CBSE Class 12 Maths Sample Paper 1: marking scheme and solutions

Answer: (d) 144 + y² + 16 = 169, so y² = 9 and, as y > 0, y = 3. Stopping at y² gives 9.

Answer: (d) The conditions x ≥ 0, y ≥ 0 are the non-negative restrictions. x + y ≤ 12 and x − 2y ≥ 0 are other constraints (x − 2y ≥ 0 allows negative values of y), and Z = 5x + 3y is an objective function.

Answer: (d) An anti derivative is F(x) = x² + x. F(3) − F(1) = 12 − 2 = 10.

Answer: (c) (sin x)⁻¹ is the reciprocal 1/sin x = cosec x. It must not be confused with sin⁻¹ x, the inverse function.

Answer: (d) Theorem 2: if R is bounded, Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R. The possibility of no maximum or no minimum arises only for an unbounded region.

Answer: (b) cot⁻¹ : R → (0, π). cos⁻¹ has the closed range [0, π]; sec⁻¹ has [0, π] − {π/2}, which contains 0 and π; tan⁻¹ has (−π/2, π/2).

Answer: (c) Equal matrices have equal corresponding elements: x + y = 9 and x − y = 1, so x = 5, y = 4 and xy = 20.

Answer: (a) Split at x = 2. On [0, 2] the region is a triangle of area 2²2 = 2; on [2, k] it is a triangle of area (k − 2)²2. So 2 + (k − 2)²2 = 10, (k − 2)² = 16 and k = 6. 2 + 2√5 ignores the triangle on [0, 2].

Answer: (c) The equation is a polynomial in y′ and y″. The highest order derivative is y″, and its highest power is 2, so the degree is 2. Option (b) takes the highest power of any derivative; (d) adds the two powers.

Answer: (c) Equal vectors have equal components: p − 2 = 5, 3 = r + 1, −q = 4. So p = 7, r = 2, q = −4 and p + q + r = 5. Taking q = 4 gives 13.

Answer: (c) Mutually exclusive means A ∩ B = φ, so P(A ∩ B) = 0 and P(A|B) = 0/0.4 = 0. Since P(A) P(B) = 0.12 ≠ 0, A and B are not independent, which rules out (a) and (b); P(A ∪ B) = 0.3 + 0.4 = 0.7, not 0.58, which would be the value for independent events.

Answer: (c) The three pairs (1, 1), (2, 2), (3, 3) must be present. Symmetry makes the other six pairs come in three couples {(1, 2), (2, 1)}, {(1, 3), (3, 1)}, {(2, 3), (3, 2)}, each wholly in or wholly out: 2³ = 8. 64 counts reflexive relations without the symmetry condition.

Answer: (b) OP = √(1 + 1 + 2) = 2, so cos γ = √22 = 1√2 and γ = 45°. Option (a) takes cos γ = 12 from the x or y coordinate.

Answer: (c) For x ≠ 2, (x³ − 8)/(x − 2) = x² + 2x + 4, so lim_(x→2) f(x) = 4 + 4 + 4 = 12. Continuity at 2 needs f(2) = 12, so k = 12.

Answer: (c) P(first wins) = 2/n and P(second wins | first wins) = 1/(n − 1), so 2/(n(n − 1)) = 1/15, n(n − 1) = 30 and n = 6. Check: 2/6 × 1/5 = 1/15. Option (a) gives 1/10; (b) treats the draws as one draw of a winner.

Answer: (a) For independent events P(A ∩ B) = P(A) P(B), so P(B) = 0.12/0.3 = 0.4 and P(B′) = 1 − 0.4 = 0.6. Option (b) is P(B); (c) is 1 − P(A ∩ B).

Answer: (b) f′(x) = 3x² − 27 = 3(x − 3)(x + 3) = 0 at x = ±3; f is differentiable everywhere, so these are the only critical points.

Answer: (b) Lines in space which are neither parallel nor intersecting are skew lines; they lie in different planes, so (a) is wrong. Perpendicular lines may intersect, and concurrent lines meet.

Answer: (a) ∫₀³ (−x²) dx = −9; the curve is below the x-axis, so the area is |−9| = 9. R is the rule that gives A.

Answer: (a) The highest derivative is y‴, so the order is 3. log (y′) is not a polynomial in y′, so by R the degree is not defined. R is exactly the reason for the second half of A.

Separating the variables, 3y² dy = sec² x dx. Integrating, ∫ 3y² dy = ∫ sec² x dx, so y³ = tan x + C, the general solution.

• separates: 3y² dy = sec² x dx (1)
• y³ = tan x + C (1)

Separating, e^2y dy = x dx. Integrating, e^2y/2 = x²/2 + C/2, so e^2y = x² + C, the general solution.

• separates: e^2y dy = x dx (1)
• e^2y = x² + C (or an equivalent form) (1)

[1] = {b ∈ R : f(b) = f(1)} and f(1) = 1 − 4 = −3. So b² − 4b = −3, b² − 4b + 3 = 0, (b − 1)(b − 3) = 0, b = 1 or 3. Hence [1] = {1, 3}.

• sets up f(b) = f(1) = −3 (1)
• [1] = {1, 3} (1)

[5] = {b ∈ Z : |b − 2| = |5 − 2| = 3}. So b − 2 = 3 or b − 2 = −3, b = 5 or b = −1. Hence [5] = {−1, 5}.

• sets up |b − 2| = 3 (1)
• [5] = {−1, 5} (1)

P(A′ ∩ B′) = P((A ∪ B)′) = 1 − 0.7 = 0.3 and P(B′) = 1 − 0.4 = 0.6. So P(A′|B′) = P(A′ ∩ B′)/P(B′) = 0.3/0.6 = 0.5.

• P(A′ ∩ B′) = 1 − P(A ∪ B) = 0.3 (1)
• P(B′) = 0.6 and P(A′|B′) = 0.5 (1)

cos² γ = 1 − cos² 120° − cos² 60° = 1 − 14 − 14 = 12; γ is acute, so cos γ = 1√2. The direction cosines are −12, 12, 1√2, and P = 4(l, m, n) = (−2, 2, 2√2).

• cos γ = 1√2 from l² + m² + n² = 1 (1)
• P = (−2, 2, 2√2) (1)

Left side: x² − 12. Right side: 8 − (−5) = 13. So x² − 12 = 13, x² = 25, x = ±5.

• both determinants expanded: x² − 12 and 13 (1)
• x² = 25, x = ±5 (1)

Let α = tan⁻¹(1/3). sin 2α = 2 tan α/(1 + tan² α) = (2/3)/(10/9) = 3/5. Let β = tan⁻¹(2√2) ∈ (0, π/2), so tan β = 2√2, sec² β = 1 + 8 = 9 and cos β = 1/3 (positive in (0, π/2)). Sum = 3/5 + 1/3 = 14/15.

• sin(2 tan⁻¹(1/3)) = 3/5 (1)
• cos(tan⁻¹ 2√2) = 1/3 with the sign justified (1)
• sum 14/15 (1)

Let α = tan⁻¹(1/4). sin 2α = 2 tan α/(1 + tan² α) = (1/2)/(17/16) = 8/17. Let β = tan⁻¹ √15 ∈ (0, π/2), so sec² β = 1 + 15 = 16 and cos β = 1/4 (positive in (0, π/2)). Value = 8/17 − 1/4 = (32 − 17)/68 = 15/68.

• sin(2 tan⁻¹(1/4)) = 8/17 (1)
• cos(tan⁻¹ √15) = 1/4 with the sign justified (1)
• value 15/68 (1)

The feasible region is bounded, with corners (0, 0), (5, 0), (5, 3) [x = 5 meets x + y = 8], (2, 6) [y = 6 meets x + y = 8] and (0, 6). Z there: 0, 20, 29, 26, 18; the maximum is 29 at (5, 3). Without x ≤ 5 the corners are (0, 0), (8, 0), (2, 6), (0, 6), with Z = 0, 32, 26, 18: the maximum becomes 32 at (8, 0). The increase is 3.

• corners of the region found (1)
• maximum 29 at (5, 3) (1)
• new maximum 32 at (8, 0); increase 3 (1)

The feasible region is bounded, with corners (0, 0), (7, 0), (7, 2) [x = 7 meets x + y = 9], (3, 6) [y = 6 meets x + y = 9] and (0, 6). Z there: 0, 14, 24, 36, 30; the maximum is 36 at (3, 6). Without y ≤ 6 the corners are (0, 0), (7, 0), (7, 2), (0, 9), with Z = 0, 14, 24, 45: the maximum becomes 45 at (0, 9). The increase is 9.

• corners of the region found (1)
• maximum 36 at (3, 6) (1)
• new maximum 45 at (0, 9); increase 9 (1)

Continuity: f(1) = 1, left hand limit 1, right hand limit 2 − 1 = 1. Left hand derivative: lim_(h→0⁻) [(1 + h)² − 1]/h = lim (2 + h) = 2. Right hand derivative: lim_(h→0⁺) [2(1 + h) − 1 − 1]/h = 2. They are finite and equal, so f is differentiable at 1 with f′(1) = 2.

• continuity at x = 1 checked (1)
• left hand derivative = 2 (1)
• right hand derivative = 2, so differentiable (1)

Continuity: f(1) = 1, left hand limit 1, right hand limit 3 − 2 = 1, so f is continuous at 1. Left hand derivative: lim_(h→0⁻) [(1 + h)² − 1]/h = 2. Right hand derivative: lim_(h→0⁺) [3(1 + h) − 2 − 1]/h = 3. Since 2 ≠ 3, f is not differentiable at 1.

• continuity at x = 1 shown (1)
• left hand derivative = 2 (1)
• right hand derivative = 3, so not differentiable (1)

Let the rhombus be OACB with →OA = a and →OB = b, so |a| = |b|. The diagonals are →OC = a + b and →BA = a − b. (a + b) · (a − b) = |a|² − |b|² = 0. Both diagonals are non-zero, so they are perpendicular.

• diagonals written as a + b and a − b (1)
• (a + b) · (a − b) = |a|² − |b|² (1)
• = 0 as |a| = |b|, hence perpendicular (1)

A′ = A and B′ = −B. (AB + BA)′ = (AB)′ + (BA)′ = B′A′ + A′B′ = (−B)A + A(−B) = −(BA + AB) = −(AB + BA), so AB + BA is skew symmetric. (AB − BA)′ = B′A′ − A′B′ = −BA + AB = AB − BA, so AB − BA is symmetric.

• uses A′ = A, B′ = −B and (AB)′ = B′A′ (1)
• (AB + BA)′ = −(AB + BA) (1)
• (AB − BA)′ = AB − BA (1)

Let Fᵢ be 'attempt i fails'. P(F₁) = 9/10; given F₁, 9 digits are left, one correct, so P(F₂|F₁) = 8/9; P(F₃|F₁F₂) = 7/8. By the multiplication rule, P(all three fail) = 9/10 × 8/9 × 7/8 = 7/10. So P(unlocks within three attempts) = 1 − 7/10 = 3/10.

• P(F₁) = 9/10, P(F₂|F₁) = 8/9, P(F₃|F₁F₂) = 7/8 (1)
• product 7/10 by the multiplication rule (1)
• answer 3/10 (1)

(a) 2 cos x − 1 = 0 gives cos x = 12, x = π3; the curve is above the x-axis on (0, π3) and below on (π3, π). (b) An anti derivative is 2 sin x − x. ∫₀^π/3 = √3 − π3; ∫_π/3^π = (0 − π) − (√3 − π3) = −√3 − 2π3, numerical value √3 + 2π3. Area = 2√3 + π3 square units. (c) ∫₀^π = (√3 − π3) + (−√3 − 2π3) = −π: the area above the axis less the area below it, not the area of the region.

• (a) x = π3 with the sign on each side (1)
• anti derivative 2 sin x − x (1)
• pieces √3 − π3 and −√3 − 2π3 (1)
• (b) area 2√3 + π3 square units (1)
• (c) −π as area above less area below (1)

(a) sin x = 12 gives x = π6 and 5π6; the curve is above the x-axis on (0, π6) and (5π6, π), below on (π6, 5π6). (b) An anti derivative is x + 2 cos x. ∫₀^π/6 = π6 + √3 − 2; ∫_π/6^5π/6 = 2π3 − 2√3, numerical value 2√3 − 2π3; ∫_5π/6^π = π6 + √3 − 2. Area = 4√3 − 4 − π3 square units. (c) Adding the three signed pieces gives π3 + 2√3 − 4 + 2π3 − 2√3 = π − 4: the area above the axis less the area below it.

• (a) x = π6, 5π6 with the signs (1)
• anti derivative x + 2 cos x (1)
• the three pieces (1)
• (b) area 4√3 − 4 − π3 square units (1)
• (c) π − 4 as area above less area below (1)

(a) Let I = ∫₀^π x g(sin x) dx. By P₄, I = ∫₀^π (π − x) g(sin (π − x)) dx = ∫₀^π (π − x) g(sin x) dx = π ∫₀^π g(sin x) dx − I. So 2I = π ∫₀^π g(sin x) dx. (b) sin x cos² x = sin x (1 − sin² x) is a function of sin x, so the integral is π2 ∫₀^π sin x cos² x dx. With cos x = t, ∫₀^π sin x cos² x dx = ∫₋₁¹ t² dt = 23. The value is π2 · 23 = π3.

• (a) applies P₄ and uses sin (π − x) = sin x (1)
• (a) adds to reach 2I = π ∫ g(sin x) dx (1)
• (b) recognises cos² x as a function of sin x and applies (a) (1)
• (b) ∫₀^π sin x cos² x dx = 23 by cos x = t (1)
• (b) value π3 (1)

Let I be the integral. By P₄, I = ∫₀^π/2 cos² xcos x + sin x dx. Adding, 2I = ∫₀^π/2 dxsin x + cos x. Now sin x + cos x = √2 sin (x + π4), so 2I = 1√2 ∫₀^π/2 cosec (x + π4) dx = 1√2 [log |cosec (x + π4) − cot (x + π4)|]₀^π/2. At π2: cosec 3π4 − cot 3π4 = √2 + 1; at 0: √2 − 1. So 2I = 1√2 log √2 + 1√2 − 1 = 2√2 log (√2 + 1), and I = 1√2 log (√2 + 1).

• applies P₄ to get the cos² x form (1)
• adds to reach 2I = ∫ dxsin x + cos x (1)
• writes sin x + cos x = √2 sin (x + π4) (1)
• integrates cosec and substitutes the limits (1)
• value 1√2 log (√2 + 1) (1)

(a) f′(x) = 4x³ − 12x² − 16x + 48 = 4(x² − 4)(x − 3) = 4(x + 2)(x − 2)(x − 3), zero at −2, 2, 3. Signs: (−∞, −2) negative; (−2, 2) positive; (2, 3) negative; (3, ∞) positive. Decreasing in (−∞, −2) and (2, 3); increasing in (−2, 2) and (3, ∞). (b) f″(x) = 12x² − 24x − 16: f″(−2) = 80 > 0, f″(2) = −16 < 0, f″(3) = 20 > 0. Local minima at x = −2, value f(−2) = 16 + 32 − 32 − 96 + 1 = −79; local maxima at x = 2, value f(2) = 16 − 32 − 32 + 96 + 1 = 49; local minima at x = 3, value f(3) = 81 − 108 − 72 + 144 + 1 = 46.

• f′(x) = 4(x + 2)(x − 2)(x − 3) (1)
• sign of f′ and the four intervals (1)
• classifies −2, 2, 3 (by f″ or the sign change) (1)
• local minimum values −79 and 46 (1)
• local maximum value 49 (1)

A(2, 0, 0), C(0, 2, 0), D(0, 0, 2), F(2, 2, 2). (a) OF has direction ratios 1, 1, 1 and AC has −2, 2, 0: −2 + 2 + 0 = 0, so the angle is 90°. (b) OF: r = λ(î + ĵ + k̂); edge: r = 2î + μk̂. b₁ × b₂ = î − ĵ, of magnitude √2; (a₂ − a₁) = 2î, dot product 2. d = 2√2 = √2 units. (c) AC: r = 2î + μ(−2î + 2ĵ). b₁ × b₂ = |î ĵ k̂; 1 1 1; −2 2 0| = −2î − 2ĵ + 4k̂, of magnitude 2√6; dot product with 2î is −4. d = 42√6 = 2√6 = √63 units.

• angle 90° from a₁a₂ + b₁b₂ + c₁c₂ = 0 (1)
• (b) b₁ × b₂ = î − ĵ and a₂ − a₁ = 2î (1)
• (b) d = √2 (1)
• (c) b₁ × b₂ = −2î − 2ĵ + 4k̂ (1)
• (c) d = 2√6 (1)