Class 12 Mathematics Practice Paper Set 10 with Solutions

Apna CareerApna CareerApna Career

Class 12 Mathematics · Practice Paper SET-10 with Solutions

CBSE PRACTICE PAPER 2026–27 · MATHEMATICS (SET – 10)

www.careerinformationportal.in

CLASS : XIISUBJECT : MATHEMATICSROLL NO. : ..................

Time : 3 Hrs.Session : 2026–27 · Full Syllabus · High LevelM.M. : 80

General Instructions

1. This question paper contains 38 questions. All questions are compulsory.
2. The paper is divided into five sections – A, B, C, D and E. The paper is of higher difficulty level (HOTS / competitive pattern).
3. Section A: Q. 1 to 18 are MCQs and Q. 19–20 are Assertion–Reason questions, 1 mark each.
4. Section B: Q. 21 to 25 are short answer questions of 2 marks each.
5. Section C: Q. 26 to 31 are long-type short answer questions of 3 marks each.
6. Section D: Q. 32 to 35 are long answer questions of 5 marks each.
7. Section E: Q. 36 to 38 are case-based questions of 4 marks each, with sub-parts.
8. There is no overall choice. Internal choice is provided in some questions of Sections B, C, D and E.
9. Use of calculator is not permitted. Draw neat figures wherever required.
10. The detailed Answer Key & Solutions are given at the end of this paper.

SECTION A (20 × 1 = 20 Marks)

Q. 1 to Q. 18 : Multiple Choice Questions (choose the correct option)

1. The number of onto functions from the set {1, 2, 3} to the set {a, b} is

(a) 2(b) 8(c) 6(d) 4

2. On the set Z, a binary operation * is defined by a * b = a + b − 3. The inverse of 4 is

(a) 2(b) −2(c) 3(d) 6

3. The value of sin⁻¹(cos(3π/5)) is

(a) π/10(b) 3π/10(c) −3π/10(d) −π/10

4. If sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2, then x + y + z is equal to

(a) 1(b) 3(c) 2(d) 0

5. If A =

cos α	sin α
−sin α	cos α

and A + A^T = I, then α (0 < α < π/2) is

(a) π/6(b) π/2(c) π/3(d) 2π/3

6. The number of all possible 2 × 2 matrices, each of whose entries is either 0 or 1, is

(a) 4(b) 8(c) 9(d) 16

7. If the points (a, b), (a′, b′) and (a − a′, b − b′) are collinear, then

(a) ab′ = a′b(b) ab = a′b′(c) aa′ = bb′(d) ab′ + a′b = 0

8. If f(x) = sin 5x3x for x ≠ 0 is continuous at x = 0, then f(0) is

(a) 3/5(b) 5/3(c) 5(d) 1

9. If y = log(x + √(x² + 1)), then dy/dx is

(a) 1x² + 1(b) x√(x² + 1)(c) 1√(x² + 1)(d) √(x² + 1)

10. The minimum value of f(x) = |x − 1| + |x − 3| is

(a) 2(b) 0(c) 1(d) 4

11. ∫ dxsin²x cos²x =

(a) tan x + cot x + C(b) cot x − tan x + C(c) −tan x − cot x + C(d) tan x − cot x + C

12. ∫₀^π/4 tan²x dx =

(a) π/4(b) 1 − π/4(c) 1 + π/4(d) π/4 − 1

13. The general solution of the differential equation dy/dx = x²y is

(a) y = Ce^x2(b) y = x³/3 + C(c) y = Ce^x3/3(d) y = Ce^−x3/3

14. If |a × b|² + |a · b|² = 144 and |a| = 4, then |b| is

(a) 4(b) 3(c) 9(d) 12

15. The direction cosines of the vector 2i − 3j + 6k are

(a) 2/7, −3/7, 6/7(b) 2, −3, 6(c) 2/49, −3/49, 6/49(d) 1/7, −1/7, 1/7

16. The distance of the origin from the plane 2x + 3y + 6z = 35 is

(a) 35(b) 7(c) 1/5(d) 5

17. If A and B are independent events with P(A) = 0.3 and P(B) = 0.6, then P(A′ ∩ B) is

(a) 0.18(b) 0.7(c) 0.42(d) 0.12

18. A random variable X takes the values 1, 2, 3 with probabilities 1/2, 1/3, 1/6 respectively. Then E(X) is

(a) 2(b) 5/3(c) 11/6(d) 1

Q. 19 and Q. 20 : Assertion–Reason Questions

Options: (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.

19. Assertion (A): If A is a square matrix such that A² = A, then |A| = 0 or |A| = 1.
Reason (R): |A²| = |A|² for every square matrix A.

20. Assertion (A): If A and B are independent events, then P(A ∩ B) = P(A)P(B).
Reason (R): Mutually exclusive events having non-zero probabilities are always independent.

SECTION B (5 × 2 = 10 Marks) — Short Answer Questions

21. Prove that 2 sin⁻¹(3/5) − tan⁻¹(17/31) = π/4.(2)

OR

Write in the simplest form : tan⁻¹√(1 − cos x)√(1 + cos x), 0 < x < π.

22. Differentiate x^√x with respect to x.(2)

OR

If x√(1 + y) + y√(1 + x) = 0 and x ≠ y, show that dy/dx = −1(1 + x)².

23. The volume of a sphere is increasing at the rate of 8π cm³/s. Find the rate at which its radius is increasing when the radius is 2 cm.(2)

OR

Show that f(x) = cos x is strictly decreasing on (0, π).

24. Evaluate ∫ log xx² dx.(2)

OR

Evaluate ∫ dx√(x + 1) + √x.

25. Find the scalar triple product [a b c] where a = i + j + k, b = i − j + k and c = i + j − k.(2)

OR

Show that the vectors a − b, b − c and c − a are always coplanar.

SECTION C (6 × 3 = 18 Marks) — Long-Type Short Answer Questions

26. Evaluate ∫ x + 1x(1 + x e^x) dx.(3)

27. Evaluate ∫₀^π x sin x cos²x dx.(3)

OR

Evaluate ∫_−π/2^π/2 (sin |x| + cos |x|) dx.

28. Solve the differential equation x dy/dx + 2y = x² log x, x > 0.(3)

29. A family has two children. Find the probability that both are boys, given that (i) at least one of the children is a boy, (ii) the elder child is a boy.(3)

30. Find the vector equation of the line passing through the point A(1, 2, −1) and parallel to the line 5x − 25 = 14 − 7y = 35z.(3)

31. If A =

1	2
3	4

and B =

5	6
7	8

, find the matrix X such that XA = B.(3)

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

32. Using integration, find the area of the region {(x, y) : 0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}.(5)

OR

Using integration, find the area of the region enclosed by the ellipse x²/9 + y²/4 = 1.

33. Solve the following system of equations using matrices:
2x + 3y + 10z = 4,   4x − 6y + 5z = 1,   6x + 9y − 20z = 2.(5)

OR

Show that the relation R on the set of real numbers defined by R = {(a, b) : a ≤ b²} is neither reflexive, nor symmetric, nor transitive.

34. The sum of two non-negative numbers is 24. Find the numbers if the product of the first number and the cube of the second number is maximum.(5)

OR

Find the local maximum and local minimum values and the absolute maximum and absolute minimum values of f(x) = sin 2x − x on the interval [−π/2, π/2].

35. Solve the following linear programming problem graphically: Minimise Z = 4x + 3y subject to 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.(5)

OR

Show that the lines x + 13 = y + 35 = z + 57 and x − 21 = y − 43 = z − 65 are coplanar. Also find the equation of the plane containing them.

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

36. Case Study – Melting ice ball. A spherical ball of ice of radius 3 cm melts so that its volume decreases at the constant rate of 2 cm³/min. Let r cm be the radius of the ball at time t minutes.

(i) Write the volume V of the ball in terms of r. (1)
(ii) Find the rate at which the radius is decreasing when r = 3 cm. (1)
(iii) Find the rate at which the surface area of the ball is decreasing when r = 3 cm. (2)

OR

(iii) Find the time taken for the ball (of radius 3 cm) to melt completely. (2)

37. Case Study – Cost function. A small workshop has fixed cost of ₹ 100. The marginal cost of producing x units is MC(x) = 6x + 4 (in ₹).

(i) How is the total cost C(x) related to MC(x)? Write it as an integral with a constant. (1)
(ii) Find C(x). (1)
(iii) Find the total cost of producing 10 units and the increase in total cost when the production is increased from 10 to 15 units. (2)

OR

(iii) Find the average cost per unit when 10 units are produced, and the cost of the 11th unit alone, i.e. C(11) − C(10). (2)

38. Case Study – Triangular plot. A farmer has a triangular plot of land with vertices A(1, 1), B(5, 2) and C(3, 6) (coordinates in hundreds of metres, so 1 unit² = 1 hectare).

(i) Write the area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) in determinant form. (1)
(ii) Find the area of the plot. (1)
(iii) A well is dug at P(k, 3). Find k if A, B and P are collinear. (2)

OR

(iii) Find the values of k for which the area of triangle ABP, with P(k, 3), is 5 square units. (2)

ANSWER KEY & SOLUTIONS · MATHEMATICS PRACTICE PAPER (SET – 10)

www.careerinformationportal.in

CLASS : XIISUBJECT : MATHEMATICSROLL NO. : ..................

Time : 3 Hrs.Class XII · Session : 2026–27M.M. : 80

SECTION A — Answer Key (1 mark each)

Q.1	(c)	Q.2	(a)	Q.3	(d)	Q.4	(b)
Q.5	(c)	Q.6	(d)	Q.7	(a)	Q.8	(b)
Q.9	(c)	Q.10	(a)	Q.11	(d)	Q.12	(b)
Q.13	(c)	Q.14	(b)	Q.15	(a)	Q.16	(d)
Q.17	(c)	Q.18	(b)	Q.19	(a)	Q.20	(c)

Hints for Q.1–20. (1) Total functions 2³ = 8, minus the 2 constant functions ⇒ 6. (2) Identity e: a + e − 3 = a ⇒ e = 3. For 4: 4 + y − 3 = 3 ⇒ y = 2. (3) cos(3π/5) = sin(π/2 − 3π/5) = sin(−π/10), and −π/10 ∈ [−π/2, π/2]. (4) Each sin⁻¹ is at most π/2, so x = y = z = 1 and the sum is 3. (5) A + A^T = 2cos α · I = I ⇒ cos α = ½ ⇒ α = π/3. (6) Four entries, 2 choices each: 2⁴ = 16. (7) The determinant |a b 1; a′ b′ 1; a − a′ b − b′ 1| = 0; R₃ → R₃ − R₁ + R₂ gives ab′ − a′b = 0. (8) lim sin 5x/(3x) = 5/3. (9) dy/dx = [1 + x/√(x² + 1)]/[x + √(x² + 1)] = 1/√(x² + 1). (10) The sum of distances from x to 1 and 3 is at least 2 (attained for 1 ≤ x ≤ 3). (11) 1/(sin²x cos²x) = sec²x + cosec²x. (12) ∫(sec²x − 1)dx = [tan x − x]₀^π/4 = 1 − π/4. (13) dy/y = x²dx ⇒ log y = x³/3 + c. (14) |a × b|² + |a·b|² = |a|²|b|² = 144 ⇒ |b|² = 9. (15) |vector| = √(4 + 9 + 36) = 7. (16) d = 35/√(4 + 9 + 36) = 35/7 = 5. (17) P(A′)P(B) = 0.7 × 0.6 = 0.42. (18) 1/2 + 2/3 + 3/6 = 5/3. (19) (a): A² = A ⇒ |A|² = |A| ⇒ |A|(|A| − 1) = 0; R explains A. (20) (c): A is the definition of independence (true); R is false, since for mutually exclusive events P(A ∩ B) = 0 ≠ P(A)P(B).

SECTION B — Solutions (2 marks each)

21. Let θ = sin⁻¹(3/5); tan θ = 3/4. tan 2θ = (3/2)/(1 − 9/16) = 24/7 (1). tan(2θ − tan⁻¹(17/31)) = (24/7 − 17/31)/(1 + (24/7)(17/31)) = (625/217)/(625/217) = 1 (½), so the expression is π/4 (½). OR: √(1 − cos x)/√(1 + cos x) = √(2sin²(x/2))/√(2cos²(x/2)) = tan(x/2) for 0 < x < π (1) ⇒ tan⁻¹(tan(x/2)) = x/2 (1).

22. y = x^√x ⇒ log y = √x log x (½). (1/y)y′ = log x/(2√x) + √x · (1/x) = (log x + 2)/(2√x) (1). dy/dx = x^√x(log x + 2)/(2√x) (½). OR: x√(1 + y) = −y√(1 + x) ⇒ x²(1 + y) = y²(1 + x) ⇒ (x − y)(x + y + xy) = 0 (1). Since x ≠ y: y(1 + x) = −x ⇒ y = −x/(1 + x) and dy/dx = −1/(1 + x)² (1).

23. V = (4/3)πr³ ⇒ dV/dt = 4πr² dr/dt (1). 8π = 4π(4) dr/dt ⇒ dr/dt = 1/2 cm/s (1). OR: f′(x) = −sin x (1); for x ∈ (0, π), sin x > 0 ⇒ f′(x) < 0, so f is strictly decreasing (1).

24. By parts: ∫log x · x⁻²dx = −(log x)/x + ∫(1/x)(1/x)dx (1) = −(log x + 1)/x + C (1). OR: 1/(√(x + 1) + √x) = √(x + 1) − √x (1) ⇒ (2/3)[(x + 1)^3/2 − x^3/2] + C (1).

25. [a b c] =

1	1	1
1	−1	1
1	1	−1

= 1(1 − 1) − 1(−1 − 1) + 1(1 + 1) (1) = 0 + 2 + 2 = 4 (1). OR: [a − b, b − c, c − a] = (a − b)·[(b − c) × (c − a)] = (a − b)·(b×c − b×a + c×a) (1); expanding, every term cancels in pairs, so the triple product is 0 ⇒ coplanar (1).

SECTION C — Solutions (3 marks each)

26. Multiply numerator and denominator by e^x: I = ∫(x + 1)e^x/[xe^x(1 + xe^x)]dx (1). Put xe^x = t ⇒ dt = (x + 1)e^xdx (1). I = ∫dt/[t(1 + t)] = ∫[1/t − 1/(1 + t)]dt = log|t/(1 + t)| + C = log|xe^x/(1 + xe^x)| + C (1).

27. I = ∫₀^πx sin x cos²x dx = ∫₀^π(π − x) sin x cos²x dx (1). Adding: 2I = π∫₀^πsin x cos²x dx = π[−cos³x/3]₀^π = 2π/3 (1). So I = π/3 (1).

OR

sin|x| + cos|x| is an even function (1). I = 2∫₀^π/2(sin x + cos x)dx (1) = 2[−cos x + sin x]₀^π/2 = 2(1 + 1) = 4 (1).

28. dy/dx + (2/x)y = x log x: linear with IF = e^{∫2/x dx} = x² (1). yx² = ∫x³ log x dx = (x⁴/4) log x − ∫x³/4 dx = (x⁴/4) log x − x⁴/16 + C (1½). So y = (x²/4) log x − x²/16 + C/x² (½).

29. Sample space {BB, BG, GB, GG} (elder child first) (½). (i) P(BB | at least one boy) = P(BB)/P(BB, BG, GB) = (1/4)/(3/4) = 1/3 (1½). (ii) P(BB | elder is a boy) = (1/4)/(2/4) = 1/2 (1).

30. Rewrite 5x − 25 = 14 − 7y = 35z by dividing by 35: (x − 5)/7 = (2 − y)/5 = z/1, i.e. (x − 5)/7 = (y − 2)/(−5) = z/1 (1½). The direction ratios are (7, −5, 1) (½). Required line: r = (i + 2j − k) + λ(7i − 5j + k) (1).

31. XA = B ⇒ X = BA⁻¹ (|A| = −2 ≠ 0) (½). A⁻¹ = (1/−2)

4	−2
−3	1

=

−2	1
3/2	−1/2

(1). X =

5	6
7	8

−2	1
3/2	−1/2

=

−10 + 9	5 − 3
−14 + 12	7 − 4

=

−1	2
−2	3

(1½). (Check: XA = [(5, 6), (7, 8)] ✓.)

SECTION D — Solutions (5 marks each)

32. x² + 1 = x + 1 ⇒ x = 0, 1 (1). For 0 ≤ x ≤ 1, x² + 1 ≤ x + 1; for 1 ≤ x ≤ 2, x + 1 ≤ x² + 1. So the region is bounded above by y = x² + 1 on [0, 1] and by y = x + 1 on [1, 2] (1½). Area = ∫₀¹(x² + 1)dx + ∫₁²(x + 1)dx (1) = (1/3 + 1) + [x²/2 + x]₁² = 4/3 + (4 − 3/2) (1) = 23/6 square units (½).

OR

y = (2/3)√(9 − x²) in the first quadrant; by symmetry Area = 4∫₀³(2/3)√(9 − x²)dx (2) = (8/3)[x/2 √(9 − x²) + (9/2) sin⁻¹(x/3)]₀³ (1½) = (8/3)(9π/4) = 6π square units (1½).

33. Put 1/x = u, 1/y = v, 1/z = w: 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2, i.e. AX = B with A =

2	3	10
4	−6	5
6	9	−20

(1). |A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200 ≠ 0 (1). Cofactors: C₁₁ = 75, C₁₂ = 110, C₁₃ = 72, C₂₁ = 150, C₂₂ = −100, C₂₃ = 0, C₃₁ = 75, C₃₂ = 30, C₃₃ = −24; adj A =

75	150	75
110	−100	30
72	0	−24

(1). X = (1/1200) adj A · (4, 1, 2)^T: u = (300 + 150 + 150)/1200 = 1/2; v = (440 − 100 + 60)/1200 = 1/3; w = (288 + 0 − 48)/1200 = 1/5 (1½). Hence x = 2, y = 3, z = 5 (½).

OR

Not reflexive: for a = 1/2, a ≤ a² means 1/2 ≤ 1/4, which is false, so (1/2, 1/2) ∉ R (1½). Not symmetric: (1, 4) ∈ R as 1 ≤ 16, but (4, 1) ∉ R as 4 ≤ 1 is false (1½). Not transitive: (10, 4) ∈ R (10 ≤ 16) and (4, 2) ∈ R (4 ≤ 4), but (10, 2) ∉ R (10 ≤ 4 is false) (2).

34. Let the numbers be x and y with x + y = 24, and let P = xy³ = (24 − y)y³ = 24y³ − y⁴ (1½). dP/dy = 72y² − 4y³ = 4y²(18 − y) = 0 ⇒ y = 18 (y = 0 gives P = 0) (1½). d²P/dy² = 144y − 12y² = 12y(12 − y) < 0 at y = 18 ⇒ maximum (1). The numbers are x = 6, y = 18 (1).

OR

f′(x) = 2cos 2x − 1 = 0 ⇒ 2x = ±π/3 ⇒ x = ±π/6 (1). f″(x) = −4 sin 2x: f″(π/6) = −2√3 < 0 ⇒ local maximum f(π/6) = √3/2 − π/6 ≈ 0.342; f″(−π/6) > 0 ⇒ local minimum f(−π/6) = π/6 − √3/2 ≈ −0.342 (2). End points: f(π/2) = −π/2, f(−π/2) = π/2 (1). Comparing: absolute maximum = π/2 (at x = −π/2), absolute minimum = −π/2 (at x = π/2) (1).

35. Draw 2x + y = 8 (through (4, 0), (0, 8)) and x + 2y = 10 (through (10, 0), (0, 5)); they meet at (2, 4). The feasible region is unbounded with corner points A(0, 8), B(2, 4) and C(10, 0) (2). Z = 4x + 3y: A = 24, B = 8 + 12 = 20, C = 40 (1½). The smallest value is 20 at B. As the region is unbounded, the open half-plane 4x + 3y < 20 has no point in common with the feasible region, so the minimum value of Z is 20 at (2, 4) (1½).

OR

a₁ = (−1, −3, −5), a₂ = (2, 4, 6); b₁ = (3, 5, 7), b₂ = (1, 3, 5). a₂ − a₁ = (3, 7, 11) (1). Determinant

3	7	11
3	5	7
1	3	5

= 3(25 − 21) − 7(15 − 7) + 11(9 − 5) = 12 − 56 + 44 = 0 ⇒ the lines are coplanar (2). Normal = b₁ × b₂ = (4, −8, 4) ∝ (1, −2, 1). Plane through (−1, −3, −5): (x + 1) − 2(y + 3) + (z + 5) = 0 ⇒ x − 2y + z = 0 (2).

SECTION E — Solutions (4 marks each)

36. (i) V = (4/3)πr³ (1). (ii) dV/dt = 4πr² dr/dt = −2; at r = 3: dr/dt = −2/(36π) = −1/(18π) cm/min (decreasing at 1/(18π) cm/min) (1). (iii) S = 4πr² ⇒ dS/dt = 8πr dr/dt = 8π(3)(−1/(18π)) = −4/3 cm²/min, i.e. decreasing at 4/3 cm²/min (2). OR (iii): Initial volume = (4/3)π(27) = 36π cm³ (1); time = 36π/2 = 18π minutes (≈ 56.5 min) (1).

37. (i) C(x) = ∫MC(x) dx + K, where K is the fixed cost (1). (ii) C(x) = ∫(6x + 4)dx + K = 3x² + 4x + K; C(0) = 100 ⇒ K = 100. So C(x) = 3x² + 4x + 100 (1). (iii) C(10) = 300 + 40 + 100 = ₹ 440 (1). Increase from 10 to 15 units = ∫₁₀¹⁵(6x + 4)dx = [3x² + 4x]₁₀¹⁵ = (675 + 60) − (300 + 40) = ₹ 395 (1). OR (iii): Average cost = 440/10 = ₹ 44 (1); C(11) = 363 + 44 + 100 = 507, so C(11) − C(10) = ₹ 67 (1).

38. (i) Area = ½|

x1	y1	1
x2	y2	1
x3	y3	1

| (1). (ii)

1	1	1
5	2	1
3	6	1

= 1(2 − 6) − 1(5 − 3) + 1(30 − 6) = −4 − 2 + 24 = 18, so area = ½ × 18 = 9 hectares (1). (iii)

1	1	1
5	2	1
k	3	1

= 1(2 − 3) − 1(5 − k) + 1(15 − 2k) = 9 − k (1). A, B, P collinear ⇒ 9 − k = 0 ⇒ k = 9 (1). OR (iii): ½|9 − k| = 5 ⇒ |9 − k| = 10 (1) ⇒ k = −1 or k = 19 (1).