CBSE Class 12 Maths Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 12 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours, span
TL;DR: This is a full-length CBSE Class 12 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours,…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 12 Mathematics paper built to the current board blueprint: 38 questions across five sections for 80 marks in 3 hours, spanning relations and functions, matrices and determinants, calculus, vectors and three-dimensional geometry, linear programming and probability. Section A is objective, Section E is case-based, and every question is followed by a fully worked solution with each step of the algebra shown.
Board: CBSE · Class: Class 12 · Subject: Mathematics · Total: 80 marks · Time: 3 hours
Section A — Objective (1 mark each) — 20 marks
Questions 1–18 are multiple choice. Questions 19 and 20 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. If A is a square matrix of order 3 and |A| = 5, then |2A| equals (1 mark)
Answer: 40. For an n × n matrix, |kA| = kⁿ|A|, so |2A| = 2³ × 5 = 40.
Q2. If A is a 3 × 3 matrix with |A| = 4, then |adj A| equals (1 mark)
Answer: 16. |adj A| = |A|ⁿ⁻¹ = 4² = 16 for n = 3.
Q3. The principal value of sin⁻¹(−1/2) is (1 mark)
Answer: −π/6. The principal branch of sin⁻¹ is [−π/2, π/2], and sin(−π/6) = −1/2.
Q4. The value of tan⁻¹(1) + cos⁻¹(−1/2) is (1 mark)
Answer: 11π/12. tan⁻¹(1) = π/4 and cos⁻¹(−1/2) = 2π/3, so the sum is 3π/12 + 8π/12 = 11π/12.
Q5. The function f(x) = |x| is (1 mark)
Answer: Continuous everywhere but not differentiable at x = 0, where the left-hand derivative is −1 and the right-hand derivative is +1.
Q6. If y = log(sin x), then dy/dx equals (1 mark)
Answer: cot x. By the chain rule dy/dx = (1/sin x)·cos x = cot x.
Q7. The derivative of x^x with respect to x is (1 mark)
Answer: x^x(1 + log x). Taking logs, log y = x log x; differentiating, (1/y)y' = log x + 1, so y' = x^x(1 + log x).
Q8. The interval in which f(x) = x² − 4x + 6 is strictly increasing is (1 mark)
Answer: (2, ∞). f'(x) = 2x − 4 > 0 when x > 2.
Q9. ∫ (1/x) dx equals (1 mark)
Answer: log|x| + C. The modulus is required because the integrand is defined for negative x as well.
Q10. ∫₀^(π/2) sin x dx equals (1 mark)
Answer: 1. The antiderivative is −cos x, and [−cos x]₀^(π/2) = −0 − (−1) = 1.
Q11. The area bounded by the curve y = x², the x-axis and the lines x = 0 and x = 3 is (1 mark)
Answer: 9 square units. ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9.
Q12. The order and degree of the differential equation (d²y/dx²)³ + (dy/dx)² = 0 are (1 mark)
Answer: Order 2 and degree 3. The order is that of the highest derivative, and the degree is its power once the equation is polynomial in the derivatives.
Q13. The integrating factor of dy/dx + y = e^x is (1 mark)
Answer: e^x. For dy/dx + Py = Q the integrating factor is e^∫P dx, and here P = 1.
Q14. If vectors a and b are perpendicular, then (1 mark)
Answer: Their dot product is zero, a·b = 0, since a·b = |a||b|cos 90° = 0.
Q15. The magnitude of the vector 2i − 3j + 6k is (1 mark)
Answer: 7. √(4 + 9 + 36) = √49 = 7.
Q16. The angle between the lines with direction ratios (1, 0, 0) and (0, 1, 0) is (1 mark)
Answer: 90°. cos θ = (1·0 + 0·1 + 0·0)/(1 × 1) = 0, so θ = π/2.
Q17. If P(A) = 0.3, P(B) = 0.4 and A and B are independent, then P(A ∩ B) equals (1 mark)
Answer: 0.12. For independent events P(A ∩ B) = P(A)·P(B) = 0.3 × 0.4 = 0.12.
Q18. A die is thrown twice. The probability of getting a sum of 9 is (1 mark)
Answer: 1/9. The favourable outcomes are (3,6), (4,5), (5,4) and (6,3) — four out of 36, which is 1/9.
Q19. Assertion (A): The relation R on the set {1, 2, 3} given by R = {(1,1), (2,2), (3,3)} is an equivalence relation. Reason (R): A relation is an equivalence relation if it is reflexive, symmetric and transitive. (1 mark)
Answer: (a) Both true and R explains A. R contains every (a, a) so it is reflexive, and having no other pairs it is vacuously symmetric and transitive — it satisfies exactly the three conditions in the reason.
Q20. Assertion (A): The function f(x) = x³ has a local maximum at x = 0. Reason (R): f'(0) = 0 for this function. (1 mark)
Answer: (d) A is false, R is true. f'(0) = 0 indeed, but the sign of f' does not change at 0, so x = 0 is a point of inflection, not a maximum — a vanishing derivative alone does not establish an extremum.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Show that the relation R on the set of integers defined by aRb if a − b is even, is symmetric. (2 marks)
Answer: Suppose aRb, so a − b is even, say a − b = 2k for an integer k. Then b − a = −2k, which is also even, so bRa holds. Since aRb implies bRa for all integers a and b, the relation is symmetric.
Q2. Find dy/dx if y = sin⁻¹(2x√(1 − x²)). (2 marks)
Answer: Put x = sin θ, so √(1 − x²) = cos θ and 2x√(1 − x²) = 2 sin θ cos θ = sin 2θ. Then y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹x, and dy/dx = 2/√(1 − x²).
Q3. Evaluate ∫ x·e^x dx. (2 marks)
Answer: Integrate by parts with u = x and dv = e^x dx, so du = dx and v = e^x. Then ∫x e^x dx = x e^x − ∫e^x dx = x e^x − e^x + C = e^x(x − 1) + C.
Q4. Find a unit vector in the direction of a = 3i + 4j. (2 marks)
Answer: |a| = √(9 + 16) = 5, so the unit vector is â = a/|a| = (3i + 4j)/5, that is (3/5)i + (4/5)j.
Q5. If A and B are events with P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.2, find P(A | B). (2 marks)
Answer: P(A | B) = P(A ∩ B)/P(B) = 0.2/0.3 = 2/3 ≈ 0.667.
Section C — Short Answer (3 marks each) — 18 marks
Answer all 6 questions. Each carries 3 marks.
Q1. Prove that the function f(x) = 2x³ − 6x² + 6x + 5 is increasing for all real x. (3 marks)
Answer: f'(x) = 6x² − 12x + 6 = 6(x² − 2x + 1) = 6(x − 1)². A square is never negative, so f'(x) ≥ 0 for every real x, with equality only at the single point x = 1. A function whose derivative is non-negative throughout, vanishing only at isolated points, is increasing on the whole real line.
Q2. Evaluate ∫ (2x + 3)/((x + 1)(x + 2)) dx using partial fractions. (3 marks)
Answer: Write (2x + 3)/((x+1)(x+2)) = A/(x+1) + B/(x+2), so 2x + 3 = A(x+2) + B(x+1). Putting x = −1 gives 1 = A, and x = −2 gives −1 = −B, so B = 1. The integral is ∫dx/(x+1) + ∫dx/(x+2) = log|x+1| + log|x+2| + C = log|(x+1)(x+2)| + C.
Q3. Solve the differential equation dy/dx = (1 + y²)/(1 + x²). (3 marks)
Answer: The variables separate: dy/(1 + y²) = dx/(1 + x²). Integrating both sides gives tan⁻¹y = tan⁻¹x + C. Taking the tangent of both sides, the solution can also be written as y = tan(tan⁻¹x + C).
Q4. Find the area of the triangle with vertices A(1, 1, 1), B(1, 2, 3) and C(2, 3, 1) using vectors. (3 marks)
Answer: AB = (0, 1, 2) and AC = (1, 2, 0). Their cross product is AB × AC = i(1·0 − 2·2) − j(0·0 − 2·1) + k(0·2 − 1·1) = (−4, 2, −1). Its magnitude is √(16 + 4 + 1) = √21. The area of the triangle is half of this, √21/2 square units.
Q5. Find the shortest distance between the lines r = i + j + λ(2i − j + k) and r = 2i + j − k + μ(3i − 5j + 2k). (3 marks)
Answer: Here a₂ − a₁ = (1, 0, −1), b₁ = (2, −1, 1) and b₂ = (3, −5, 2). Now b₁ × b₂ = i((−1)(2) − (1)(−5)) − j((2)(2) − (1)(3)) + k((2)(−5) − (−1)(3)) = (3, −1, −7), with magnitude √(9 + 1 + 49) = √59. The shortest distance is |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂| = |3 + 0 + 7|/√59 = 10/√59 units.
Q6. A bag contains 4 red and 6 black balls. Two balls are drawn at random without replacement. Find the probability that both are red. (3 marks)
Answer: P(first red) = 4/10. After removing one red ball, 3 red remain among 9, so P(second red | first red) = 3/9. By the multiplication rule the probability that both are red is (4/10) × (3/9) = 12/90 = 2/15.
Section D — Long Answer (5 marks each) — 20 marks
Answer all 4 questions. Each carries 5 marks.
Q1. Solve the system of equations using matrices: x + 2y + z = 7, x + 3z = 11, 2x − 3y = 1. (5 marks)
Answer: Write AX = B with A = [[1,2,1],[1,0,3],[2,−3,0]], X = [x,y,z]ᵀ and B = [7,11,1]ᵀ. Expanding along the first row, |A| = 1(0·0 − 3·(−3)) − 2(1·0 − 3·2) + 1(1·(−3) − 0·2) = 9 + 12 − 3 = 18, which is non-zero, so a unique solution exists. Solving by elimination: from the second equation x = 11 − 3z. Substituting into the third, 2(11 − 3z) − 3y = 1 gives 22 − 6z − 3y = 1, so 3y = 21 − 6z and y = 7 − 2z. Substituting both into the first, (11 − 3z) + 2(7 − 2z) + z = 7 gives 11 − 3z + 14 − 4z + z = 7, so 25 − 6z = 7 and z = 3. Then x = 11 − 9 = 2 and y = 7 − 6 = 1. Hence x = 2, y = 1, z = 3.
Q2. Find the area of the region bounded by the parabola y² = 4x and the line y = 2x. (5 marks)
Answer: Find the intersections: substituting y = 2x into y² = 4x gives 4x² = 4x, so x² = x and x = 0 or x = 1. The points are (0, 0) and (1, 2). For a given x between 0 and 1 the parabola gives y = 2√x and the line gives y = 2x, and since √x ≥ x on this interval the parabola lies above. The area is ∫₀¹ (2√x − 2x) dx = [2·(2/3)x^(3/2) − x²]₀¹ = 4/3 − 1 = 1/3 square units.
Q3. A manufacturer produces two products A and B. Each unit of A needs 2 hours on machine I and 1 hour on machine II; each unit of B needs 1 hour on machine I and 2 hours on machine II. Machine I is available for 10 hours and machine II for 8 hours. Profit is ₹50 per unit of A and ₹40 per unit of B. Formulate as an LPP and find the maximum profit. (5 marks)
Answer: Let x units of A and y units of B be produced. Maximise Z = 50x + 40y subject to 2x + y ≤ 10, x + 2y ≤ 8, x ≥ 0 and y ≥ 0. The corner points of the feasible region are (0, 0), (5, 0), (0, 4) and the intersection of the two constraint lines. Solving 2x + y = 10 with x + 2y = 8: doubling the second gives 2x + 4y = 16, and subtracting the first gives 3y = 6, so y = 2 and x = 4. Evaluating Z: at (0,0) it is 0, at (5,0) it is 250, at (0,4) it is 160, and at (4,2) it is 200 + 80 = 280. The maximum profit is ₹280, obtained by making 4 units of A and 2 units of B.
Q4. A factory has three machines producing 25%, 35% and 40% of the total output, with defect rates of 5%, 4% and 2% respectively. An item drawn at random is found defective. Find the probability that it was produced by the third machine. (5 marks)
Answer: Let E₁, E₂ and E₃ be the events that the item came from machines 1, 2 and 3, with P(E₁) = 0.25, P(E₂) = 0.35 and P(E₃) = 0.40, and let D be the event that it is defective, with P(D|E₁) = 0.05, P(D|E₂) = 0.04 and P(D|E₃) = 0.02. By the law of total probability, P(D) = 0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.014 + 0.008 = 0.0345. By Bayes' theorem, P(E₃|D) = 0.008/0.0345 = 80/345 = 16/69 ≈ 0.232. So there is about a 23.2% chance the defective item came from the third machine.
Section E — Case-based questions (4 marks each) — 12 marks
Answer all 3 case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A window is in the shape of a rectangle surmounted by a semicircle, and its total perimeter is 10 m. Let the width of the rectangle be 2x and its height h. (i) Express the perimeter in terms of x and h. (ii) Express the area of the window in terms of x. (iii) Find the value of x that maximises the area. (4 marks)
Answer: (i) The perimeter consists of the two vertical sides, the base and the semicircular arc: P = 2h + 2x + πx = 10. (ii) From that, h = (10 − 2x − πx)/2. The area is the rectangle plus the semicircle: A = 2x·h + ½πx² = x(10 − 2x − πx) + ½πx² = 10x − 2x² − πx² + ½πx² = 10x − 2x² − ½πx². (iii) dA/dx = 10 − 4x − πx, which vanishes when x = 10/(4 + π) ≈ 1.40 m. Since d²A/dx² = −(4 + π) < 0, this gives a maximum.
Q2. CASE STUDY 2 — The number of bacteria in a culture grows at a rate proportional to the number present. The culture starts with 1000 bacteria and doubles in 2 hours. (i) Write the differential equation. (ii) Solve it for N(t). (iii) Find the growth constant k. (iv) How many bacteria are present after 6 hours? (4 marks)
Answer: (i) dN/dt = kN, where N is the number of bacteria and k the growth constant. (ii) Separating variables, dN/N = k dt, so log N = kt + C and N = N₀e^(kt), with N₀ = 1000. (iii) Doubling in 2 hours means 2000 = 1000e^(2k), so e^(2k) = 2, giving 2k = log 2 and k = (log 2)/2 ≈ 0.3466 per hour. (iv) After 6 hours — three doubling periods — N = 1000 × 2³ = 8000 bacteria.
Q3. CASE STUDY 3 — Two friends each throw a fair die once. (i) Write the sample space size. (ii) Find the probability that both get the same number. (iii) Find the probability that the sum is at least 10. (iv) Find the probability that at least one gets a six. (4 marks)
Answer: (i) Each die has 6 outcomes, so the sample space has 6 × 6 = 36 equally likely outcomes. (ii) The matching outcomes are (1,1) through (6,6), six in all, so the probability is 6/36 = 1/6. (iii) A sum of at least 10 arises from (4,6), (5,5), (6,4), (5,6), (6,5) and (6,6) — six outcomes, giving 6/36 = 1/6. (iv) It is easier to use the complement: neither shows a six in 5 × 5 = 25 outcomes, so P(at least one six) = 1 − 25/36 = 11/36.
FAQs
What is the total marks for Class 12 Maths board exam?
100 marks total: 80 marks for theory and 20 marks for internal assessment.
Which chapters have highest weightage?
Calculus (integration and differentiation), linear algebra, and probability have significant weightage.
Is a calculator allowed in the exam?
Non-programmable scientific calculators are allowed only for certain questions.
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