CBSE Class 11 Maths Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 11 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, covering
TL;DR: This is a full-length CBSE Class 11 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, cov…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 11 Mathematics paper built to the board's own pattern: 38 questions across five sections for 80 marks in 3 hours, covering sets and functions, trigonometry, complex numbers, permutations and combinations, sequences and series, straight lines and conic sections, three-dimensional geometry, limits and statistics. Section A is objective and Section E is case-based, and every question is followed by a fully worked solution showing each step of the algebra.
Board: CBSE · Class: Class 11 · 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 = {1, 2, 3}, the number of subsets of A is (1 mark)
Answer: 8. A set with n elements has 2ⁿ subsets, and 2³ = 8.
Q2. If n(A) = 3 and n(B) = 4, then n(A × B) equals (1 mark)
Answer: 12. The cartesian product has n(A) × n(B) = 3 × 4 = 12 ordered pairs.
Q3. The domain of the function f(x) = 1/(x − 2) is (1 mark)
Answer: All real numbers except 2, written R − {2}, since the denominator vanishes at x = 2.
Q4. The value of sin 75° is (1 mark)
Answer: (√6 + √2)/4. Using sin(45° + 30°) = sin45 cos30 + cos45 sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
Q5. The general solution of sin x = 0 is (1 mark)
Answer: x = nπ, where n is any integer.
Q6. The value of i⁴ⁿ⁺³ for any positive integer n is (1 mark)
Answer: −i. Since i⁴ = 1, i⁴ⁿ⁺³ = i³ = −i.
Q7. The modulus of the complex number 3 + 4i is (1 mark)
Answer: 5. |z| = √(3² + 4²) = √25 = 5.
Q8. The solution set of the inequality 2x + 3 < 7 is (1 mark)
Answer: x < 2, that is the interval (−∞, 2).
Q9. The number of ways of arranging the letters of the word 'MATHS' is (1 mark)
Answer: 120. All five letters are distinct, so the count is 5! = 120.
Q10. The value of ⁿC₀ + ⁿC₁ + … + ⁿCₙ is (1 mark)
Answer: 2ⁿ. Putting x = 1 in the binomial expansion of (1 + x)ⁿ gives exactly this sum.
Q11. The 10th term of the AP 3, 7, 11, … is (1 mark)
Answer: 39. aₙ = a + (n − 1)d = 3 + 9 × 4 = 39.
Q12. The sum of an infinite GP with first term 4 and common ratio 1/2 is (1 mark)
Answer: 8. S∞ = a/(1 − r) = 4/(1 − 0.5) = 8, valid because |r| < 1.
Q13. The slope of the line joining (2, 3) and (4, 7) is (1 mark)
Answer: 2. m = (y₂ − y₁)/(x₂ − x₁) = (7 − 3)/(4 − 2) = 4/2 = 2.
Q14. The equation of a circle with centre (0, 0) and radius 5 is (1 mark)
Answer: x² + y² = 25 — the standard form x² + y² = r² with r = 5.
Q15. The eccentricity of a parabola is (1 mark)
Answer: Exactly 1. An ellipse has e < 1 and a hyperbola e > 1.
Q16. The distance of the point (1, 2, 3) from the origin is (1 mark)
Answer: √14. √(1² + 2² + 3²) = √(1 + 4 + 9) = √14.
Q17. The value of lim(x→0) (sin x)/x is (1 mark)
Answer: 1 — a standard limit, with x measured in radians.
Q18. The derivative of x⁵ with respect to x is (1 mark)
Answer: 5x⁴, by the power rule d(xⁿ)/dx = nxⁿ⁻¹.
Q19. Assertion (A): The empty set is a subset of every set. Reason (R): A set A is a subset of B if every element of A belongs to B. (1 mark)
Answer: (a) Both true and R explains A. The empty set has no elements, so the condition in R is satisfied vacuously for any B.
Q20. Assertion (A): The function f(x) = x² is one-one on the set of real numbers. Reason (R): A function is one-one if distinct inputs give distinct outputs. (1 mark)
Answer: (d) A is false, R is true. The definition in R is correct, but f(−2) = f(2) = 4, so x² is not one-one on all of R — it is one-one only when restricted to non-negative reals.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. If A = {1, 2, 3, 4} and B = {3, 4, 5}, find A ∪ B and A ∩ B. (2 marks)
Answer: The union collects every element of either set without repetition: A ∪ B = {1, 2, 3, 4, 5}. The intersection keeps only the common elements: A ∩ B = {3, 4}.
Q2. Prove that sin²x + cos²x = 1 using the unit circle. (2 marks)
Answer: Take a point P on the unit circle at angle x from the positive x-axis; by definition its coordinates are (cos x, sin x). The distance of P from the origin is the radius, 1, so by the distance formula cos²x + sin²x = 1² = 1, which is the required identity.
Q3. Find the conjugate and modulus of z = 2 − 5i. (2 marks)
Answer: The conjugate is obtained by reversing the sign of the imaginary part: z̄ = 2 + 5i. The modulus is |z| = √(2² + (−5)²) = √(4 + 25) = √29.
Q4. How many three-digit numbers can be formed from the digits 1, 2, 3, 4, 5 without repetition? (2 marks)
Answer: The hundreds place can be filled in 5 ways, the tens in 4 ways from those left, and the units in 3 ways, giving 5 × 4 × 3 = 60. Equivalently this is the permutation ⁵P₃ = 60.
Q5. Find the equation of the line passing through (1, 2) with slope 3. (2 marks)
Answer: Using the point-slope form y − y₁ = m(x − x₁): y − 2 = 3(x − 1), so y − 2 = 3x − 3 and the equation is y = 3x − 1, or 3x − y − 1 = 0.
Section C — Short Answer (3 marks each) — 18 marks
Answer all 6 questions. Each carries 3 marks.
Q1. In a class of 60 students, 30 play cricket, 25 play football and 10 play both. Find how many play neither. (3 marks)
Answer: By the inclusion–exclusion principle, the number playing at least one game is n(C ∪ F) = n(C) + n(F) − n(C ∩ F) = 30 + 25 − 10 = 45. Therefore those who play neither number 60 − 45 = 15.
Q2. Prove that cos 3x = 4cos³x − 3cos x. (3 marks)
Answer: Write cos 3x = cos(2x + x) = cos 2x cos x − sin 2x sin x. Substituting cos 2x = 2cos²x − 1 and sin 2x = 2 sin x cos x gives cos 3x = (2cos²x − 1)cos x − 2 sin²x cos x. Replacing sin²x by 1 − cos²x, this becomes 2cos³x − cos x − 2cos x(1 − cos²x) = 2cos³x − cos x − 2cos x + 2cos³x = 4cos³x − 3cos x.
Q3. Solve the quadratic equation x² + 4 = 0 in the set of complex numbers. (3 marks)
Answer: Rearranging, x² = −4. Since i² = −1, we may write −4 = 4i², so x² = 4i² and x = ±2i. The roots are 2i and −2i, a conjugate pair, as expected for a real quadratic with negative discriminant.
Q4. Find the sum of the first 20 terms of the AP 5, 9, 13, … (3 marks)
Answer: Here a = 5 and d = 4, with n = 20. Using Sₙ = n/2[2a + (n − 1)d], S₂₀ = 10[10 + 19 × 4] = 10[10 + 76] = 10 × 86 = 860.
Q5. Find the coordinates of the focus and the length of the latus rectum of the parabola y² = 12x. (3 marks)
Answer: Comparing with the standard form y² = 4ax gives 4a = 12, so a = 3. The parabola opens rightwards, so the focus is at (a, 0) = (3, 0), the directrix is the line x = −3, and the length of the latus rectum is 4a = 12 units.
Q6. Evaluate lim(x→2) (x² − 4)/(x − 2). (3 marks)
Answer: Direct substitution gives 0/0, so factorise the numerator: x² − 4 = (x − 2)(x + 2). For x ≠ 2 the expression simplifies to x + 2, and the limit is the value this approaches as x → 2, namely 4. The cancellation is valid because a limit concerns values near 2 and not at 2 itself.
Section D — Long Answer (5 marks each) — 20 marks
Answer all 4 questions. Each carries 5 marks.
Q1. Prove by the principle of mathematical induction that 1 + 2 + 3 + … + n = n(n + 1)/2 for all natural numbers n. (5 marks)
Answer: Let P(n) be the statement that the sum of the first n natural numbers is n(n + 1)/2. Base step: for n = 1 the left side is 1 and the right side is 1 × 2/2 = 1, so P(1) holds. Inductive step: assume P(k) is true, that is 1 + 2 + … + k = k(k + 1)/2. Adding (k + 1) to both sides gives 1 + 2 + … + k + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)[k/2 + 1] = (k + 1)(k + 2)/2. This is precisely the statement P(k + 1). Since P(1) is true and P(k) implies P(k + 1), by the principle of mathematical induction P(n) holds for every natural number n.
Q2. Find the sum of the series 1 + 3 + 5 + … up to n terms, and hence find the sum of the first 25 odd numbers. (5 marks)
Answer: The series is an AP with first term a = 1 and common difference d = 2. Using Sₙ = n/2[2a + (n − 1)d], Sₙ = n/2[2 + 2(n − 1)] = n/2[2n] = n². So the sum of the first n odd numbers is n², a neat result worth remembering. For the first 25 odd numbers the sum is 25² = 625. As a check, the 25th odd number is 1 + 24 × 2 = 49, and the average of the first and last terms is (1 + 49)/2 = 25, which multiplied by 25 terms gives 625.
Q3. Find the equation of the circle passing through the points (0, 0), (4, 0) and (0, 6). (5 marks)
Answer: Take the general equation x² + y² + 2gx + 2fy + c = 0. Substituting (0, 0) gives c = 0. Substituting (4, 0): 16 + 8g = 0, so g = −2. Substituting (0, 6): 36 + 12f = 0, so f = −3. The equation is therefore x² + y² − 4x − 6y = 0. Completing the squares, (x − 2)² + (y − 3)² = 4 + 9 = 13, so the circle has centre (2, 3) and radius √13. As a check, the distance from (2, 3) to the origin is √(4 + 9) = √13, as required.
Q4. Calculate the mean and standard deviation of the data 6, 8, 10, 12, 14. (5 marks)
Answer: The mean is x̄ = (6 + 8 + 10 + 12 + 14)/5 = 50/5 = 10. The deviations from the mean are −4, −2, 0, 2 and 4, and their squares are 16, 4, 0, 4 and 16, summing to 40. The variance is σ² = Σ(x − x̄)²/n = 40/5 = 8, so the standard deviation is σ = √8 = 2√2 ≈ 2.83. The data are evenly spaced about the mean, which is why the deviations are symmetric and the mean sits exactly at the middle value.
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 school of 200 students was surveyed on reading habits. 120 read newspapers, 90 read magazines and 40 read both. (i) How many read at least one? (ii) How many read only newspapers? (iii) How many read neither? (4 marks)
Answer: (i) n(N ∪ M) = n(N) + n(M) − n(N ∩ M) = 120 + 90 − 40 = 170 students read at least one. (ii) Those reading only newspapers are n(N) − n(N ∩ M) = 120 − 40 = 80. (iii) Those reading neither are 200 − 170 = 30 students.
Q2. CASE STUDY 2 — A ball is dropped from a height of 80 m and on each bounce rises to half the height from which it fell. (i) Identify the type of progression formed by the heights. (ii) Write its first term and common ratio. (iii) Find the height after the fourth bounce. (iv) Find the total distance the ball would travel before coming to rest. (4 marks)
Answer: (i) The successive heights form a geometric progression, since each is a fixed fraction of the one before. (ii) The first term is 80 m and the common ratio is 1/2. (iii) After the fourth bounce the height is 80 × (1/2)⁴ = 80/16 = 5 m. (iv) The ball first falls 80 m; thereafter it rises and falls through each bounce height. The total is 80 + 2(40 + 20 + 10 + …) = 80 + 2 × [40/(1 − 1/2)] = 80 + 2 × 80 = 240 m.
Q3. CASE STUDY 3 — A committee of 5 is to be formed from 6 boys and 4 girls. (i) In how many ways can it be formed with no restriction? (ii) In how many ways if it must contain exactly 2 girls? (iii) In how many ways if it must contain at least 3 boys? (4 marks)
Answer: (i) Choosing 5 from 10 gives ¹⁰C₅ = 252 ways. (ii) Choose 2 girls from 4 and 3 boys from 6: ⁴C₂ × ⁶C₃ = 6 × 20 = 120 ways. (iii) At least 3 boys means 3, 4 or 5 boys: ⁶C₃×⁴C₂ + ⁶C₄×⁴C₁ + ⁶C₅×⁴C₀ = 20×6 + 15×4 + 6×1 = 120 + 60 + 6 = 186 ways.
FAQs
What is the total marks for Class 11 Maths?
100 marks total: 80 for theory and 20 for internal assessment.
Which chapters are most important?
Complex numbers, sequences and series, trigonometry, and conic sections have significant weightage.
Are proofs necessary for trigonometric identities?
Yes, formal proofs of important identities are often asked and carry 3-4 marks.
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