CBSE Class 12 Mathematics Sample Paper 2026 Standard Set 1 with Solutions

Practicing with the CBSE Class 12 Mathematics Sample Paper is the single most effective way to understand the latest board examination blueprint, internal choice distribution, and step-marking scheme. Mathematics (Subject Code 041) demands rigorous conceptual clarity, high-speed calculation accuracy, and precise mathematical presentation. This complete CBSE Mathematics sample paper with solutions follows the latest 2025–2026 standard curriculum, featuring full step-by-step working for all sections from single-mark objective questions to comprehensive five-mark calculus and 3D geometry proofs.

Sample Paper Format and Examination Blueprint

The CBSE Class 12 Mathematics question paper is designed for a total duration of 3 Hours carrying a maximum weightage of 80 Marks (with the remaining 20 marks allocated to internal assessment and practical lab work). The paper contains 38 mandatory questions structured into five distinct sections:

There is no overall choice in the paper; however, internal choices are provided in 2 questions of Section B, 3 questions of Section C, 2 questions of Section D, and in the 2-mark sub-questions of 2 questions in Section E.

Complete Sample Paper with Step-by-Step Solutions

Section A (1 Mark Each — Objective & Assertion-Reason)

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

Solution:
For any square matrix A of order n, the standard property states that |adj(A)| = |A|^{n − 1}.
Here, order n = 3 and |A| = 5.
Therefore, |adj(A)| = 5^{3 − 1} = 5² = 25.
Correct Option: (b) 25

Question 2: Find the value of λ for which the vectors a⃗ = 2î − 3ĵ + 4k̂ and b⃗ = λî + 6ĵ − 8k̂ are collinear.
(a) 4
(b) −4
(c) 2
(d) −2

Solution:
Two vectors a⃗ = a₁î + a₂ĵ + a₃k̂ and b⃗ = b₁î + b₂ĵ + b₃k̂ are collinear if their direction ratios are proportional:
a₁ / b₁ = a₂ / b₂ = a₃ / b₃
2 / λ = −3 / 6 = 4 / (−8)
2 / λ = −1 / 2 ⇒ λ = −4.
Correct Option: (b) −4

Question 3: The value of ∫ e^x (tan x + sec² x) dx is:
(a) e^x sec² x + C
(b) e^x tan x + C
(c) e^x sec x + C
(d) −e^x tan x + C

Solution:
Using the standard integration identity ∫ e^x [f(x) + f'(x)] dx = e^x f(x) + C.
Let f(x) = tan x, then f'(x) = sec² x.
Therefore, ∫ e^x (tan x + sec² x) dx = e^x tan x + C.
Correct Option: (b) e^x tan x + C

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

Solution:
The highest order derivative present in the equation is d²y/dx², so the order is 2. Because the equation cannot be written as a polynomial in its derivatives due to the term sin(dy/dx), the degree is Not Defined.
Correct Option: (c) Order = 2, Degree = Not Defined

Question 5: If P(A) = 0.4, P(B) = 0.8, and P(B|A) = 0.6, then P(A ∪ B) is equal to:
(a) 0.96
(b) 0.24
(c) 0.56
(d) 0.72

Solution:
From the definition of conditional probability:
P(A ∩ B) = P(A) × P(B|A) = 0.4 × 0.6 = 0.24.
Using the addition theorem of probability:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(A ∪ B) = 0.4 + 0.8 − 0.24 = 1.20 − 0.24 = 0.96.
Correct Option: (a) 0.96

Question 6 (Assertion-Reason):
Assertion (A): The function f(x) = |x − 1| is continuous at x = 1 but not differentiable at x = 1.
Reason (R): The left-hand derivative of f(x) at x = 1 is −1 and the right-hand derivative of f(x) at x = 1 is 1.
(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:
For f(x) = |x − 1|:
Continuity: lim_{x → 1} |x − 1| = 0 = f(1), so it is continuous.
Differentiability at x = 1:
LHD = lim_{h → 0⁻} (|1 − h − 1| − 0) / (−h) = lim_{h → 0} h / (−h) = −1.
RHD = lim_{h → 0⁺} (|1 + h − 1| − 0) / h = lim_{h → 0} h / h = 1.
Since LHD ≠ RHD, the function is non-differentiable at x = 1. Hence, Reason correctly explains Assertion.
Correct Option: (a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Section B (2 Marks Each — Very Short Answer)

Question 7: Show that the relation R in the set of integers Z defined by R = {(a, b) : 5 divides (a − b)} is an equivalence relation.

Solution:
To establish that R is an equivalence relation, we verify three fundamental properties:

• Reflexivity: For any a ∈ Z, a − a = 0 = 5 × 0. Since 0 is divisible by 5, (a, a) ∈ R. Thus, R is reflexive.
• Symmetry: Let (a, b) ∈ R. Then a − b = 5k for some integer k.
  ⇒ b − a = −(a − b) = 5(−k), where −k ∈ Z.
  Since (b − a) is divisible by 5, (b, a) ∈ R. Thus, R is symmetric.
• Transitivity: Let (a, b) ∈ R and (b, c) ∈ R. Then a − b = 5k and b − c = 5m for integers k, m.
  Adding the equations: (a − b) + (b − c) = 5k + 5m ⇒ a − c = 5(k + m).
  Since (k + m) ∈ Z, (a − c) is divisible by 5 ⇒ (a, c) ∈ R. Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation on Z.

Question 8: Find the principal value of tan⁻¹(√3) − sec⁻¹(−2).

Solution:
1. Let tan⁻¹(√3) = θ₁, where θ₁ ∈ (−π/2, π/2).
Since tan(π/3) = √3, we have tan⁻¹(√3) = π/3.
2. Let sec⁻¹(−2) = θ₂, where θ₂ ∈ [0, π] − {π/2}.
Using the identity sec⁻¹(−x) = π − sec⁻¹(x):
sec⁻¹(−2) = π − sec⁻¹(2) = π − π/3 = 2π/3.
3. Substitute both values into the expression:
tan⁻¹(√3) − sec⁻¹(−2) = π/3 − 2π/3 = −π/3.

Question 9: Find the projection of the vector a⃗ = î + 3ĵ + 7k̂ on the vector b⃗ = 7î − ĵ + 8k̂.

Solution:
The scalar projection of vector a⃗ on vector b⃗ is given by the formula:
Projection = (a⃗ · b⃗) / |b⃗|
Calculate dot product a⃗ · b⃗:
a⃗ · b⃗ = (1)(7) + (3)(−1) + (7)(8) = 7 − 3 + 56 = 60.
Calculate magnitude |b⃗|:
|b⃗| = √(7² + (−1)² + 8²) = √(49 + 1 + 64) = √114.
Therefore, the projection is 60 / √114 (or 10√114 / 19).

Section C (3 Marks Each — Short Answer)

Question 10: Find the values of a and b such that the function f(x) defined below is continuous everywhere on R:

Solution:
For f(x) to be continuous everywhere, it must be continuous at x = 2 and x = 10.

Continuity at x = 2:
LHL = lim_{x → 2⁻} f(x) = 5
RHL = lim_{x → 2⁺} f(x) = lim_{x → 2⁺} (ax + b) = 2a + b
f(2) = 5
For continuity: LHL = RHL ⇒ 2a + b = 5   — (Equation 1)

Continuity at x = 10:
LHL = lim_{x → 10⁻} f(x) = lim_{x → 10⁻} (ax + b) = 10a + b
RHL = lim_{x → 10⁺} f(x) = 21
f(10) = 21
For continuity: LHL = RHL ⇒ 10a + b = 21   — (Equation 2)

Subtracting Equation (1) from Equation (2):
(10a + b) − (2a + b) = 21 − 5
8a = 16 ⇒ a = 2.
Substituting a = 2 into Equation (1):
2(2) + b = 5 ⇒ 4 + b = 5 ⇒ b = 1.
Final Answer: a = 2, b = 1

Question 11: Evaluate the indefinite integral ∫ [x / ((x − 1)(x² + 1))] dx.

Solution:
We resolve the integrand using partial fractions decomposition:
x / [(x − 1)(x² + 1)] = A / (x − 1) + (Bx + C) / (x² + 1)
Multiplying both sides by (x − 1)(x² + 1):
x = A(x² + 1) + (Bx + C)(x − 1)
x = (A + B)x² + (−B + C)x + (A − C)

Equating coefficients:
For x²: A + B = 0 ⇒ B = −A
For x: −B + C = 1 ⇒ A + C = 1
Constant term: A − C = 0 ⇒ C = A

Substituting C = A into A + C = 1 gives 2A = 1 ⇒ A = 1/2.
Then B = −1/2 and C = 1/2.

Now, perform integration:
∫ [x / ((x − 1)(x² + 1))] dx = (1/2) ∫ [1 / (x − 1)] dx − (1/2) ∫ [x / (x² + 1)] dx + (1/2) ∫ [1 / (x² + 1)] dx
= (1/2) ln|x − 1| − (1/4) ∫ [2x / (x² + 1)] dx + (1/2) tan⁻¹(x) + C
= (1/2) ln|x − 1| − (1/4) ln(x² + 1) + (1/2) tan⁻¹(x) + C

Question 12: Solve the linear differential equation dy/dx + y cot x = 2x + x² cot x, given that y(π/2) = 0.

Solution:
The given equation is in the standard first-order linear form dy/dx + P(x)y = Q(x), where P(x) = cot x and Q(x) = 2x + x² cot x.
1. Integrating Factor (I.F.):
I.F. = e^{∫ P(x) dx} = e^{∫ cot x dx} = e^{ln(sin x)} = sin x.
2. General Solution:
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
Recognizing that d/dx [x² sin x] = 2x sin x + x² cos x:
y sin x = x² sin x + C ⇒ y = x² + C cosec x.
3. Applying Initial Condition y(π/2) = 0:
0 = (π/2)² + C cosec(π/2) ⇒ 0 = π²/4 + C(1) ⇒ C = −π²/4.
Particular Solution: y = x² − (π²/4) cosec x.

Section D (5 Marks Each — Long Answer)

Question 13: Solve the following system of linear equations using the matrix inversion method:

Solution:
The given system can be represented in matrix form as AX = B, where:

A =

Step 1: Evaluate Determinant |A|
|A| = 1(12 − 5) − (−1)(9 − (−10)) + 2(−3 − 8)
|A| = 1(7) + 1(19) + 2(−11) = 7 + 19 − 22 = 4 ≠ 0.
Since |A| ≠ 0, A is non-singular and A⁻¹ exists with a unique solution X = A⁻¹B.

Step 2: Calculate Cofactors of Matrix A
C₁₁ = +(12 − 5) = 7
C₁₂ = −(9 + 10) = −19
C₁₃ = +(−3 − 8) = −11
C₂₁ = −(−3 + 2) = 1
C₂₂ = +(3 − 4) = −1
C₂₃ = −(−1 + 2) = −1
C₃₁ = +(5 − 8) = −3
C₃₂ = −(−5 − 6) = 11
C₃₃ = +(4 + 3) = 7

Step 3: Form Adjoint and Inverse Matrix
adj(A) = [C_ij]^T =

Step 4: Compute X = A⁻¹B
x = (1/4) [7(7) + 1(−5) + (−3)(12)] = (1/4) [49 − 5 − 36] = (1/4)(8) = 2.
y = (1/4) [−19(7) + (−1)(−5) + 11(12)] = (1/4) [−133 + 5 + 132] = (1/4)(4) = 1.
z = (1/4) [−11(7) + (−1)(−5) + 7(12)] = (1/4) [−77 + 5 + 84] = (1/4)(12) = 3.
Final Answer: x = 2, y = 1, z = 3

Question 14: Find the shortest distance between the pair of skew lines given by the vector equations:
r⃗ = (î + 2ĵ + 3k̂) + λ(î − 3ĵ + 2k̂)
r⃗ = (4î + 5ĵ + 6k̂) + μ(2î + 3ĵ + k̂)

Solution:
Comparing with r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂:
a⃗₁ = î + 2ĵ + 3k̂, b⃗₁ = î − 3ĵ + 2k̂
a⃗₂ = 4î + 5ĵ + 6k̂, b⃗₂ = 2î + 3ĵ + k̂

1. Compute (a⃗₂ − a⃗₁):
a⃗₂ − a⃗₁ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)k̂ = 3î + 3ĵ + 3k̂.

2. Compute (b⃗₁ × b⃗₂):
b⃗₁ × b⃗₂ =

3. Calculate magnitude |b⃗₁ × b⃗₂|:
|b⃗₁ × b⃗₂| = √[(−9)² + 3² + 9²] = √[81 + 9 + 81] = √171 = 3√19.

4. Calculate scalar product (a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂):
(3î + 3ĵ + 3k̂) · (−9î + 3ĵ + 9k̂) = 3(−9) + 3(3) + 3(9) = −27 + 9 + 27 = 9.

5. Shortest Distance Formula:
d = |(a⃗₂ − a⃗₁) · (b⃗₁ × b⃗₂)| / |b⃗₁ × b⃗₂|
d = |9| / (3√19) = 3 / √19 = (3√19 / 19) units.

Section E (4 Marks Each — Case-Study Integrated Questions)

Question 15 (Case Study 1 — Application of Derivatives):
An environmental engineering team is constructing an open-top rectangular water filtration reservoir with a square base of side length x meters and vertical height y meters. The required storage volume is fixed at 216 m³. The cost of material for the base is ₹100 per m², while the cost for the vertical side walls is ₹50 per m².

Solution:
(i) Given volume V = x²y = 216 ⇒ y = 216 / x².
Total Cost C(x) = (Area of Base × Rate) + (Area of 4 Vertical Sides × Rate)
C(x) = (x² × 100) + (4xy × 50) = 100x² + 200x(216 / x²)
C(x) = 100x² + 43200 / x.

(ii) To minimize cost, set dC/dx = 0:
dC/dx = 200x − 43200 / x² = 0
200x³ = 43200 ⇒ x³ = 216 ⇒ x = 6 meters.
Second derivative test:
d²C/dx² = 200 + 86400 / x³
At x = 6: d²C/dx² = 200 + 86400 / 216 = 200 + 400 = 600 > 0.
Since the second derivative is strictly positive, the cost is minimized when the base side is 6 meters (and height y = 216/36 = 6 meters).

(iii) Minimum total cost:
C(6) = 100(6)² + 43200 / 6 = 100(36) + 7200 = 3600 + 7200 = ₹10,800.

Question 16 (Case Study 2 — Probability and Bayes' Theorem):
An aerospace sensor manufacturing firm operates three precision robotic production lines: Line A, Line B, and Line C. Line A produces 50% of the daily component output, Line B produces 30%, and Line C produces 20%. Extensive testing records show that 2% of items from Line A, 3% from Line B, and 5% from Line C are defective. A component is drawn at random from a finished batch.

Solution:
Let E₁, E₂, E₃ denote the events that the component was manufactured by Line A, Line B, and Line C respectively. Let D be the event that the chosen component is defective.
P(E₁) = 0.50, P(E₂) = 0.30, P(E₃) = 0.20
P(D|E₁) = 0.02, P(D|E₂) = 0.03, P(D|E₃) = 0.05

(i) Total Probability of Defect:
P(D) = P(E₁)P(D|E₁) + P(E₂)P(D|E₂) + P(E₃)P(D|E₃)
P(D) = (0.50 × 0.02) + (0.30 × 0.03) + (0.20 × 0.05)
P(D) = 0.010 + 0.009 + 0.010 = 0.029 (or 2.9%).

(ii) Bayes' Theorem for Line B:
P(E₂|D) = [P(E₂) · P(D|E₂)] / P(D)
P(E₂|D) = (0.30 × 0.03) / 0.029 = 0.009 / 0.029 = 9 / 29 (≈ 0.3103 or 31.03%).

Difficulty Analysis & Chapter-Wise Weightage Breakdown

Analyzing the balance between direct computational items and competency-based questions helps students allocate revision time efficiently. The standard distribution across units aligns with official CBSE curriculum guidelines:

How to Practice with This Sample Paper

To maximize score gains from this CBSE practice paper, follow these structured simulation strategies:

• Simulate Exact 3-Hour Exam Conditions: Sit in an uninterrupted environment from 10:15 AM to 1:30 PM (utilizing the standard 15-minute question paper reading window to select your internal choice questions).
• Time Management Benchmark:
  • Section A (20 Marks): Finish within 25–30 minutes. Do not spend more than 90 seconds on any single MCQ.
  • Section B (10 Marks): Complete within 20 minutes.
  • Section C (18 Marks): Complete within 35 minutes.
  • Section D (20 Marks): Allocate 45 minutes for rigorous algebraic verification.
  • Section E (12 Marks): Spend 30 minutes reading scenarios and setting up objective models.
  • Revision: Reserve the remaining 15–20 minutes to double-check arithmetic signs and units.
• Master Step-Marking Protocols: In CBSE evaluation, full marks are awarded only when intermediate steps (formula declaration, substitution, intermediate simplifications, and concluding units) are explicitly shown. Avoid jumping directly to answers in 3-mark and 5-mark problems.
• Maintain an Error Log: Categorize mistakes made during mock tests into conceptual gaps, algebraic slip-ups, or timing bottlenecks, and revise targeted NCERT chapters immediately.

More Sample Papers & Practice Resources

Consistent practice with diverse difficulty levels is the key to securing 95+ in Class 12 Board Exams. Generate customized board-standard question papers, download chapter-wise NCERT question banks, and access full-length mock tests with step-by-step marking rubrics at the QPTool Question Paper Generator.

Explore more full-length sets and subject resources: Download additional CBSE sample paper standard sets and mock papers on Theorify QPTool today.