CBSE Class 12 Maths Sample Paper 2: marking scheme and solutions

Answer: (b) With R the event 'it rains' and L 'the bus is late', P(R ∩ L) = P(R) P(L|R) = 0.3 × 0.6 = 0.18. Option (a) adds; (c) divides; (d) uses P(R′) = 0.7.

Answer: (b) Below the x-axis every strip has y < 0, so ∫_a^b f(x) dx is negative and the area is its numerical value, |∫_a^b f(x) dx| = −∫_a^b f(x) dx. −∫_b^a f(x) dx equals ∫_a^b f(x) dx, which is negative.

Answer: (d) a · b = a₁b₁ + a₂b₂ + a₃b₃ = 6 − 8 − 1 = −3. Taking the k̂ term as +1 gives −1; taking (−2)(4) as +8 gives 13.

Answer: (a) P(E) = 1/2 and P(G) = 4/8 = 1/2. If G contains m even numbers, P(E ∩ G) = m/8. Independence needs m/8 = 1/2 × 1/2 = 1/4, so m = 2; for example G = {1, 2, 3, 4}. Option (b) would make G = E, with P(E ∩ G) = 1/2.

Answer: (c) Separating, dy/y = 2x dx/(1 + x²). Integrating: log y = log (1 + x²) + log C, so y = C(1 + x²). Option (a) forgets that the constant multiplies after taking exponentials.

Answer: (c) The classes are the six remainder classes on division by 6. Since 20 = 6 × 3 + 2, the remainders 1 and 2 occur four times ({1, 7, 13, 19}, {2, 8, 14, 20}) and the remainders 3, 4, 5, 0 three times. So 2 classes have 4 elements. 6 is the number of classes; 3 is the size of the others.

Answer: (d) A general point of the line is (3 + 2λ, −1 − 5λ, 4 + 6λ); λ = 1 gives (5, −6, 10). For (b) the three ratios are 0, −25, 0; (c) reads the direction ratios as a point; for (a) the ratios are 1, 35, 1.

Answer: (a) By definition a feasible region is bounded if it can be enclosed within a circle; otherwise it is unbounded. A region in the first quadrant, or one containing the origin, can still extend indefinitely, and a region with only two corner points can be unbounded.

Answer: (b) Theorem 3: differentiability at c implies continuity at c. The converse fails: |x| is continuous at 0 but not differentiable there, which rules out (a) and (c).

Answer: (b) If x₁ < x₂, then x₁³ < x₂³, so −2x₁³ > −2x₂³ and f(x₁) > f(x₂): f is strictly decreasing on R. f′(x) = −6x² is 0 at x = 0 only, which does not spoil strictness.

Answer: (d) For order n, |kA| = kⁿ|A|. Here n = 3, so |2A| = 2³ × 4 = 32. Writing 2|A| gives 8; 2²|A| gives 16.

Answer: (b) ∫ dxx² + a² = 1a tan^-1 xa + C with a = 5. Option (d) is the form for x² − 25.

Answer: (a) Z at the corners: (2, 1) → 11, (5, 1) → 23, (5, 4) → 32, (1, 3) → 13. For a bounded region the smallest corner value is the minimum: 11 at (2, 1). 13 is the value at (1, 3), the corner with the smallest x, and 32 is the maximum.

Answer: (a) Perpendicular means the scalar product is 0: 8 − λ − 6 = 0, so λ = 2. Adding −6 as +6 gives 14.

Answer: (b) The largest sum of two elements of A is 5 + 5 = 10, so no pair has a + b = 11 and R = φ. (a) holds for every pair, so it is the universal relation; (c) contains (5, 1); (d) contains (1, 5) and (5, 1).

Answer: (b) x = cos(5π/6) = −√3/2. sin⁻¹(−√3/2) must lie in [−π/2, π/2], and sin(−π/3) = −√3/2, so sin⁻¹ x = −π/3.

Answer: (d) By the multiplication rule for three events, P(E ∩ F ∩ G) = P(E) P(F|E) P(G|EF) = 0.9 × 0.8 × 0.75 = 0.54. Option (b) stops after two hubs; (c) multiplies the three complements.

Answer: (c) Perpendicular to the z-axis: γ = 90°, n = 0. l = cos 30° = √32. Then m² = 1 − 34 − 0 = 14 and m = 12 (acute angle). Option (b) swaps l and m; (d) puts n = cos 0°.

Answer: (a) f is a polynomial, and a polynomial is continuous at every real number: lim_(x→2) f(x) = 8 − 8 + 7 = 7 = f(2). R is true and is the reason A holds.

Answer: (b) A: F(λx, λy) = λ(y − x)/(λ(y + x)) = λ⁰ F(x, y), so F has degree zero and the equation is homogeneous. R: each term has degree 2, so R is true. But A rests on F being of degree zero, a fact about a different function; R does not explain A.

X and Y form a partition. By the theorem of total probability, P(G) = 0.7 × 0.8 + 0.3 × 0.6 = 0.56 + 0.18 = 0.74, so P(does not germinate) = 1 − 0.74 = 0.26. (Equally, 0.7 × 0.2 + 0.3 × 0.4 = 0.26.)

• P(G) = P(X) P(G|X) + P(Y) P(G|Y) set up (1)
• 0.74, so the answer is 0.26 (1)

Fiction (F) and non-fiction (N) form a partition. P(B) = P(F) P(B|F) + P(N) P(B|N) = 0.4 × 0.3 + 0.6 × 0.1 = 0.12 + 0.06 = 0.18.

• total probability set up with P(N) = 0.6 (1)
• 0.18 (1)

π/2 < 2 < π, and tan 2 = tan(2 − π) with 2 − π ∈ (−π/2, π/2), so tan⁻¹(tan 2) = 2 − π. 3π/2 < 5 < 2π, and cos 5 = cos(2π − 5) with 2π − 5 ∈ [0, π], so cos⁻¹(cos 5) = 2π − 5. Sum = (2 − π) + (2π − 5) = π − 3.

• tan⁻¹(tan 2) = 2 − π, with the branch stated (1)
• cos⁻¹(cos 5) = 2π − 5, and sum π − 3 (1)

|a − b|² = 25 + 9 − 2→a · b = 49 gives a · b = −152. Then |a + b|² = 25 + 9 + 2(−152) = 19, so |a + b| = √19.

• a · b = −152 (1)
• |a + b| = √19 (1)

dx/dt = e^t − e^−t = y and dy/dt = e^t + e^−t = x, so dy/dx = (dy/dt)/(dx/dt) = x/y (for t ≠ 0, where y ≠ 0).

• dx/dt = y and dy/dt = x (1)
• dy/dx = x/y (1)

For a skew symmetric matrix a_ji = −a_ij. From a₂₁ = 3: a₁₂ = 2x − 1 = −3, so x = −1. From a₃₂ = −5: a₂₃ = y + 2 = 5, so y = 3. From a₁₃ = 4: a₃₁ = z = −4. The diagonal elements are already 0.

• uses a_ji = −a_ij (1)
• x = −1, y = 3, z = −4 (1)

For a skew symmetric matrix a_ji = −a_ij, and the diagonal is zero. a + 1 = −4, so a = −5. b is a diagonal element, so b = 0. c − 3 = −6, so c = −3. The pair a₁₃ = −2, a₃₁ = 2 already agrees.

• uses a_ji = −a_ij and zero diagonal (1)
• a = −5, b = 0, c = −3 (1)

Maximise Z = 300x + 500y subject to x + y ≤ 3, y ≤ x, x ≥ 0, y ≥ 0. The line y = x passes through the origin; testing (1, 0) shows y ≤ x is the side below it. The bounded region has corners (0, 0), (3, 0) and (3/2, 3/2), where y = x meets x + y = 3. Z there: 0, 900, 1200. The maximum is 1200 kcal, by walking 1.5 hours and cycling 1.5 hours.

• formulation: Z = 300x + 500y; x + y ≤ 3, y ≤ x, x, y ≥ 0 (1)
• corners (0, 0), (3, 0), (3/2, 3/2) (1)
• maximum 1200 kcal at 1.5 h walking, 1.5 h cycling (1)

f(x) = 3x for x ≥ 0 and f(x) = x for x < 0; f(x) ≥ 0 when x ≥ 0 and f(x) < 0 when x < 0. One-one: if f(x₁) = f(x₂), the common value has one sign, so x₁, x₂ are on the same side of 0; then 3x₁ = 3x₂ or x₁ = x₂, giving x₁ = x₂. Onto: for y ≥ 0, x = y/3 ≥ 0 and f(x) = y; for y < 0, x = y < 0 and f(x) = y. So f is one-one and onto.

• f written piecewise, with the sign of f(x) noted (1)
• one-one by cases (1)
• onto: preimages y/3 and y (1)

f(x) = 2x for x ≥ 0 and f(x) = 0 for x < 0. f(−1) = f(−2) = 0, so f is not one-one. f(x) ≥ 0 for every x, so −1 is not the image of any x: f is not onto. On D = [0, ∞), f(x) = 2x: 2x₁ = 2x₂ gives x₁ = x₂, and every y ≥ 0 is f(y/2) with y/2 ∈ D, so f : [0, ∞) → [0, ∞) is one-one and onto.

• not one-one: f(−1) = f(−2) (1)
• not onto: negative numbers are not images (1)
• D = [0, ∞), with one-one and onto shown (1)

Let A, B, C be the hits; their complements are independent too. P(A B C′) = 1/2 × 1/3 × 3/4 = 3/24; P(A B′ C) = 1/2 × 2/3 × 1/4 = 2/24; P(A′ B C) = 1/2 × 1/3 × 1/4 = 1/24. These are mutually exclusive, so P(exactly two) = 6/24 = 1/4.

• the three mutually exclusive cases written (1)
• each probability by independence: 3/24, 2/24, 1/24 (1)
• sum 1/4 (1)

P(at least one valve works) = 1 − P(both fail) = 1 − 0.2 × 0.5 = 0.9, using the independence of the complements. The motor is independent of the valves, so P(pump works) = 0.9 × 0.9 = 0.81.

• P(both valves fail) = 0.2 × 0.5 = 0.1 (1)
• P(at least one valve works) = 0.9 (1)
• × 0.9 for the motor: 0.81 (1)

General points: (1 + 2λ, −1 + 3λ, 2 − λ) and (2 + μ, 4 − 2μ, −2 + 3μ). Equating x and y: 2λ − μ = 1 and 3λ + 2μ = 5, so λ = 1, μ = 1. Then z: 2 − 1 = 1 and −2 + 3 = 1, equal, so the lines intersect. The point is (3, 2, 1).

• general points of both lines (1)
• λ = 1, μ = 1 from two coordinates, third coordinate verified (1)
• point of intersection (3, 2, 1) (1)

Let each side perpendicular to the canal be w m; the side along the canal is 800/w m. Fencing L(w) = 2w + 800/w. L′(w) = 2 − 800/w² = 0 at w = 20 (w > 0). L″(w) = 1600/w³ > 0, a minimum. Dimensions 20 m by 40 m (40 m along the canal); least fencing 40 + 40 = 80 m.

• L(w) = 2w + 800/w (1)
• L′(w) = 0 at w = 20 with L″ > 0 (1)
• 20 m × 40 m, fencing 80 m (1)

Base side x m, depth 32/x² m. Area S(x) = x² + 4x(32/x²) = x² + 128/x. S′(x) = 2x − 128/x² = 0 gives x³ = 64, x = 4. S″(x) = 2 + 256/x³ > 0, a minimum. Base 4 m × 4 m, depth 2 m (sheet needed 48 m²).

• S(x) = x² + 128/x (1)
• S′(x) = 0 at x = 4 with S″ > 0 (1)
• base 4 m by 4 m, depth 2 m (1)

A = 12−12−11310, B = 316. |A| = 1(0 − 1) − 2(0 − 3) + (−1)(2 + 3) = −1 + 6 − 5 = 0, so A is singular. Cofactors: A₁₁ = −1, A₁₂ = 3, A₁₃ = 5, A₂₁ = −1, A₂₂ = 3, A₂₃ = 5, A₃₁ = 1, A₃₂ = −3, A₃₃ = −5, so adj A = −1−1133−355−5. (adj A)B = −3 − 1 + 69 + 3 − 1815 + 5 − 30 = 2−6−10 ≠ O, so the system is inconsistent.

• |A| = 0 (1)
• adj A (1)
• (adj A)B ≠ O, so no solution (1)

(a) y = (x − 3)(x + 1) meets the x-axis at x = −1 and x = 3; it is an upward parabola with vertex (1, −4), above the axis on (−2, −1) and (3, 4) and below it on (−1, 3). (b) F(x) = x³3 − x² − 3x: F(−2) = −23, F(−1) = 53, F(3) = −9, F(4) = −203. The pieces are 73, −323 and 73, so the area is 73 + 323 + 73 = 463 square units. (c) ∫₋₂⁴ = F(4) − F(−2) = −6 = 73 − 323 + 73: the integral counts the part below the x-axis as negative, while an area takes its numerical value.

• (a) zeros −1 and 3 and a sketch showing the vertex (1, −4) (1)
• anti derivative and the split at −1 and 3 (1)
• the three pieces 73, −323, 73 (1)
• (b) area 463 square units (1)
• (c) −6, with the reason that the part below the axis is subtracted (1)

(a) y = (x − 2)(x + 1) meets the x-axis at x = −1 and x = 2; vertex (12, −94); above the axis on (−2, −1) and (2, 3), below on (−1, 2). (b) F(x) = x³3 − x²2 − 2x: F(−2) = −23, F(−1) = 76, F(2) = −103, F(3) = −32. The pieces are 116, −92 and 116, so the area is 116 + 92 + 116 = 496 square units. (c) ∫₋₂³ = F(3) − F(−2) = −56: the part below the x-axis enters the integral as negative, while an area takes its numerical value.

• (a) zeros −1 and 2 and a sketch showing the vertex (12, −94) (1)
• anti derivative and the split at −1 and 2 (1)
• the three pieces 116, −92, 116 (1)
• (b) area 496 square units (1)
• (c) −56, with the reason (1)

(a) By parts with 1 as the second function: ∫ log (x² + 1) dx = x log (x² + 1) − ∫ 2x²x² + 1 dx = x log (x² + 1) − ∫ (2 − 2x² + 1) dx = x log (x² + 1) − 2x + 2 tan^-1 x + C. (b) With f(x) = cot x, f′(x) = − cosec² x, so the integrand is e^x[f(x) + f′(x)] and the integral is e^x cot x + C.

• (a) takes 1 as the second function and integrates by parts (1)
• (a) splits 2x²x² + 1 as 2 − 2x² + 1 (1)
• (a) x log (x² + 1) − 2x + 2 tan^-1 x + C (1)
• (b) identifies f(x) = cot x and f′(x) = − cosec² x (1)
• (b) e^x cot x + C (1)

(a) With x² first: ∫ x² cos x dx = x² sin x − ∫ 2x sin x dx, and ∫ 2x sin x dx = − 2x cos x + 2 sin x. So the integral is x² sin x + 2x cos x − 2 sin x + C. (b) The integrand has the form e^x[f(x) + f′(x)] with f(x) = cosec x, since f′(x) = − cosec x cot x; so the integral is e^x cosec x + C.

• (a) first integration by parts with x² as first function (1)
• (a) second integration by parts on 2x sin x (1)
• (a) x² sin x + 2x cos x − 2 sin x + C (1)
• (b) recognises e^x[f(x) + f′(x)] with f = cosec x (1)
• (b) e^x cosec x + C (1)

dy/dx = (x² + 3xy + y²)/x² = F(x, y). F(λx, λy) = λ²(x² + 3xy + y²)/(λ² x²) = λ⁰ F(x, y): homogeneous. Put y = vx: v + x dv/dx = 1 + 3v + v², so x dv/dx = v² + 2v + 1 = (v + 1)². Separating, dv/(v + 1)² = dx/x; integrating, −1/(v + 1) = log x + C. At x = 1, y = 0, so v = 0: −1 = C. Hence 1/(v + 1) = 1 − log x, v + 1 = 1/(1 − log x), and v = log x/(1 − log x). So y = x log x/(1 − log x).

• shows F(λx, λy) = λ⁰ F(x, y) (1)
• y = vx gives x dv/dx = (v + 1)² (1)
• −1v + 1 = log x + C (1)
• C = −1 from the condition (1)
• y = x log x1 − log x (1)

(a) →AB = 2î + 2ĵ + k̂ and →DC = 2î + 2ĵ + k̂, so →AB = →DC and ABCD is a parallelogram; |→AB| = 3 and →AD = î + 2ĵ − 2k̂ has |→AD| = 3, so adjacent sides are equal: a rhombus. (b) →AC = 3î + 4ĵ − k̂, →BD = −î − 3k̂; →AC · →BD = −3 + 0 + 3 = 0. (c) →AB × →AD = |î ĵ k̂; 2 2 1; 1 2 −2| = −6î + 5ĵ + 2k̂, area √65. From the diagonals, 12|→AC||→BD| = 12√26 √10 = √65, the same. (d) cos ∠DAB = →AB · →AD9 = 2 + 4 − 29 = 49.

• →AB = →DC and |→AB| = |→AD| = 3 (1)
• →AC · →BD = 0 (1)
• →AB × →AD = −6î + 5ĵ + 2k̂, area √65 (1)
• 12|→AC||→BD| = √65 (1)
• cos ∠DAB = 49 (1)