CBSE Class 12 Maths Board Paper 2025 (65/1/1): Solutions

A is the 3 × 3 diagonal matrix with diagonal entries −1, 1, 1. Find A⁻¹.

1. For a diagonal matrix, the inverse is found by taking the reciprocal of each diagonal entry.
2. The reciprocals are 1/(−1) = −1, 1/1 = 1, and 1/1 = 1, so A⁻¹ = diag(−1, 1, 1), which is equal to A.

Answer: Each diagonal entry is replaced by its reciprocal, and 1/(−1) = −1, so A⁻¹ = A = diag(−1, 1, 1) — option (D).

For a = 3î + 2ĵ − k̂ and b = î − ĵ + k̂, which holds: a ∥ b, a ⊥ b, |b| > |a| or |a| = |b|?

1. Calculate the dot product a · b as (3)(1) + (2)(−1) + (−1)(1) = 3 − 2 − 1 = 0.
2. Since the dot product is zero, the vectors are perpendicular, so a ⊥ b is the correct option.

Answer: a · b = 3 − 2 − 1 = 0, so a ⊥ b — option (B).

Evaluate the integral of |x|/x from x = −1 to 1 (x ≠ 0).

1. Identify that |x|/x equals -1 for x in the interval (-1, 0) and 1 for x in the interval (0, 1).
2. Split the integral into two parts: ∫ from -1 to 0 of -1 dx plus ∫ from 0 to 1 of 1 dx.
3. Evaluate the first part as -1 × [0 - (-1)] = -1 and the second part as 1 × [1 - 0] = 1.
4. Add the results to get -1 + 1 = 0, which corresponds to option (B).

Answer: |x|/x is −1 on (−1, 0) and 1 on (0, 1), so the integral is −1 + 1 = 0 — option (B).

Among y² − xy, x − 3y, sin²(y/x) + y/x and tan x − sec y, which one is NOT homogeneous in x and y?

1. Recall that a function f(x, y) is homogeneous of degree n if f(λx, λy) = λⁿ f(x, y) for any constant λ.
2. Test the function tan x − sec y by substituting x with λx and y with λy to get tan(λx) − sec(λy).
3. Observe that tan(λx) − sec(λy) cannot be factored as λⁿ(tan x − sec y) for any constant n, so it is not homogeneous.

Answer: tan(λx) − sec(λy) cannot be written as λⁿ(tan x − sec y), so tan x − sec y is not homogeneous — option (D).

f(x) = |x| + |x − 1|. Decide its continuity and differentiability at x = 0 and x = 1.

1. Since f(x) is a sum of continuous modulus functions, it is continuous at x = 0 and x = 1.
2. The function has corners at x = 0 and x = 1, so the left and right derivatives are not equal, making f not differentiable at these points.

Answer: A sum of modulus functions is continuous everywhere but has corners at x = 0 and x = 1, so f is continuous but not differentiable there — option (C).

A is a 2 × 2 matrix with det A = 4. Find det(4 adj A).

1. Use the property det(kA) = k^n det(A) for an n × n matrix, where n = 2 and k = 4.
2. Apply the property det(adj A) = |A|^(n−1) = 4^(2−1) = 4.
3. Calculate det(4 adj A) = 4² × 4 = 16 × 4 = 64.

Answer: det(4 adj A) = 4² · |A|^(2−1) = 16 × 4 = 64 — option (B).

E and F are independent with P(E) = 2/3 and P(F) = 3/7. Find P(E | not F).

1. Since E and F are independent events, E is also independent of the complement of F (not F).
2. Therefore, P(E | not F) = P(E) = 2/3.

Answer: E is independent of F and hence of not-F, so P(E | F̄) = P(E) = 2/3 — option (C).

Find the absolute maximum of f(x) = x³ − 3x + 2 on [0, 2].

1. Find the derivative: f'(x) = 3x² − 3.
2. Set f'(x) = 0 to find critical points: 3x² − 3 = 0, so x² = 1, giving x = 1 (since x is in [0, 2]).
3. Evaluate f(x) at the critical point and endpoints: f(0) = 2, f(1) = 0, and f(2) = 4.
4. Compare the values: 4 is the largest, so the absolute maximum is 4.

Answer: f′(x) = 3x² − 3 = 0 gives x = 1; f(0) = 2, f(1) = 0, f(2) = 4, so the absolute maximum is 4 — option (C).

A is 3 × 3, B is the 3 × 1 column (−2, −5, −7) and C is the 1 × 3 row [9 8 7]. Which of AB, AC, BA is defined?

1. For matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix.
2. In AB, A is 3 × 3 and B is 3 × 1, so the inner orders (3 and 3) match, making AB defined.
3. In AC, A is 3 × 3 and C is 1 × 3, so the inner orders (3 and 1) do not match, making AC undefined.
4. In BA, B is 3 × 1 and A is 3 × 3, so the inner orders (1 and 3) do not match, making BA undefined.
5. Therefore, only AB is defined, which corresponds to option (A).

Answer: Only AB (3×3 times 3×1) has matching inner orders; AC and BA do not — option (A).

If the integral of 2^(1/x)/x² dx equals k · 2^(1/x) + C, find k.

1. Let t = 1/x, then dt = -1/x² dx, so dx/x² = -dt.
2. Substitute into the integral: ∫ 2^(1/x) (dx/x²) = ∫ 2^t (-dt) = -∫ 2^t dt.
3. Integrate: -∫ 2^t dt = -2^t / log 2 + C.
4. Substitute back t = 1/x: -2^(1/x) / log 2 + C.
5. Compare with k · 2^(1/x) + C to find k = -1/log 2.

Answer: Put t = 1/x, dt = −dx/x²: the integral is −2ᵗ/log 2 + C, so k = −1/log 2 — option (A).

a + b + c = 0 with |a| = √37, |b| = 3, |c| = 4. Find the angle between b and c.

1. Since a + b + c = 0, we have b + c = -a, which implies |b + c|² = |a|² = 37.
2. Expanding the left side gives |b|² + |c|² + 2|b||c|cos θ = 37, substituting the magnitudes yields 9 + 16 + 24cos θ = 37.
3. Solving for cos θ gives 24cos θ = 12, so cos θ = 1/2, which means the angle θ is π/3.

Answer: |b + c|² = |a|² gives 9 + 16 + 2(3)(4)cos θ = 37, so cos θ = 1/2 and θ = π/3 — option (C).

Find the integrating factor of (x + 2y³) dy/dx = 2y.

1. Rewrite the equation as dx/dy − x/(2y) = y², which is a linear differential equation in x.
2. Identify the coefficient of x as P = −1/(2y) and calculate the integrating factor as e^(∫P dy) = e^(−∫dy/(2y)).
3. Evaluate the integral to get e^(−(1/2)ln y) = e^(ln y^(−1/2)) = 1/√y, which corresponds to option (B).

Answer: Write it as dx/dy − x/(2y) = y², a linear equation in x; IF = e^(−∫dy/(2y)) = 1/√y — option (B).

The matrix with rows (7, 0, x), (0, 7, 0), (0, 0, y) is a scalar matrix. Find yˣ.

1. A scalar matrix has all off-diagonal elements as zero and all diagonal elements equal.
2. Comparing the given matrix with the definition, the off-diagonal element x must be 0.
3. The diagonal elements must be equal, so y must be 7.
4. Substituting these values, yˣ = 7⁰ = 1.

Answer: A scalar matrix needs x = 0 and y = 7, so yˣ = 7⁰ = 1 — option (B).

An LPP has corner points (2, 72), (15, 20) and (40, 15), with Z = 18x + 9y. Where are the maximum and minimum of Z?

1. Evaluate Z = 18x + 9y at (2, 72): 18(2) + 9(72) = 36 + 648 = 684.
2. Evaluate Z = 18x + 9y at (15, 20): 18(15) + 9(20) = 270 + 180 = 450.
3. Evaluate Z = 18x + 9y at (40, 15): 18(40) + 9(15) = 720 + 135 = 855.
4. Compare values: 450 < 684 < 855, so minimum is 450 at (15, 20) and maximum is 855 at (40, 15).

Answer: Z = 684, 450, 855 at the three corners: maximum 855 at (40, 15), minimum 450 at (15, 20) — option (C).

A and B are invertible. Which is false: (A + B)⁻¹ = B⁻¹ + A⁻¹, (AB)⁻¹ = B⁻¹A⁻¹, adj A = |A| A⁻¹, or |A|⁻¹ = |A⁻¹|?

1. Recall that the inverse of a product follows the rule (AB)⁻¹ = B⁻¹A⁻¹, which is a valid property.
2. Recall that for an invertible matrix A, the adjoint is given by adj A = |A|A⁻¹, which is also a valid property.
3. Recall that the determinant of an inverse is the reciprocal of the determinant, so |A|⁻¹ = |A⁻¹| is a valid property.
4. Observe that there is no general sum rule for matrix inverses, meaning (A + B)⁻¹ is not equal to B⁻¹ + A⁻¹.
5. Conclude that the statement (A + B)⁻¹ = B⁻¹ + A⁻¹ is false, identifying it as the correct option.

Answer: There is no sum rule for inverses, so (A + B)⁻¹ = B⁻¹ + A⁻¹ is not correct — option (A).

For Z = ax + by, what can be said about its optimum values when the feasible region is bounded?

1. According to the Fundamental Theorem of Linear Programming, if the feasible region is bounded, the objective function Z attains both a maximum and a minimum value.
2. These optimum values occur specifically at the corner points (vertices) of the feasible region, which corresponds to option (C).

Answer: On a bounded feasible region Z attains both a maximum and a minimum (at corner points) — option (C).

Which integral gives the area between y² = x, the line x = 4 and the x-axis, in the first quadrant (shown shaded)?

Uses a figure printed in CBSE's paper — what the figure shows: The figure shades only the region above the x-axis, under the upper branch y = √x, from x = 0 to x = 4.

1. The shaded region is bounded above by the curve y = √x, below by the x-axis (y = 0), and on the right by the line x = 4.
2. The area is given by the integral of the upper function minus the lower function from x = 0 to x = 4, which is ∫₀⁴ √x dx.

Answer: The shaded part lies above the x-axis under y = √x from 0 to 4, so area = ∫₀⁴ √x dx (= 16/3) — option (D).

A graph of a trigonometric function is shown; pick the graph of its inverse from four options.

Uses a figure printed in CBSE's paper — what the figure shows: The given curve is cos x: peak 1 at x = 0, crosses the x-axis at −π/2 and π/2, reaches −1 at π. Option (C) is a decreasing curve on [−1, 1] from height π at x = −1 through π/2 at x = 0 to 0 at x = 1; (A), (B) are increasing curves through the origin and (D) is a decreasing curve with horizontal asymptotes at π and 0.

1. The given curve equals 1 at x = 0, is 0 at x = ±π/2 and −1 at x = π, so it is y = cos x.
2. On its principal branch [0, π], cos x takes values from 1 down to −1, so y = cos⁻¹x has domain [−1, 1] and range [0, π] and is decreasing.
3. Its graph runs from (−1, π) through (0, π/2) to (1, 0); only option (C) matches.

Answer: The curve is y = cos x (value 1 at x = 0, zeros at ±π/2, −1 at π); its inverse y = cos⁻¹x runs from (−1, π) down to (1, 0) — option (C).

Assertion: f: Z → Z, f(x) = 3x − 5 is bijective. Reason: a function that is one-one and onto is bijective.

1. Check if f is one-one: If f(x1) = f(x2), then 3x1 − 5 = 3x2 − 5, which implies x1 = x2, so f is one-one.
2. Check if f is onto: For y = 0, 3x − 5 = 0 gives x = 5/3, which is not an integer, so f is not onto Z.
3. Since f is one-one but not onto, the assertion is false, while the reason is a true definition, making option (D) correct.

Answer: f is one-one but not onto Z (e.g. 0 = 3x − 5 has no integer solution), so the assertion is false; the reason is a true definition — option (D).

Assertion: f(x) = 3x − 8 for x ≤ 5 and 2k for x > 5 is continuous at x = 5 when k = 5/2. Reason: continuity at a needs both one-sided limits equal to f(a).

1. Calculate the left-hand limit as x approaches 5 using f(x) = 3x − 8, which gives 3(5) − 8 = 7.
2. Calculate the right-hand limit as x approaches 5 using f(x) = 2k, which is 2k.
3. Equate the limits for continuity: 2k = 7, which implies k = 7/2.
4. Since the assertion claims k = 5/2, it is false, while the reason correctly states the condition for continuity, making option (D) correct.

Answer: Continuity needs 2k = 3(5) − 8 = 7, so k = 7/2, not 5/2; the assertion is false and the reason true — option (D).

Find the derivative of 2^(cos²x) with respect to cos²x.

1. Let u = 2^(cos²x) and v = cos²x.
2. Differentiate u with respect to x: du/dx = 2^(cos²x) × log 2 × d(cos²x)/dx = 2^(cos²x) log 2 × 2 cos x sin x.
3. Differentiate v with respect to x: dv/dx = 2 cos x sin x.
4. Find du/dv by dividing du/dx by dv/dx: du/dv = [2^(cos²x) log 2 × 2 cos x sin x] / [2 cos x sin x] = 2^(cos²x) log 2.

Answer: With u = 2^(cos²x) and v = cos²x, du/dv = (du/dx)/(dv/dx) = 2^(cos²x) log 2.

Given tan⁻¹(x² + y²) = a², find dy/dx.

1. Take the tangent of both sides of the given equation to remove the inverse tangent function.
2. This gives x² + y² = tan(a²), which is a constant value.
3. Differentiate both sides with respect to x, treating y as a function of x.
4. The derivative of x² is 2x, and the derivative of y² is 2y(dy/dx), while the derivative of the constant is 0.
5. Equating these gives 2x + 2y(dy/dx) = 0.
6. Solving for dy/dx yields dy/dx = -x/y.

Answer: x² + y² = tan a² (a constant), so 2x + 2y dy/dx = 0 and dy/dx = −x/y.

Find the value of tan⁻¹[2 sin(2 cos⁻¹(√3/2))].

1. Let θ = cos⁻¹(√3/2), which implies cos θ = √3/2.
2. Since θ lies in the range [0, π], we identify θ = π/6.
3. Substitute θ into the expression to get tan⁻¹[2 sin(2 × π/6)].
4. Simplify the argument inside the sine function: 2 × π/6 = π/3.
5. Evaluate the sine term: 2 sin(π/3) = 2 × (√3/2) = √3.
6. Calculate the final value: tan⁻¹(√3) = π/3.

Answer: cos⁻¹(√3/2) = π/6, so 2 sin(π/3) = √3 and tan⁻¹√3 = π/3.

A parallelogram has diagonals a = 2î − ĵ + k̂ and b = î + 3ĵ − k̂. Find its area.

1. a × b = î[(−1)(−1) − (1)(3)] − ĵ[(2)(−1) − (1)(1)] + k̂[(2)(3) − (−1)(1)] = î(1 − 3) − ĵ(−2 − 1) + k̂(6 + 1) = −2î + 3ĵ + 7k̂.
2. |a × b| = √((−2)² + 3² + 7²) = √(4 + 9 + 49) = √62.
3. When a and b are the diagonals, area of the parallelogram = ½|a × b| = √62/2 square units.

Answer: a × b = −2î + 3ĵ + 7k̂, |a × b| = √62, and area = ½|a × b| = √62/2 square units.

On which intervals is f(x) = 5x^(3/2) − 3x^(5/2) (i) increasing, (ii) decreasing?

1. f′(x) = (15/2)x^(1/2) − (15/2)x^(3/2) = (15/2)√x (1 − x), defined for x ≥ 0.
2. f′(x) = 0 at x = 0 and x = 1.
3. For 0 < x < 1, √x > 0 and 1 − x > 0, so f′(x) > 0; for x > 1, 1 − x < 0, so f′(x) < 0.
4. (i) f is increasing on [0, 1]; (ii) f is decreasing on [1, ∞).

Answer: f′(x) = (15/2)√x (1 − x), zero at x = 0, 1. (i) f is increasing on [0, 1]; (ii) f is decreasing on [1, ∞).

Note: Writing the open intervals (0, 1) and (1, ∞) is also accepted.

Two crossing kite strings lie along a = 3î + ĵ + 2k̂ and b = 2î − 2ĵ + 4k̂ (taut). Find the angle between them.

1. Calculate the dot product of the vectors: a · b = (3)(2) + (1)(−2) + (2)(4) = 6 − 2 + 8 = 12.
2. Find the magnitude of vector a: |a| = √(3² + 1² + 2²) = √(9 + 1 + 4) = √14.
3. Find the magnitude of vector b: |b| = √(2² + (−2)² + 4²) = √(4 + 4 + 16) = √24.
4. Use the formula cos θ = (a · b) / (|a| |b|) to find the cosine of the angle: cos θ = 12 / (√14 × √24) = 12 / √336.
5. Simplify the expression: √336 = √(16 × 21) = 4√21, so cos θ = 12 / (4√21) = 3 / √21.
6. The angle between the strings is θ = cos⁻¹(3/√21).

Answer: a · b = 12, |a| = √14, |b| = √24, so cos θ = 12/√336 = 3/√21 ≈ 0.6547 and θ = cos⁻¹(3/√21).

A = (2, 1, 3), B = (8, −1, 0). Find the vector of length 21 pointing opposite to AB.

1. AB = (8 − 2)î + (−1 − 1)ĵ + (0 − 3)k̂ = 6î − 2ĵ − 3k̂.
2. |AB| = √(36 + 4 + 9) = √49 = 7.
3. The required vector = −21 × (6î − 2ĵ − 3k̂)/7 = −3(6î − 2ĵ − 3k̂) = −18î + 6ĵ + 9k̂.

Answer: AB = 6î − 2ĵ − 3k̂ with |AB| = 7, so the vector is −(21/7)(6î − 2ĵ − 3k̂) = -18î + 6ĵ + 9k̂.

An equilateral triangle's side grows at 3 cm/s. How fast is its area growing when the side is 15 cm?

1. Let a be the side of the equilateral triangle and A be its area. The formula for the area is A = (√3/4)a².
2. Differentiate the area with respect to time t to find the rate of change: dA/dt = (√3/4) × 2a × (da/dt) = (√3/2)a(da/dt).
3. Substitute the given values a = 15 cm and da/dt = 3 cm/s into the equation: dA/dt = (√3/2)(15)(3).
4. Calculate the final result: dA/dt = 45√3/2 cm²/s.

Answer: A = (√3/4)a², so dA/dt = (√3/2)a · da/dt = (√3/2)(15)(3) = 45√3/2 cm²/s.

Graphically maximise Z = x + 2y with x − y ≥ 0, x − 2y ≥ −2, x ≥ 0, y ≥ 0.

1. Draw x − y = 0 (the line y = x) and x − 2y = −2 (through (−2, 0) and (0, 1)); the region lies on or below both lines, in the first quadrant.
2. The two lines meet where x = y and x − 2y = −2, so −y = −2, giving (2, 2). The feasible region is unbounded, with corner points O(0, 0) and (2, 2).
3. Z at O(0, 0) = 0 and Z at (2, 2) = 2 + 2(2) = 6.
4. Since the region is unbounded, check x + 2y > 6: it has points in the region, e.g. (8, 0) gives Z = 8 > 6.
5. So Z = x + 2y has no maximum value on this region.

Answer: The feasible region is unbounded with corners O(0, 0) (Z = 0) and (2, 2) (Z = 6). The half-plane x + 2y > 6 meets the region, so Z has no maximum value.

Integrate (x + sin x)/(1 + cos x) with respect to x.

1. Use 1 + cos x = 2cos²(x/2) and sin x = 2 sin(x/2) cos(x/2): (x + sin x)/(1 + cos x) = x/(2cos²(x/2)) + 2 sin(x/2)cos(x/2)/(2cos²(x/2)) = (x/2)sec²(x/2) + tan(x/2).
2. Integrate the first part by parts with u = x and dv = (1/2)sec²(x/2) dx, so v = tan(x/2): ∫(x/2)sec²(x/2) dx = x tan(x/2) − ∫tan(x/2) dx.
3. Add the second part: I = x tan(x/2) − ∫tan(x/2) dx + ∫tan(x/2) dx.
4. The two tan(x/2) integrals cancel, so I = x tan(x/2) + C.

Answer: Using 1 + cos x = 2cos²(x/2) and sin x = 2 sin(x/2) cos(x/2), the integrand is (x/2)sec²(x/2) + tan(x/2); by parts the result is x tan(x/2) + C.

Evaluate the integral from 0 to π/4 of dx / (cos³x · √(2 sin 2x)).

1. √(2 sin 2x) = √(4 sin x cos x) = 2√(sin x cos x) = 2cos x √(tan x), since cos x > 0 on [0, π/4].
2. So the integrand is 1/(2cos⁴x √(tan x)) = sec⁴x/(2√(tan x)) = (1 + tan²x)sec²x/(2√(tan x)).
3. Put t = tan x, dt = sec²x dx; x = 0 gives t = 0 and x = π/4 gives t = 1. The integral becomes ½∫₀¹ (1 + t²)/√t dt = ½∫₀¹ (t^(−1/2) + t^(3/2)) dt.
4. = ½[2t^(1/2) + (2/5)t^(5/2)] from 0 to 1 = ½(2 + 2/5) = ½ × 12/5 = 6/5 = 1.2.

Answer: The integrand is sec⁴x / (2√tan x); with t = tan x the integral is ½∫₀¹ (t^(−1/2) + t^(3/2)) dt = ½(2 + 2/5) = 6/5 = 1.2.

Show that r = (1 − λ)î + (λ − 2)ĵ + (3 − 2λ)k̂ and r = (μ + 1)î + (2μ − 1)ĵ − (2μ + 1)k̂ are skew, then find the shortest distance between them.

1. Identify the direction vectors of the two lines as b₁ = (−1, 1, −2) and b₂ = (1, 2, −2).
2. Calculate the cross product b₁ × b₂ to get (2, −4, −3), which is not zero, so the lines are not parallel.
3. Find the vector connecting a point on the first line, a₁ = (1, −2, 3), to a point on the second line, a₂ = (1, −1, −1), which is a₂ − a₁ = (0, 1, −4).
4. Compute the scalar triple product (a₂ − a₁) · (b₁ × b₂) = 0(2) + 1(−4) + (−4)(−3) = 8.
5. Since the scalar triple product is 8, which is not zero, the lines are skew.
6. Calculate the shortest distance using the formula |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|, which is 8 / √(2² + (−4)² + (−3)²) = 8 / √29.

Answer: a₁ = (1, −2, 3), b₁ = (−1, 1, −2); a₂ = (1, −1, −1), b₂ = (1, 2, −2). b₁ × b₂ = (2, −4, −3) (not parallel), and (a₂ − a₁)·(b₁ × b₂) = 8 ≠ 0, so the lines are skew. Shortest distance = 8/√29.

Bowler B = 2î + 8ĵ, wicket keeper W = 6î + 12ĵ, leg slip F = 12î + 18ĵ are collinear. In what ratio does W divide BF?

1. Let W divide BF in the ratio k : 1. Then W = (k(12î + 18ĵ) + (2î + 8ĵ))/(k + 1).
2. Compare the î parts: (12k + 2)/(k + 1) = 6, so 12k + 2 = 6k + 6, giving 6k = 4 and k = 2/3.
3. Check with the ĵ parts: (18 × 2/3 + 8)/(2/3 + 1) = 20/(5/3) = 12, which matches W.
4. So W divides BF in the ratio 2/3 : 1 = 2 : 3.

Answer: Let W divide BF in k : 1; then (12k + 2)/(k + 1) = 6 gives k = 2/3, so the ratio is 2 : 3.

X (students absent on a Saturday) takes 0, 2, 4, 5 with probabilities p, 2p, 3p, p. (i) Find p (1 mark). (ii) Find the mean of X (2 marks).

1. The sum of probabilities for all values of X must equal 1.
2. p + 2p + 3p + p = 1, which simplifies to 7p = 1.
3. Solving for p gives p = 1/7.
4. The mean of X is calculated as the sum of each value multiplied by its probability: E(X) = Σ [x · P(x)].
5. Substituting the values: E(X) = 0(p) + 2(2p) + 4(3p) + 5(p).
6. Simplifying the expression: E(X) = 0 + 4p + 12p + 5p = 21p.
7. Substituting p = 1/7 into the mean: E(X) = 21 × (1/7) = 3.

Answer: (i) 7p = 1, so p = 1/7; mean = 3 students for (ii), since ΣX·P(X) = 0(p) + 2(2p) + 4(3p) + 5(p) = 21p = 21/7.

Of 3000 applicants, two-thirds are women. A man gets a distinction in the written test with probability 0.4, a woman with 0.35. Find P(a random applicant gets a distinction).

1. Calculate the number of men and women: Men = 3000 - (2/3) × 3000 = 1000, so P(M) = 1/3; Women = (2/3) × 3000 = 2000, so P(W) = 2/3.
2. Apply the Law of Total Probability: P(Distinction) = P(M)P(D|M) + P(W)P(D|W).
3. Substitute the given values: P(Distinction) = (1/3)(0.4) + (2/3)(0.35).
4. Calculate the final result: (1/3)(2/5) + (2/3)(7/20) = 2/15 + 7/30 = 4/30 + 7/30 = 11/30.

Answer: P = (1/3)(0.4) + (2/3)(0.35) = 11/30.

Sketch y = |x + 3| and use integration to find the area between it and the x-axis from x = −6 to x = 0.

1. Sketch the graph of y = |x + 3|, which is a V-shaped curve with its vertex at (−3, 0).
2. Identify that the function is symmetric about the line x = −3, so the area from x = −6 to x = 0 is twice the area from x = −3 to x = 0.
3. Set up the integral for the right half: Area = 2 ∫ from −3 to 0 of (x + 3) dx.
4. Integrate the function: 2 [ (x²/2) + 3x ] evaluated from −3 to 0.
5. Substitute the limits: 2 [ (0 + 0) − (9/2 − 9) ] = 2 [ 0 − (−9/2) ] = 2 × 9/2.
6. Calculate the final area: 2 × 9/2 = 9 square units.

Answer: The graph is a V with vertex (−3, 0); by symmetry area = 2∫₋₃⁰ (x + 3) dx = 2 × 9/2 = 9 square units.

Given √(1 − x²) + √(1 − y²) = a(x − y), prove dy/dx = √((1 − y²)/(1 − x²)).

1. Put x = sin A and y = sin B, so √(1 − x²) = cos A and √(1 − y²) = cos B. The equation becomes cos A + cos B = a(sin A − sin B).
2. Use cos A + cos B = 2cos((A + B)/2)cos((A − B)/2) and sin A − sin B = 2cos((A + B)/2)sin((A − B)/2).
3. Cancelling 2cos((A + B)/2) gives cos((A − B)/2) = a sin((A − B)/2), so cot((A − B)/2) = a.
4. Hence A − B = 2cot⁻¹a, that is, sin⁻¹x − sin⁻¹y = 2cot⁻¹a, a constant.
5. Differentiate with respect to x: 1/√(1 − x²) − (1/√(1 − y²)) dy/dx = 0.
6. So dy/dx = √(1 − y²)/√(1 − x²) = √((1 − y²)/(1 − x²)).

Answer: Put x = sin A, y = sin B: cos A + cos B = a(sin A − sin B) gives cot((A − B)/2) = a, so sin⁻¹x − sin⁻¹y = 2cot⁻¹a. Differentiating, 1/√(1 − x²) − (1/√(1 − y²)) dy/dx = 0, hence dy/dx = √(1 − y²)/√(1 − x²).

x = a(cos θ + log tan(θ/2)) and y = sin θ. Find d²y/dx² at θ = π/4.

1. d/dθ[log tan(θ/2)] = (1/tan(θ/2)) × sec²(θ/2) × ½ = 1/(2 sin(θ/2)cos(θ/2)) = 1/sin θ.
2. So dx/dθ = a(−sin θ + 1/sin θ) = a(1 − sin²θ)/sin θ = a cos²θ/sin θ, and dy/dθ = cos θ.
3. dy/dx = cos θ ÷ (a cos²θ/sin θ) = sin θ/(a cos θ) = tan θ/a.
4. d²y/dx² = [d/dθ(tan θ/a)] ÷ (dx/dθ) = (sec²θ/a) × sin θ/(a cos²θ) = sec³θ tan θ/a².
5. At θ = π/4: sec θ = √2 and tan θ = 1, so d²y/dx² = (√2)³ × 1/a² = 2√2/a².

Answer: dx/dθ = a cos²θ/sin θ and dy/dθ = cos θ, so dy/dx = tan θ / a and d²y/dx² = sec³θ tan θ / a². At θ = π/4 this is 2√2/a².

Find the absolute maximum and minimum of f(x) = 2x³ − 15x² + 36x + 1 on [1, 5].

1. Differentiate the function to find the derivative: f'(x) = 6x² − 30x + 36.
2. Set the derivative equal to zero to find critical points: 6x² − 30x + 36 = 0.
3. Simplify the equation by dividing by 6: x² − 5x + 6 = 0.
4. Factor the quadratic equation: (x − 2)(x − 3) = 0, giving critical points x = 2 and x = 3.
5. Evaluate f(x) at the critical points: f(2) = 2(8) − 15(4) + 36(2) + 1 = 29 and f(3) = 2(27) − 15(9) + 36(3) + 1 = 28.
6. Evaluate f(x) at the endpoints of the interval [1, 5]: f(1) = 2(1) − 15(1) + 36(1) + 1 = 24 and f(5) = 2(125) − 15(25) + 36(5) + 1 = 56.
7. Compare the values 24, 29, 28, and 56 to determine the absolute maximum is 56 at x = 5 and the absolute minimum is 24 at x = 1.

Answer: f′(x) = 6(x − 2)(x − 3), critical points 2 and 3; f(1) = 24, f(2) = 29, f(3) = 28, f(5) = 56. Absolute maximum 56 (at x = 5); absolute minimum 24 (at x = 1).

Find the image A′ of A(1, 6, 3) in the line x/1 = (y − 1)/2 = (z − 2)/3, and the equation of line AA′.

1. Any point on the line is M(λ, 2λ + 1, 3λ + 2); let M be the foot of the perpendicular from A(1, 6, 3).
2. AM has direction ratios λ − 1, 2λ − 5, 3λ − 1, and the line has direction ratios 1, 2, 3.
3. AM ⊥ line: 1(λ − 1) + 2(2λ − 5) + 3(3λ − 1) = 14λ − 14 = 0, so λ = 1 and M = (1, 3, 5).
4. M is the midpoint of AA′: A′ = (2 × 1 − 1, 2 × 3 − 6, 2 × 5 − 3) = (1, 0, 7).
5. AA′ has direction ratios 1 − 1, 0 − 6, 7 − 3 = 0, −6, 4, which are proportional to 0, −3, 2.
6. Line AA′: (x − 1)/0 = (y − 6)/(−3) = (z − 3)/2.

Answer: A general point is M(λ, 2λ + 1, 3λ + 2); AM ⊥ line gives λ = 1, so the foot is M(1, 3, 5) and A′ = (1, 0, 7). Line AA′: (x − 1)/0 = (y − 6)/(−3) = (z − 3)/2.

On the line (x + 5)/1 = (y + 3)/4 = (z − 6)/(−9), find P with PQ = 7 where Q = (2, 4, −1); then find the equation of PQ.

1. Let P be a point on the line, so its coordinates are (λ − 5, 4λ − 3, −9λ + 6).
2. Using the distance formula with Q(2, 4, −1), PQ² = (λ − 7)² + (4λ − 7)² + (−9λ + 7)².
3. Expanding and simplifying gives PQ² = 98λ² − 196λ + 147.
4. Given PQ = 7, set PQ² = 49 to get 98λ² − 196λ + 147 = 49.
5. Rearranging yields 98(λ − 1)² = 0, so λ = 1.
6. Substituting λ = 1 gives P = (−4, 1, −3).
7. The direction ratios of PQ are (2 − (−4), 4 − 1, −1 − (−3)) = (6, 3, 2), so the equation is (x − 2)/6 = (y − 4)/3 = (z + 1)/2.

Answer: P = (λ − 5, 4λ − 3, −9λ + 6); PQ² = 49 gives 98(λ − 1)² = 0, so λ = 1 and P = (-4, 1, -3). Line PQ: (x − 2)/6 = (y − 4)/3 = (z + 1)/2.

180 students go into Sports, Music and Drama clubs. Sports = Music + Drama, and Music = half of Sports plus 20. Use the matrix method to find each club's size.

1. Let Sports = x, Music = y, Drama = z. Then x = y + z, y = x/2 + 20 and x + y + z = 180, i.e. x − y − z = 0, x − 2y = −40, x + y + z = 180.
2. In the form AX = B: A has rows (1, −1, −1), (1, −2, 0), (1, 1, 1) and B = (0, −40, 180).
3. |A| = 1(−2 − 0) − (−1)(1 − 0) + (−1)(1 + 2) = −2 + 1 − 3 = −4 ≠ 0, so A⁻¹ exists.
4. Cofactors: C₁₁ = −2, C₁₂ = −1, C₁₃ = 3; C₂₁ = 0, C₂₂ = 2, C₂₃ = −2; C₃₁ = −2, C₃₂ = −1, C₃₃ = −1. So adj A has rows (−2, 0, −2), (−1, 2, −1), (3, −2, −1).
5. adj A × B = (0 + 0 − 360, 0 − 80 − 180, 0 + 80 − 180) = (−360, −260, −100).
6. X = A⁻¹B = (1/|A|)(adj A)B = (−1/4)(−360, −260, −100) = (90, 65, 25).
7. Sports has 90 students, Music 65 and Drama 25.

Answer: x − y − z = 0, x − 2y = −40, x + y + z = 180; |A| = −4 and X = A⁻¹B gives x, y, z = 90, 65, 25: Sports 90, Music 65, Drama 25.

A rectangular solar installation on a roof uses 300 m of boundary material, including one partition parallel to one pair of sides. Sides perpendicular to the partition are x m, sides parallel to it y m. Write the material equation in x and y.

1. Identify the sides perpendicular to the partition as x, so there are 2 such sides.
2. Identify the sides parallel to the partition as y, which includes 2 outer sides and 1 internal partition, totaling 3 lengths of y.
3. Sum the lengths of all boundary material to form the equation: 2x + 3y = 300.

Answer: Two sides of length x and three lengths y (two sides + partition): 2x + 3y = 300.

Note: CBSE's marking scheme also gives full credit for other readings of the layout, such as 2x + 2y = 300 or 4x + 3y = 300, with the later parts worked to match.

For the same installation (2x + 3y = 300), write the area as a function of x.

1. From the constraint 2x + 3y = 300, solve for y to get y = (300 − 2x)/3.
2. Substitute this expression for y into the area formula A = xy to obtain A(x) = x(300 − 2x)/3.

Answer: y = (300 − 2x)/3, so A(x) = x(300 − 2x)/3.

For A(x) = x(300 − 2x)/3, find the critical point, confirm a maximum by the second derivative test, and find the maximum area.

1. Differentiate A(x) = (300x - 2x²)/3 to get A'(x) = (300 - 4x)/3.
2. Set A'(x) = 0, which gives 300 - 4x = 0, so x = 75.
3. Differentiate again to get A''(x) = -4/3.
4. Since A''(75) = -4/3 < 0, x = 75 is a point of maximum.
5. Substitute x = 75 into A(x): A(75) = 75(300 - 150)/3 = 75(150)/3 = 3750.

Answer: A′(x) = (300 − 4x)/3 = 0 at x = 75; A″(x) = −4/3 < 0, so maximum. Maximum area = 75 × 150/3 = 3750 m².

Using the first derivative test, find the largest area possible with the 300 m of material and the partition.

1. From 2x + 3y = 300, A(x) = x(300 − 2x)/3 = 100x − (2/3)x², so A′(x) = 100 − (4/3)x = (300 − 4x)/3.
2. A′(x) = 0 gives x = 75.
3. For x slightly less than 75, A′(x) > 0; for x slightly more than 75, A′(x) < 0. A′ changes sign from + to −, so x = 75 gives a maximum.
4. Then y = (300 − 150)/3 = 50 and the largest area = 75 × 50 = 3750 m².

Answer: A′(x) = (300 − 4x)/3 changes sign from + to − at x = 75, so it is a maximum; maximum area = 3750 m².

On A = {1, 2, 3}: R₁ = {(2,3),(3,2)}, R₂ = {(1,2),(1,3),(3,2)}, R₃ = {(1,2),(2,1),(1,1)}, R₄ = {(1,1),(1,2),(3,3),(2,2)}, R₅ = {(1,1),(1,2),(3,3),(2,2),(2,1),(2,3),(3,2)}. Which is reflexive and transitive but not symmetric?

1. R₄ contains (1,1), (2,2) and (3,3), so it is reflexive.
2. R₄ contains (1,2) but not (2,1), so it is not symmetric.
3. Transitive: the only chains are (1,1),(1,2) → (1,2) and (1,2),(2,2) → (1,2), and the rest involve pairs (a,a); every result is in R₄, so R₄ is transitive. Answer: R₄.

Answer: R₄: it contains all (a, a), is transitive, but has (1, 2) without (2, 1).

From R₁–R₅ on {1, 2, 3} above, which is reflexive and symmetric but not transitive?

1. R5 is reflexive because it contains (1, 1), (2, 2), and (3, 3), and it is symmetric because for every (a, b) in R5, (b, a) is also in R5.
2. R5 is not transitive because (1, 2) and (2, 3) are in R5, but (1, 3) is not in R5.

Answer: R₅: reflexive and symmetric, but (1, 2), (2, 3) are in it while (1, 3) is not.

From R₁–R₅ on {1, 2, 3} above, which are symmetric yet neither reflexive nor transitive?

1. Symmetric ones: R₁, R₃ and R₅ (R₂ has (1,2) without (2,1); R₄ has (1,2) without (2,1)).
2. R₅ is also reflexive, so it is excluded. R₁ and R₃ lack (2,2), so they are not reflexive.
3. R₁ has (2,3) and (3,2) but not (2,2); R₃ has (2,1) and (1,2) but not (2,2). So neither is transitive.
4. Hence R₁ and R₃ are symmetric but neither reflexive nor transitive.

Answer: R₁ and R₃ (e.g. R₁ has (2,3), (3,2) but not (2,2); R₃ has (2,1), (1,2) but not (2,2)).

Which pairs must be added to R₂ = {(1,2),(1,3),(3,2)} on {1, 2, 3} to make it an equivalence relation?

1. Reflexive: R₂ lacks (1,1), (2,2) and (3,3), so add them.
2. Symmetric: R₂ has (1,2), (1,3) and (3,2), so add (2,1), (3,1) and (2,3).
3. Now the relation contains all 9 pairs, i.e. A × A, which is also transitive.
4. So add (1,1), (2,2), (3,3), (2,1), (3,1) and (2,3).

Answer: Add (1,1), (2,2), (3,3), (2,1), (3,1) and (2,3) (giving all of A × A).

Bank loans are fixed, floating or variable rate with probabilities 10%, 20%, 70%; default rates are 5%, 3%, 1% respectively. Find P(a borrower defaults).

1. Let F, V, and R be the events for fixed, floating, and variable rate loans, and D be the event of default.
2. Apply the law of total probability: P(D) = P(D|F)P(F) + P(D|V)P(V) + P(D|R)P(R).
3. Substitute the given probabilities: P(D) = 0.1(0.05) + 0.2(0.03) + 0.7(0.01).
4. Calculate the result: P(D) = 0.005 + 0.006 + 0.007 = 0.018 = 9/500.

Answer: P(default) = 0.1(0.05) + 0.2(0.03) + 0.7(0.01) = 0.018 = 9/500.

With the same data, a borrower has defaulted. Find the probability the loan was at the variable rate.

1. Let E₃ be a variable-rate loan and D a default. From part (i), P(D) = 0.018.
2. By Bayes' theorem, P(E₃ | D) = P(E₃)P(D | E₃)/P(D) = (0.7 × 0.01)/0.018 = 0.007/0.018.
3. 0.007/0.018 = 7/18.

Answer: P(variable | default) = 0.007/0.018 = 7/18.