CBSE Class 12 Maths Sample Paper 2027 with Solutions PDF — Complete Guide

In short: CBSE Class 12 Maths Sample Paper 2027 has 38 questions across 5 sections — 80 marks in 3 hours. Calculus (Integration + Differentiation) carries 35 marks (44%), making it the highest-weightage unit. This guide covers complete solutions with step-by-step working.

Paper Pattern 2027

Section-wise Breakdown

Chapter-wise Weightage

High-Weightage Topics

Section A — MCQs (20 × 1 = 20 Marks)

Relations and Functions

Q1. Let f: R → R be defined by f(x) = x². Then f is:

• (a) One-one but not onto
• (b) Onto but not one-one
• (c) Neither one-one nor onto
• (d) Both one-one and onto

Solution:

• Not one-one: f(2) = f(-2) = 4 (different inputs, same output)
• Not onto: No x gives f(x) = -1 (negative numbers not in range)

Answer: (c) Neither one-one nor onto

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Q2. If A = {1, 2, 3} and relation R = {(1,1), (2,2), (3,3), (1,2), (2,1)}, then R is:

• (a) Reflexive only
• (b) Symmetric only
• (c) Reflexive and symmetric
• (d) Equivalence relation

Solution:

• Reflexive: (1,1), (2,2), (3,3) ∈ R ✓
• Symmetric: (1,2) ∈ R and (2,1) ∈ R ✓
• Transitive: (1,2), (2,1) ∈ R but (1,1) ∈ R ✓

Answer: (d) Equivalence relation

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Matrices and Determinants

Q3. If A is a 3×3 matrix such that |A| = 5, then |adj A| is:

• (a) 5
• (b) 25
• (c) 125
• (d) 1/5

Solution: For n×n matrix: |adj A| = |A|^(n-1) |adj A| = 5^(3-1) = 5² = 25

Answer: (b) 25

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Q4. If A = [2 3; 1 2], then A⁻¹ is:

• (a) [2 -3; -1 2]
• (b) [-2 3; 1 -2]
• (c) [2 3; 1 2]
• (d) Does not exist

Solution: |A| = 4 - 3 = 1 ≠ 0, so inverse exists. A⁻¹ = (1/|A|) × adj(A) adj(A) = [2 -3; -1 2] A⁻¹ = [2 -3; -1 2]

Answer: (a) [2 -3; -1 2]

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Calculus

Q5. ∫ eˣ(1 + tan x + tan²x) dx equals:

• (a) eˣ tan x + C
• (b) eˣ sec x + C
• (c) eˣ sec²x + C
• (d) eˣ cot x + C

Solution: 1 + tan²x = sec²x So integral = ∫ eˣ(tan x + sec²x) dx Using ∫ eˣ(f(x) + f'(x)) dx = eˣf(x) + C Here f(x) = tan x, f'(x) = sec²x

Answer: (a) eˣ tan x + C

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Q6. The value of ∫₀^(π/2) log(tan x) dx is:

• (a) 0
• (b) 1
• (c) π/4
• (d) π/2

Solution: Let I = ∫₀^(π/2) log(tan x) dx Using property: ∫₀^a f(x) dx = ∫₀^a f(a-x) dx I = ∫₀^(π/2) log(tan(π/2 - x)) dx = ∫₀^(π/2) log(cot x) dx I = ∫₀^(π/2) log(1/tan x) dx = -∫₀^(π/2) log(tan x) dx = -I 2I = 0, so I = 0

Answer: (a) 0

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Q7. If y = sin⁻¹x, then (1-x²)y₂ - xy₁ equals:

• (a) 0
• (b) 1
• (c) -1
• (d) 2

Solution: y = sin⁻¹x y₁ = 1/√(1-x²) √(1-x²) · y₁ = 1 Squaring: (1-x²)y₁² = 1 Differentiating: (1-x²)·2y₁y₂ + y₁²·(-2x) = 0 (1-x²)y₂ - xy₁ = 0

Answer: (a) 0

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Vectors and 3D

Q8. If |a⃗| = 3, |b⃗| = 4, and a⃗·b⃗ = 6, then |a⃗ × b⃗| is:

• (a) 6
• (b) 6√3
• (c) 12
• (d) 6√2

Solution: a⃗·b⃗ = |a⃗||b⃗|cos θ 6 = 3 × 4 × cos θ cos θ = 1/2, so θ = 60° sin θ = √3/2

|a⃗ × b⃗| = |a⃗||b⃗|sin θ = 3 × 4 × √3/2 = 6√3

Answer: (b) 6√3

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Q9. The direction cosines of the line joining (1, 2, 3) and (4, 6, 3) are:

• (a) 3/5, 4/5, 0
• (b) 3/5, 4/5, 1
• (c) 3, 4, 0
• (d) 1/3, 1/4, 0

Solution: Direction ratios: (4-1, 6-2, 3-3) = (3, 4, 0) Magnitude = √(9 + 16 + 0) = 5 Direction cosines = (3/5, 4/5, 0)

Answer: (a) 3/5, 4/5, 0

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Probability

Q10. If P(A) = 0.4, P(B) = 0.5, and P(A∩B) = 0.2, then P(A|B) is:

• (a) 0.4
• (b) 0.5
• (c) 0.2
• (d) 0.8

Solution: P(A|B) = P(A∩B)/P(B) = 0.2/0.5 = 0.4

Answer: (a) 0.4

Section B — Very Short Answer (5 × 2 = 10 Marks)

Q21. Find the principal value of tan⁻¹(√3) - sec⁻¹(-2).

Solution: tan⁻¹(√3) = π/3 (since tan(π/3) = √3) sec⁻¹(-2) = π - sec⁻¹(2) = π - π/3 = 2π/3

tan⁻¹(√3) - sec⁻¹(-2) = π/3 - 2π/3 = -π/3

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Q22. If A = [1 2; 3 4], find A² - 5A + 6I.

Solution: A² = [1 2; 3 4] × [1 2; 3 4] = [7 10; 15 22]

5A = [5 10; 15 20]

6I = [6 0; 0 6]

A² - 5A + 6I = [7-5+6 10-10+0; 15-15+0 22-20+6] = [8 0; 0 8] = 8I or [8 0; 0 8]

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Q23. Evaluate: ∫ (x + 1)/(x² + 2x + 3) dx

Solution: Let u = x² + 2x + 3 du = (2x + 2) dx = 2(x + 1) dx

∫ (x + 1)/(x² + 2x + 3) dx = (1/2) ∫ du/u = (1/2) ln|u| + C = (1/2) ln|x² + 2x + 3| + C

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Q24. Find the unit vector perpendicular to both a⃗ = 2î - ĵ + k̂ and b⃗ = î + ĵ - 2k̂.

Solution: a⃗ × b⃗ = |î ĵ k̂| |2 -1 1| |1 1 -2|

= î(2-1) - ĵ(-4-1) + k̂(2+1) = î + 5ĵ + 3k̂

|a⃗ × b⃗| = √(1 + 25 + 9) = √35

Unit vector = (î + 5ĵ + 3k̂)/√35

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Section C — Short Answer (6 × 3 = 18 Marks)

Q26. Prove that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b): |a - b| is even} is an equivalence relation.

Solution:

Reflexive: For any a ∈ A, |a - a| = 0 (even) So (a, a) ∈ R for all a. ✓

Symmetric: If (a, b) ∈ R, then |a - b| is even. Since |b - a| = |a - b|, |b - a| is also even. So (b, a) ∈ R. ✓

Transitive: If (a, b) ∈ R and (b, c) ∈ R, then |a - b| = 2m and |b - c| = 2n for some integers m, n. This means a - b and b - c are both even. So (a - b) + (b - c) = a - c is even. Hence |a - c| is even, so (a, c) ∈ R. ✓

R is an equivalence relation.

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Q27. Differentiate: y = (sin x)^(cos x) + (cos x)^(sin x)

Solution: Let u = (sin x)^(cos x) and v = (cos x)^(sin x) y = u + v

For u = (sin x)^(cos x): ln u = cos x · ln(sin x) (1/u)(du/dx) = -sin x · ln(sin x) + cos x · (cos x/sin x) du/dx = (sin x)^(cos x) [cos x · cot x - sin x · ln(sin x)]

For v = (cos x)^(sin x): ln v = sin x · ln(cos x) (1/v)(dv/dx) = cos x · ln(cos x) + sin x · (-sin x/cos x) dv/dx = (cos x)^(sin x) [cos x · ln(cos x) - sin x · tan x]

dy/dx = du/dx + dv/dx

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Q28. Evaluate: ∫ x² / (x⁴ + x² + 1) dx

Solution: Divide numerator and denominator by x²: = ∫ 1 / (x² + 1 + 1/x²) dx

Note: x² + 1/x² = (x - 1/x)² + 2

Let t = x - 1/x, then dt = (1 + 1/x²) dx

We need to manipulate: ∫ x² / (x⁴ + x² + 1) dx = (1/2) ∫ (1 + 1/x² - 1 + 1/x²) / (x² + 1 + 1/x²) dx

= (1/2) ∫ (1 + 1/x²) / ((x - 1/x)² + 3) dx - (1/2) ∫ (1 - 1/x²) / ((x + 1/x)² - 1) dx

Let t = x - 1/x for first integral, s = x + 1/x for second:

= (1/2) · (1/√3) tan⁻¹(t/√3) - (1/2) · (1/2) ln|(s-1)/(s+1)| + C

= (1/2√3) tan⁻¹((x² - 1)/(√3 x)) - (1/4) ln|(x² + x + 1)/(x² - x + 1)| + C

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Section D — Long Answer (4 × 5 = 20 Marks)

Q32. Solve the differential equation: (1 + y²) dx = (tan⁻¹y - x) dy

Solution: Rewrite as: dx/dy + x/(1 + y²) = tan⁻¹y/(1 + y²)

This is linear in x: dx/dy + Px = Q where P = 1/(1 + y²), Q = tan⁻¹y/(1 + y²)

Integrating Factor: IF = e^∫P dy = e^∫1/(1+y²) dy = e^(tan⁻¹y)

Solution: x · IF = ∫ Q · IF dy x · e^(tan⁻¹y) = ∫ (tan⁻¹y/(1 + y²)) · e^(tan⁻¹y) dy

Let t = tan⁻¹y, then dt = dy/(1 + y²) = ∫ t · eᵗ dt = eᵗ(t - 1) + C (by parts) = e^(tan⁻¹y)(tan⁻¹y - 1) + C

x = (tan⁻¹y - 1) + Ce^(-tan⁻¹y)

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Q33. Find the area of the region bounded by the parabola y² = 4x and the line y = 2x - 4.

Solution: Find intersection points: y² = 4x and y = 2x - 4 y² = 4 · (y + 4)/2 = 2(y + 4) y² = 2y + 8 y² - 2y - 8 = 0 (y - 4)(y + 2) = 0 y = 4 or y = -2

Points: (4, 4) and (1, -2)

Area = ∫₋₂⁴ [x_line - x_parabola] dy = ∫₋₂⁴ [(y + 4)/2 - y²/4] dy = ∫₋₂⁴ [(2y + 8 - y²)/4] dy = (1/4) [y² + 8y - y³/3]₋₂⁴ = (1/4) [(16 + 32 - 64/3) - (4 - 16 + 8/3)] = (1/4) [(48 - 64/3) - (-12 + 8/3)] = (1/4) [48 - 64/3 + 12 - 8/3] = (1/4) [60 - 72/3] = (1/4) [60 - 24] = (1/4) × 36 = 9 sq units

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Q34. Find the shortest distance between the lines: r⃗ = (1 - t)î + (t - 2)ĵ + (3 - 2t)k̂ r⃗ = (s + 1)î + (2s - 1)ĵ - (2s + 1)k̂

Solution: Line 1: a⃗₁ = î - 2ĵ + 3k̂, b⃗₁ = -î + ĵ - 2k̂ Line 2: a⃗₂ = î - ĵ - k̂, b⃗₂ = î + 2ĵ - 2k̂

a⃗₂ - a⃗₁ = (1-1)î + (-1+2)ĵ + (-1-3)k̂ = ĵ - 4k̂

b⃗₁ × b⃗₂ = |î ĵ k̂| |-1 1 -2| |1 2 -2|

= î(-2 + 4) - ĵ(2 + 2) + k̂(-2 - 1) = 2î - 4ĵ - 3k̂

|b⃗₁ × b⃗₂| = √(4 + 16 + 9) = √29

(a⃗₂ - a⃗₁) · (b⃗₁ × b⃗₂) = 0(2) + 1(-4) + (-4)(-3) = -4 + 12 = 8

Shortest distance = |8|/√29 = 8/√29 = 8√29/29 units

Section E — Case-Based Questions (3 × 4 = 12 Marks)

Case Study 1: Linear Programming

A furniture dealer deals in tables and chairs. He has ₹50,000 to invest and a storage space for 60 pieces. A table costs ₹2,500 and a chair costs ₹500. He can sell a table at a profit of ₹250 and a chair at a profit of ₹75.

Q36.

(i) Formulate the LPP for maximum profit. (1 mark)

Solution: Let x = number of tables, y = number of chairs

Maximize: Z = 250x + 75y (Profit)

Subject to:

• 2500x + 500y ≤ 50000 (Investment constraint)
• x + y ≤ 60 (Storage constraint)
• x ≥ 0, y ≥ 0 (Non-negativity)

(ii) Find the corner points of the feasible region. (1 mark)

Solution: Simplify constraints:

• 5x + y ≤ 100
• x + y ≤ 60

Corner points:

• O(0, 0)
• A(20, 0) — from 5x + y = 100, y = 0
• B(10, 50) — intersection of 5x + y = 100 and x + y = 60
• C(0, 60) — from x + y = 60, x = 0

(iii) Find the maximum profit. (2 marks)

Solution:

Maximum profit = ₹6,250 at (10, 50) i.e., 10 tables and 50 chairs.

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Case Study 2: Probability

In a factory, machines A, B, and C produce 25%, 35%, and 40% of the total output respectively. Of their outputs, 5%, 4%, and 2% are defective.

Q37.

(i) What is the probability that a randomly selected item is defective? (2 marks)

Solution: P(A) = 0.25, P(B) = 0.35, P(C) = 0.40 P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02

P(Defective) = P(A)·P(D|A) + P(B)·P(D|B) + P(C)·P(D|C) = 0.25 × 0.05 + 0.35 × 0.04 + 0.40 × 0.02 = 0.0125 + 0.014 + 0.008 = 0.0345 or 3.45%

(ii) If an item is found defective, what is the probability it was produced by machine B? (2 marks)

Solution: Using Bayes' Theorem: P(B|D) = P(B)·P(D|B) / P(D) = (0.35 × 0.04) / 0.0345 = 0.014 / 0.0345 = 14/34.5 = 28/69 ≈ 0.406

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Case Study 3: Vectors in Real Life

A pilot is flying from city A to city B. The displacement vector from A to B is 300î + 400ĵ km. Due to wind, the plane drifts and the actual displacement becomes 320î + 380ĵ km.

Q38.

(i) Find the magnitude of the intended displacement. (1 mark)

Solution: |AB⃗| = √(300² + 400²) = √(90000 + 160000) = √250000 = 500 km

(ii) Find the drift vector (difference between actual and intended). (1 mark)

Solution: Drift = Actual - Intended = (320î + 380ĵ) - (300î + 400ĵ) = 20î - 20ĵ km

(iii) Find the magnitude of drift and its direction. (2 marks)

Solution: |Drift| = √(20² + 20²) = √800 = 20√2 ≈ 28.28 km

Direction: tan θ = -20/20 = -1 θ = -45° (i.e., 45° south of east)

Drift is 20√2 km in the direction 45° south of east.

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