 1 with Solutions

Mastering CBSE Class 12 Mathematics is the cornerstone of success for science and commerce students aiming for top scores in their 2025 board examinations. Solving the official CBSE Class 12 Sample Paper 2025 Maths Set 1 with Solutions provides crucial insights into the evolving question typology, step-marking scheme, and chapter weightage prescribed by the Central Board of Secondary Education. With the 2025 and 2026 board exam pattern emphasizing competency-based questions, conceptual clarity and rigorous practice are non-negotiable.

CBSE Class 12 Maths 2025 Exam Structure & Blueprint

The Class 12 Mathematics theory paper carries a total of 80 marks to be completed in 3 hours, supplemented by a 20-mark internal assessment. The question paper is divided systematically into five sections:

  • Section A (20 Marks): 20 Multiple Choice Questions (MCQs) and Assertion-Reason questions of 1 mark each.
  • Section B (10 Marks): 5 Very Short Answer (VSA) questions of 2 marks each.
  • Section C (18 Marks): 6 Short Answer (SA) questions of 3 marks each.
  • Section D (20 Marks): 4 Long Answer (LA) questions of 5 marks each.
  • Section E (12 Marks): 3 Case-Study/Source-based integrated units of assessment carrying 4 marks each with sub-parts.

The unit-wise mark distribution for CBSE Mathematics is structured to prioritize high-yield units:

Unit No. Unit Title Marks Weightage
I Relations and Functions (including Inverse Trigonometry) 08 Marks
II Algebra (Matrices and Determinants) 10 Marks
III Calculus (Continuity, Differentiability, Applications, Integrals, Diff. Eq.) 35 Marks
IV Vectors and Three-Dimensional Geometry 14 Marks
V Linear Programming 05 Marks
VI Probability 08 Marks
Total Marks 80 Marks

Key High-Yield Concepts & Essential Formulas

To maximize your score in board exam 12, ensure you have command over the following fundamental theoretical pillars:

1. Matrices & Systems of Linear Equations

For a non-singular square matrix A of order n:

  • Inverse Matrix: A-1 = (1 / |A|) · adj(A), provided |A| ≠ 0.
  • Adjoint Property: A · (adj A) = (adj A) · A = |A| · In.
  • Determinant of Adjoint: |adj(A)| = |A|n-1.
  • Matrix Method: For system AX = B, if |A| ≠ 0, unique solution is X = A-1B.

2. Definite Integrals & Properties

Mastering definite integral properties saves critical time in Sections C and D:

  • King's Property: ∫ab f(x) dx = ∫ab f(a + b - x) dx
  • Zero to 'a' Property: ∫0a f(x) dx = ∫0a f(a - x) dx
  • Even-Odd Function Property: ∫-aa f(x) dx = 2∫0a f(x) dx if f(-x) = f(x), and 0 if f(-x) = -f(x).

3. Vectors and 3D Geometry

The shortest distance d between two skew lines r = a1 + λb1 and r = a2 + μb2 is given by:

d = | ( (a2 - a1) · (b1 × b2) ) / |b1 × b2| |

CBSE Class 12 Sample Paper 2025 Maths Set 1: Questions with Step-by-Step Solutions

Section A: 1-Mark Questions

Question 1: If A is a square matrix of order 3 × 3 such that |A| = 5, then find the value of |adj(2A)|.

Solution:

  1. For a matrix M of order n = 3, we know |adj(M)| = |M|n-1 = |M|2.
  2. Let M = 2A. Since A is of order 3, |2A| = 23 · |A| = 8 · |A|.
  3. Given |A| = 5, we have |2A| = 8 × 5 = 40.
  4. Therefore, |adj(2A)| = |2A|2 = (40)2 = 1600.

Question 2: Find the integrating factor of the differential equation: x(dy/dx) - y = 2x2 cosec(x).

Solution:

  1. Rewrite the differential equation in standard linear form dy/dx + P(x)y = Q(x):
    dy/dx - (1/x)y = 2x cosec(x).
  2. Here, P(x) = -1/x.
  3. Integrating Factor (I.F.) = e∫ P(x) dx = e∫ (-1/x) dx = e-ln|x| = eln|x|-1 = x-1 = 1/x.

Section B: 2-Mark Questions

Question 3: Find the value of k for which the function f(x) is continuous at x = 0, where:

f(x) = (1 - cos 4x) / (8x2) for x ≠ 0 and f(0) = k.

Solution:

  1. For f(x) to be continuous at x = 0, we must have limx → 0 f(x) = f(0) = k.
  2. Evaluate the limit:
    limx → 0 (1 - cos 4x) / (8x2)
    Using identity 1 - cos 2θ = 2 sin2 θ, we have 1 - cos 4x = 2 sin2(2x).
  3. limx → 0 [ 2 sin2(2x) ] / (8x2) = limx → 0 [ sin2(2x) ] / (4x2) = limx → 0 [ sin(2x) / (2x) ]2.
  4. Since limθ → 0 (sin θ / θ) = 1, the limit equals (1)2 = 1.
  5. Thus, k = 1.

Section C: 3-Mark Questions

Question 4: Evaluate the definite integral: ∫0π/2 (sin4 x) / (sin4 x + cos4 x) dx.

Solution:

  1. Let I = ∫0π/2 [sin4 x / (sin4 x + cos4 x)] dx   — (Equation 1)
  2. Apply the property ∫0a f(x) dx = ∫0a f(a - x) dx:
    I = ∫0π/2 [sin4(π/2 - x) / (sin4(π/2 - x) + cos4(π/2 - x))] dx
  3. Using sin(π/2 - x) = cos x and cos(π/2 - x) = sin x:
    I = ∫0π/2 [cos4 x / (cos4 x + sin4 x)] dx   — (Equation 2)
  4. Adding Equations (1) and (2):
    2I = ∫0π/2 [(sin4 x + cos4 x) / (sin4 x + cos4 x)] dx
    2I = ∫0π/2 1 dx = [x]0π/2 = π/2 - 0 = π/2.
  5. Therefore, I = π / 4.

Section D: 5-Mark Questions

Question 5: Find the shortest distance between the lines whose vector equations are:

r = (i + 2j + 3k) + λ(i - 3j + 2k)
r = (4i + 5j + 6k) + μ(2i + 3j + k)

Solution:

  1. Identify vectors from line equations r = a1 + λb1 and r = a2 + μb2:
    a1 = i + 2j + 3k,   b1 = i - 3j + 2k
    a2 = 4i + 5j + 6k,   b2 = 2i + 3j + k
  2. Calculate (a2 - a1):
    a2 - a1 = (4 - 1)i + (5 - 2)j + (6 - 3)k = 3i + 3j + 3k.
  3. Calculate (b1 × b2):
    b1 × b2 = | i    j    k |
                        | 1   -3    2 |
                        | 2    3    1 |
    = i((-3)(1) - (2)(3)) - j((1)(1) - (2)(2)) + k((1)(3) - (-3)(2))
    = i(-3 - 6) - j(1 - 4) + k(3 + 6)
    = -9i + 3j + 9k.
  4. Compute the magnitude |b1 × b2|:
    |b1 × b2| = √[(-9)2 + (3)2 + (9)2] = √[81 + 9 + 81] = √171 = 3√19.
  5. Compute the dot product (a2 - a1) · (b1 × b2):
    (3i + 3j + 3k) · (-9i + 3j + 9k) = 3(-9) + 3(3) + 3(9) = -27 + 9 + 27 = 9.
  6. Substitute into the shortest distance formula d = | ((a2 - a1) · (b1 × b2)) / |b1 × b2| |:
    d = | 9 / (3√19) | = 3 / √19 = (3√19) / 19 units.

Section E: Case-Study Question (4 Marks)

Question 6 (Competency Based - Probability): In an engineering diagnostic facility, three machines M1, M2, and M3 produce 20%, 30%, and 50% of the total component output respectively. The percentage of defective components produced by these machines are 1%, 2%, and 3% respectively. A component is drawn at random from the output batch.

  • (i) Find the total probability that the selected component is defective. [2 Marks]
  • (ii) If the component drawn is found to be defective, what is the probability that it was manufactured by machine M3? [2 Marks]

Solution:

  1. Define events:
    E1: Component produced by M1 → P(E1) = 0.20 = 2/10
    E2: Component produced by M2 → P(E2) = 0.30 = 3/10
    E3: Component produced by M3 → P(E3) = 0.50 = 5/10
    A: Component is defective.
    P(A|E1) = 0.01 = 1/100,   P(A|E2) = 0.02 = 2/100,   P(A|E3) = 0.03 = 3/100.
  2. (i) Total Probability P(A):
    P(A) = P(E1)P(A|E1) + P(E2)P(A|E2) + P(E3)P(A|E3)
    P(A) = (0.20 × 0.01) + (0.30 × 0.02) + (0.50 × 0.03)
    P(A) = 0.002 + 0.006 + 0.015 = 0.023 (or 23 / 1000).
  3. (ii) Probability P(E3|A) via Bayes' Theorem:
    P(E3|A) = [ P(E3) · P(A|E3) ] / P(A)
    P(E3|A) = (0.50 × 0.03) / 0.023 = 0.015 / 0.023 = 15 / 23 ≈ 0.652.

How to Prepare for CBSE Class 12 Mathematics

Securing a 95+ score in CBSE Class 12 Mathematics requires a structured revision plan and disciplined answer-writing habits:

  1. Master NCERT Textbook & Exemplar: Every standard question and miscellaneous exercise in NCERT forms the conceptual foundation of CBSE board papers. Do not skip miscellaneous examples in Integration and 3D Geometry.
  2. Adhere to the Marking Scheme: CBSE awards marks in increments of 0.5 to 1 mark for each correct step. Always write down standard formulas (e.g., Bayes' theorem formula, integration by parts formula) before substituting values.
  3. Time Management in Mock Tests: Allocate your 180 minutes systematically:
    • Section A (MCQs): 25 minutes
    • Section B (VSA): 20 minutes
    • Section C (SA): 35 minutes
    • Section D (LA): 50 minutes
    • Section E (Case Study): 30 minutes
    • Revision & Error-checking: 20 minutes
  4. Avoid Common Calculation Errors: Always check the sign (+/-) when computing determinants and cofactors, normalize direction ratios carefully, and never forget the arbitrary constant + C in indefinite integrals.

Where to Practice More

Practicing diverse question sets under timed conditions is the single most effective way to eliminate exam anxiety and build computational accuracy for your CBSE Class 12 exams. For unlimited, topic-wise sample papers, NCERT chapter tests, and board-standard simulated test series tailored to the latest CBSE 2025 and 2026 blueprint, visit qptool.theorify.in today and elevate your board exam preparation.

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  • Target Class: Class 12
  • Subject: Mathematics
  • Curriculum: CBSE Standard
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