CBSE Class 12 Maths Board Paper 2023 (65/1/1): Solutions

A square matrix A satisfies A² − 3A + I = O, and A⁻¹ = xA + yI. Find x + y.

1. Given A² − 3A + I = O, multiply both sides by A⁻¹ to get A − 3I + A⁻¹ = O.
2. Rearrange the equation to isolate A⁻¹, yielding A⁻¹ = −A + 3I.
3. Compare this with A⁻¹ = xA + yI to identify x = −1 and y = 3.
4. Calculate the sum x + y = −1 + 3 = 2.

Answer: Multiply by A⁻¹: A − 3I + A⁻¹ = O, so A⁻¹ = −A + 3I, giving x = −1, y = 3 and x + y = 2 — option (b).

A is 2 × 2 with |A| = 2. Find the value of |4A⁻¹|.

1. Use the property |kA| = kⁿ|A| for an n×n matrix, so |4A⁻¹| = 4²|A⁻¹| = 16|A⁻¹|. Since |A⁻¹| = 1/|A|, substitute |A| = 2 to get 16 × (1/2) = 8.

Answer: |4A⁻¹| = 4² × 1/|A| = 16/2 = 8 — option (c).

For a 3 × 3 matrix A, |adj A| = 64. What can |A| be?

1. For a 3 × 3 matrix A, the determinant of its adjoint is given by |adj A| = |A|².
2. Substituting the given value, |A|² = 64, which implies |A| = 8 or |A| = −8.

Answer: |adj A| = |A|² for a 3 × 3 matrix, so |A|² = 64 and |A| = 8 or −8 — option (d).

A = [3 4; 5 2] (rows separated by ;). If 2A + B = O (zero matrix), find B.

1. From the equation 2A + B = O, we isolate B to get B = -2A.
2. Multiplying matrix A by -2 gives B = [-2(3), -2(4); -2(5), -2(2)] = [-6, -8; -10, -4].

Answer: B = −2A = [-6, -8; -10, -4] — option (b).

The derivative of f(x) is log x. Find f(x).

1. Since the derivative of f(x) is log x, f(x) is the integral of log x with respect to x.
2. Using integration by parts with u = log x and dv = dx, we get v = x and du = (1/x) dx.
3. Applying the formula ∫ u dv = uv - ∫ v du, we have x log x - ∫ x(1/x) dx.
4. Simplifying the integral gives x log x - ∫ 1 dx = x log x - x + C.
5. Factoring x yields x(log x - 1) + C, which corresponds to option (b).

Answer: By parts, ∫ log x dx = x log x − x + C = x(log x − 1) + C — option (b).

Evaluate the integral of sec²(x − π/6) from x = 0 to x = π/6.

1. The antiderivative of sec²(x − π/6) is tan(x − π/6).
2. Evaluating the definite integral from 0 to π/6 gives [tan(x − π/6)] from 0 to π/6.
3. Substituting the limits: tan(π/6 − π/6) − tan(0 − π/6) = tan(0) − tan(−π/6).
4. Since tan(0) = 0 and tan(−π/6) = −1/√3, the result is 0 − (−1/√3) = 1/√3 ≈ 0.5774.

Answer: [tan(x − π/6)] from 0 to π/6 = 0 − tan(−π/6) = 1/√3 ≈ 0.5774 — option (a).

For d²y/dx² + (dy/dx)³ = sin y, add the order and the degree.

1. The highest order derivative present in the equation is d²y/dx², so the order is 2.
2. The equation is polynomial in its derivatives, and the highest power of the highest order derivative is 1, so the degree is 1.
3. The sum of the order and degree is 2 + 1 = 3.

Answer: Highest derivative is d²y/dx² (order 2) and it appears to power 1 (degree 1), so the sum is 3 — option (c).

Find p so that 2î + pĵ + k̂ is at right angles to −4î − 6ĵ + 26k̂.

1. For two vectors to be at right angles, their dot product must be zero.
2. Calculate the dot product: (2)(−4) + (p)(−6) + (1)(26) = 0, which simplifies to −8 − 6p + 26 = 0.
3. Solve the equation: 18 − 6p = 0, so 6p = 18, giving p = 3.

Answer: Dot product −8 − 6p + 26 = 0 gives p = 3 — option (a).

Evaluate (î × ĵ)·ĵ + (ĵ × î)·k̂.

1. Calculate the cross product of the first term: î × ĵ = k̂.
2. Calculate the dot product of the result with ĵ: k̂ · ĵ = 0.
3. Calculate the cross product of the second term: ĵ × î = −k̂.
4. Calculate the dot product of the result with k̂: −k̂ · k̂ = −1.
5. Add the two results to find the total value: 0 + (−1) = −1.

Answer: î × ĵ = k̂ and k̂·ĵ = 0; ĵ × î = −k̂ and −k̂·k̂ = −1; total −1 — option (d).

a + b = î and a = 2î − 2ĵ + 2k̂. Find |b|.

1. Given a + b = î and a = 2î − 2ĵ + 2k̂, we find b by subtracting a from the sum: b = î − (2î − 2ĵ + 2k̂) = −î + 2ĵ − 2k̂.
2. The magnitude of b is calculated as |b| = √((-1)² + 2² + (-2)²) = √(1 + 4 + 4) = √9 = 3.

Answer: b = î − a = −î + 2ĵ − 2k̂, so |b| = √(1 + 4 + 4) = 3 — option (b).

Find the direction cosines of the line (x − 1)/2 = (1 − y)/3 = (2z − 1)/12.

1. The line is not in standard form yet: write (1 − y)/3 as (y − 1)/(−3), and (2z − 1)/12 = 2(z − 1/2)/12 = (z − 1/2)/6.
2. So the line is (x − 1)/2 = (y − 1)/(−3) = (z − 1/2)/6, with direction ratios 2, −3, 6.
3. Their length is √(4 + 9 + 36) = √49 = 7, so the direction cosines are 2/7, −3/7, 6/7 — option (d).

Answer: In standard form (x − 1)/2 = (y − 1)/(−3) = (z − 1/2)/6, so direction ratios 2, −3, 6 with length 7: direction cosines 2/7, −3/7, 6/7 — option (d).

P(A|B) = 0.3, P(A) = 0.4, P(B) = 0.8. Find P(B|A).

1. Using the formula P(A|B) = P(A ∩ B) / P(B), we calculate P(A ∩ B) = 0.3 × 0.8 = 0.24.
2. Now, using the formula P(B|A) = P(A ∩ B) / P(A), we substitute the values to get P(B|A) = 0.24 / 0.4 = 0.6.

Answer: P(A ∩ B) = 0.3 × 0.8 = 0.24, so P(B|A) = 0.24/0.4 = 0.6 — option (a).

f(x) = 3x + 5 for x ≥ 2 and kx² for x < 2. For which k is f continuous?

1. For f(x) to be continuous at x = 2, the left-hand limit must equal the right-hand limit and the function value.
2. Evaluate the right-hand limit as x approaches 2: lim (x→2+) (3x + 5) = 3(2) + 5 = 11.
3. Evaluate the left-hand limit as x approaches 2: lim (x→2-) (kx²) = k(2)² = 4k.
4. Equate the limits for continuity: 4k = 11.
5. Solve for k: k = 11/4.

Answer: Both pieces must agree at x = 2: 4k = 11, so k = 11/4 — option (d).

A = [0 1; −1 0] and (3I + 4A)(3I − 4A) = x²I. Find x.

1. First, calculate A² by multiplying the matrix A by itself: A² = [0 1; -1 0] × [0 1; -1 0] = [0×0 + 1×(-1) 0×1 + 1×0; -1×0 + 0×(-1) -1×1 + 0×0] = [-1 0; 0 -1] = -I.
2. Next, expand the given expression (3I + 4A)(3I - 4A) using the difference of squares formula: (3I)² - (4A)² = 9I - 16A².
3. Substitute A² = -I into the expanded expression: 9I - 16(-I) = 9I + 16I = 25I.
4. Equate the result to x²I to find x: x²I = 25I implies x² = 25, so x = ±5.

Answer: A² = −I, so the product is 9I − 16A² = 25I, giving x² = 25 and x = ±5 — option (c).

Solve x dy − (1 + x²) dx = dx in general form.

1. Rearrange the given equation x dy − (1 + x²) dx = dx to isolate the differential term x dy = (1 + x²) dx + dx.
2. Simplify the right-hand side to get x dy = (2 + x²) dx, then divide by x to find dy = (2/x + x) dx.
3. Integrate both sides with respect to x to obtain y = 2 log x + x²/2 + C, which corresponds to option (d).

Answer: dy = (2 + x²)/x dx = (2/x + x) dx, so y = 2 log x + x²/2 + C — option (d).

f(x) = a(x − cos x) decreases strictly on all of R. Where must a lie?

1. Differentiate f(x) = a(x − cos x) to get f′(x) = a(1 + sin x).
2. Since 1 + sin x ≥ 0 for all x, f′(x) ≤ 0 requires a < 0, so a ∈ (−∞, 0).

Answer: f′(x) = a(1 + sin x), and 1 + sin x ≥ 0, so f′ ≤ 0 needs a < 0, i.e. a in (−∞, 0) — option (c).

An LPP's feasible region has corners (2, 72), (15, 20), (40, 15). Where is z = 18x + 9y largest and smallest?

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), matching option (c).

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

How many corner points does the region y ≥ 0, x ≥ 0, y ≤ x, x − 2y + 2 ≥ 0 have?

1. Identify the boundary lines: y = 0 (x-axis), x = 0 (y-axis), y = x, and x - 2y + 2 = 0 (or y = x/2 + 1).
2. Find the intersection of y = x and x - 2y + 2 = 0: substituting y = x gives x - 2x + 2 = 0, so x = 2 and y = 2, giving corner (2, 2).
3. Find the intersection of y = 0 and x = 0: this gives the origin (0, 0), which is a corner.
4. Check other intersections: y = 0 meets x - 2y + 2 = 0 at (−2, 0), which is invalid since x ≥ 0; x = 0 meets y = x at (0, 0) (already counted) and meets x - 2y + 2 = 0 at (0, 1), but (0, 1) does not satisfy y ≤ x (1 ≤ 0 is false), so it is not a corner.
5. The region is unbounded in the direction where x and y increase, so there are no other corners; total corners are (0, 0) and (2, 2), hence 2 corners.

Answer: The region is unbounded with corners only at (0, 0) and (2, 2) (where y = x meets 2y = x + 2): 2 corners — option (a).

A: f(x) = 2 sin⁻¹x + 3π/2 on [−1, 1] has range [π/2, 5π/2]. R: the principal branch of sin⁻¹x has range [0, π].

1. The principal range of sin⁻¹x is [−π/2, π/2], so statement R is false.
2. Since sin⁻¹x ∈ [−π/2, π/2], then 2 sin⁻¹x ∈ [−π, π].
3. Adding 3π/2 to the interval [−π, π] gives [π/2, 5π/2], so statement A is true.
4. Therefore, A is true and R is false, which corresponds to option (c).

Answer: sin⁻¹x lies in [−π/2, π/2], so f lies in [π/2, 5π/2]: A is true. The principal range is [−π/2, π/2], not [0, π], so R is false — option (c).

A: the line joining (1, 2, 3) to (3, −1, 3) is (x − 3)/2 = (y + 1)/3 = (z − 3)/0. R: the two-point form of a line in space.

1. Calculate the direction ratios of the line joining (1, 2, 3) and (3, −1, 3) as (3 − 1), (−1 − 2), (3 − 3), which are 2, −3, 0.
2. The correct two-point form is (x − 3)/2 = (y + 1)/(−3) = (z − 3)/0, so Assertion A is false because it uses 3 instead of −3.
3. Reason R correctly states the two-point form of a line in space, so the correct option is (d).

Answer: Direction ratios are 2, −3, 0, so the middle denominator should be −3: A is false; the two-point form in R is correct — option (d).

f : A → B, f(x) = 2x, is a bijection and A = {1, 2, 3, 4}. Find B.

1. Since f is a bijection, it is onto, meaning B is the set of all images of elements in A.
2. Calculate the images: f(1) = 2, f(2) = 4, f(3) = 6, f(4) = 8.
3. Therefore, B = {2, 4, 6, 8}.

Answer: Being onto, B is exactly the set of images: B = {2, 4, 6, 8}.

Find sin⁻¹(sin 3π/4) + cos⁻¹(cos 3π/4) + tan⁻¹(1).

1. The principal value of sin⁻¹(sin 3π/4) is π/4 because 3π/4 is in the second quadrant and sin(3π/4) = sin(π/4).
2. The principal value of cos⁻¹(cos 3π/4) is 3π/4 because 3π/4 lies within the principal range [0, π] of the inverse cosine function.
3. The value of tan⁻¹(1) is π/4 since tan(π/4) = 1 and π/4 is in the principal range (-π/2, π/2).
4. Adding the values gives π/4 + 3π/4 + π/4 = 5π/4.

Answer: sin⁻¹(sin 3π/4) = π/4 (principal range), cos⁻¹(cos 3π/4) = 3π/4, tan⁻¹1 = π/4; sum = 5π/4.

Find every vector of length 3√3 parallel to î + ĵ + k̂.

1. Calculate the magnitude of the vector a = î + ĵ + k̂ as |a| = √(1² + 1² + 1²) = √3.
2. Find the unit vector in the direction of a by dividing by its magnitude: u = (î + ĵ + k̂)/√3.
3. Multiply the unit vector by the given length 3√3 to get the vector: 3√3 × (î + ĵ + k̂)/√3 = 3î + 3ĵ + 3k̂.
4. Since the vector can be parallel in either direction, the other vector is the negative: -3î - 3ĵ - 3k̂.

Answer: Unit vector (î + ĵ + k̂)/√3, so the vectors are ±3√3 times it: 3î + 3ĵ + 3k̂ and −3î − 3ĵ − 3k̂.

A, B, C have position vectors a, b, c and lie on one line in that order, with vector AC = (5/4) vector AB. Write c using a and b.

Uses a figure printed in CBSE's paper — what the figure shows: The figure shows A, B, C on a straight segment in the order A, B, C, with B nearer to C.

1. Express the vector AC in terms of position vectors: AC = c − a.
2. Express the vector AB in terms of position vectors: AB = b − a.
3. Substitute these expressions into the given relation AC = (5/4)AB to get c − a = (5/4)(b − a).
4. Expand the right side to obtain c − a = (5/4)b − (5/4)a.
5. Add a to both sides to isolate c, resulting in c = (5/4)b − (1/4)a, which simplifies to c = (5b − a)/4.

Answer: c − a = (5/4)(b − a), so c = (5b − a)/4.

Line 1: x = 2 + 2λ, y = 1 + 7λ, z = −3 − 3λ. Line 2: x = −2 − μ, y = 8 + 2μ, z = 5 + 4μ. Are they perpendicular?

1. Identify the direction ratios of Line 1 as 2, 7, -3 and Line 2 as -1, 2, 4 from their parametric equations.
2. Calculate the dot product of the direction vectors: 2(-1) + 7(2) + (-3)(4) = -2 + 14 - 12 = 0.
3. Since the dot product is zero, the lines are perpendicular.

Answer: Direction ratios 2, 7, −3 and −1, 2, 4 give 2(−1) + 7(2) + (−3)(4) = 0, so yes, the lines are perpendicular.

y = (x + √(x² − 1))². Prove (x² − 1)(dy/dx)² = 4y².

1. Given y = (x + √(x² − 1))², differentiate with respect to x using the chain rule to get dy/dx = 2(x + √(x² − 1))(1 + x/√(x² − 1)).
2. Simplify the term in the brackets by finding a common denominator: 1 + x/√(x² − 1) = (√(x² − 1) + x)/√(x² − 1).
3. Substitute this back into the derivative expression: dy/dx = 2(x + √(x² − 1))((x + √(x² − 1))/√(x² − 1)) = 2(x + √(x² − 1))²/√(x² − 1).
4. Since y = (x + √(x² − 1))², substitute y into the equation to obtain dy/dx = 2y/√(x² − 1), which implies √(x² − 1) dy/dx = 2y.
5. Square both sides of the equation to get (x² − 1)(dy/dx)² = 4y², thus proving the required identity.

Answer: dy/dx = 2(x + √(x² − 1))(1 + x/√(x² − 1)) = 2(x + √(x² − 1))²/√(x² − 1) = 2y/√(x² − 1). So √(x² − 1) dy/dx = 2y; squaring gives (x² − 1)(dy/dx)² = 4y².

Prove f(x) = 16 sin x/(4 + cos x) − x decreases strictly on (π/2, π).

1. Differentiate f(x) using the quotient rule: f'(x) = [16 cos x(4 + cos x) - 16 sin x(-sin x)] / (4 + cos x)² - 1.
2. Simplify the numerator: 16 cos x(4 + cos x) + 16 sin² x = 64 cos x + 16 cos² x + 16(1 - cos² x) = 64 cos x + 16.
3. Combine terms to get f'(x) = (64 cos x + 16)/(4 + cos x)² - 1 = [64 cos x + 16 - (16 + 8 cos x + cos² x)] / (4 + cos x)².
4. Simplify the expression to f'(x) = cos x(56 - cos x)/(4 + cos x)².
5. On (π/2, π), cos x < 0 and 56 - cos x > 0, so the numerator is negative and the denominator is positive, implying f'(x) < 0.
6. Since f'(x) < 0 on the interval, f(x) decreases strictly on (π/2, π).

Answer: f′(x) = [16 cos x(4 + cos x) + 16 sin²x]/(4 + cos x)² − 1 = cos x(56 − cos x)/(4 + cos x)². On (π/2, π), cos x < 0 while 56 − cos x > 0, so f′(x) < 0 and f strictly decreases there.

Evaluate the integral of log(sin x) − log(2 cos x) from 0 to π/2.

1. Simplify the integrand using log laws: log(sin x) − log(2 cos x) = log(sin x / (2 cos x)) = log(tan x / 2).
2. Let I = ∫₀^(π/2) log(tan x / 2) dx.
3. Apply the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a−x) dx by replacing x with π/2 − x.
4. Substitute to get I = ∫₀^(π/2) log(cot x / 2) dx.
5. Add the two expressions for I: 2I = ∫₀^(π/2) [log(tan x / 2) + log(cot x / 2)] dx.
6. Combine the logs inside the integral: log(tan x / 2) + log(cot x / 2) = log((tan x cot x) / 4) = log(1/4).
7. Evaluate the integral: 2I = ∫₀^(π/2) log(1/4) dx = (π/2) log(1/4), so I = (π/4) log(1/4) = −(π/2) log 2 ≈ −1.0888.

Answer: I = ∫ log(tan x/2) dx; replacing x by π/2 − x gives I = ∫ log(cot x/2) dx. Adding, 2I = ∫₀^(π/2) log(1/4) dx = (π/2) log(1/4), so I = −(π/2) log 2 ≈ −1.0888.

Integrate 1/[√x(√x + 1)(√x + 2)] with respect to x.

1. Substitute √x = t, which implies x = t² and dx = 2t dt.
2. Rewrite the integrand as 1/[t(t + 1)(t + 2)] and substitute dx to get ∫ [1/(t(t + 1)(t + 2))] (2t dt).
3. Simplify the expression by cancelling t to obtain 2∫ dt/[(t + 1)(t + 2)].
4. Use partial fractions to decompose 1/[(t + 1)(t + 2)] into 1/(t + 1) − 1/(t + 2).
5. Integrate the terms to get 2[log|t + 1| − log|t + 2|] + C.
6. Combine the logarithms and substitute t = √x to obtain 2 log|(√x + 1)/(√x + 2)| + C.

Answer: Put √x = t, dx/√x = 2 dt: 2∫ dt/((t + 1)(t + 2)) = 2∫ [1/(t + 1) − 1/(t + 2)] dt = 2 log|(√x + 1)/(√x + 2)| + C.

Solve dy/dx + y sec²x = tan x sec²x with y(0) = 0.

1. This is linear: dy/dx + P y = Q with P = sec²x and Q = tan x sec²x.
2. Integrating factor = e^(∫sec²x dx) = e^(tan x).
3. So d/dx[y e^(tan x)] = tan x sec²x e^(tan x). Put t = tan x, dt = sec²x dx: ∫ t eᵗ dt = eᵗ(t − 1) + C.
4. General solution: y e^(tan x) = e^(tan x)(tan x − 1) + C.
5. At x = 0, y = 0: 0 = 1 × (0 − 1) + C, so C = 1.
6. Dividing by e^(tan x): y = tan x − 1 + e^(−tan x).

Answer: IF = e^(tan x). y e^(tan x) = ∫ tan x sec²x e^(tan x) dx = e^(tan x)(tan x − 1) + C; y(0) = 0 gives C = 1. So y = tan x − 1 + e^(−tan x).

Solve x dy − y dx − √(x² + y²) dx = 0.

1. Divide by x dx: dy/dx = y/x + √(x² + y²)/x = y/x + √(1 + (y/x)²) (taking x > 0). This is homogeneous.
2. Put y = vx, so dy/dx = v + x dv/dx. Then v + x dv/dx = v + √(1 + v²), i.e. x dv/dx = √(1 + v²).
3. Separate: dv/√(1 + v²) = dx/x. Integrating, log|v + √(1 + v²)| = log|x| + log C, so v + √(1 + v²) = Cx.
4. Put v = y/x: y/x + √(x² + y²)/x = Cx. Multiplying by x gives y + √(x² + y²) = Cx².

Answer: dy/dx = y/x + √(1 + (y/x)²). With y = vx: x dv/dx = √(1 + v²), so log|v + √(1 + v²)| = log|x| + log C, i.e. y + √(x² + y²) = Cx².

Maximise z = 6x + 3y graphically subject to x + 5y ≥ 115, 4x + y ≥ 80, 3x + 2y ≤ 150 with x, y ≥ 0.

1. Corner points come from the boundary lines x + 5y = 115, 4x + y = 80 and 3x + 2y = 150.
2. x + 5y = 115 and 4x + y = 80: y = 80 − 4x, so x + 400 − 20x = 115, −19x = −285, x = 15, y = 20. Corner (15, 20).
3. 4x + y = 80 and 3x + 2y = 150: 3x + 2(80 − 4x) = 150 gives −5x = −10, x = 2, y = 72. Corner (2, 72).
4. x + 5y = 115 and 3x + 2y = 150: (3x + 15y) − (3x + 2y) = 345 − 150, so 13y = 195, y = 15 and x = 40. Corner (40, 15).
5. The feasible region is the triangle with these three corners. z = 6x + 3y: at (2, 72) z = 12 + 216 = 228; at (15, 20) z = 90 + 60 = 150; at (40, 15) z = 240 + 45 = 285.
6. Maximum z = 285 at x = 40, y = 15.

Answer: The feasible region is the triangle with corners (2, 72), (15, 20), (40, 15), where z = 228, 150, 285. Maximum z = 285 at x = 40, y = 15.

X takes 1, 2, 3 with probabilities k/2, k/3, k/6. Find (i) k, (ii) P(1 ≤ X < 3), (iii) E(X).

1. Sum of probabilities is 1, so k/2 + k/3 + k/6 = 1.
2. Combine fractions: k(3/6 + 2/6 + 1/6) = k(6/6) = k = 1.
3. P(1 ≤ X < 3) means P(X=1) + P(X=2) = 1/2 + 1/3 = 5/6.
4. E(X) = 1(1/2) + 2(1/3) + 3(1/6) = 1/2 + 2/3 + 1/2 = 5/3.

Answer: k(1/2 + 1/3 + 1/6) = 1, so k = 1; P(1 ≤ X < 3) = 5/6; E(X) = 5/3.

A, B independent with P(A ∩ B′) = 1/4 and P(A′ ∩ B) = 1/6. Find P(A) and P(B).

1. Let P(A) = x, P(B) = y. Independence gives x(1 − y) = 1/4 and (1 − x)y = 1/6.
2. So x − xy = 1/4 and y − xy = 1/6. Subtracting: x − y = 1/12, i.e. y = x − 1/12.
3. Put this in (1 − x)y = 1/6: (1 − x)(x − 1/12) = x − 1/12 − x² + x/12 = −x² + 13x/12 − 1/12 = 1/6.
4. Multiply by −12: 12x² − 13x + 1 = −2, so 12x² − 13x + 3 = 0, i.e. (3x − 1)(4x − 3) = 0.
5. x = 1/3 gives y = 1/3 − 1/12 = 1/4; x = 3/4 gives y = 3/4 − 1/12 = 2/3.
6. So P(A) = 1/3, P(B) = 1/4, or P(A) = 3/4, P(B) = 2/3.

Answer: With P(A) = x, P(B) = y: x(1 − y) = 1/4, (1 − x)y = 1/6, so x − y = 1/12 and 12x² − 13x + 3 = 0. Hence P(A) = 1/3, P(B) = 1/4 or P(A) = 3/4, P(B) = 2/3.

Evaluate the integral of eˣ sin x from 0 to π/2.

1. Let I = ∫ eˣ sin x dx. Using integration by parts with u = sin x and dv = eˣ dx, we get I = eˣ sin x − ∫ eˣ cos x dx.
2. Apply integration by parts again to ∫ eˣ cos x dx with u = cos x and dv = eˣ dx, yielding eˣ cos x + ∫ eˣ sin x dx. Substituting this back gives I = eˣ sin x − (eˣ cos x + I).
3. Rearranging the equation results in 2I = eˣ(sin x − cos x), so the indefinite integral is I = ½eˣ(sin x − cos x).
4. Evaluate the definite integral from 0 to π/2: [½eˣ(sin x − cos x)] from 0 to π/2 = ½e^(π/2)(1 − 0) − ½e⁰(0 − 1) = ½e^(π/2) + ½.
5. The final value is ½(e^(π/2) + 1) ≈ 2.9052.

Answer: By parts twice, ∫ eˣ sin x dx = ½eˣ(sin x − cos x); from 0 to π/2 this is ½(e^(π/2) + 1) ≈ 2.9052.

Integrate 1/[cos(x − a) cos(x − b)] with respect to x.

1. Use the identity sin(A − B) = sin A cos B − cos A sin B with A = x − b and B = x − a to write 1 = sin[(x − b) − (x − a)] / sin(a − b).
2. Substitute this into the integrand to get [sin(x − b)cos(x − a) − cos(x − b)sin(x − a)] / [sin(a − b)cos(x − a)cos(x − b)].
3. Simplify the numerator terms by dividing by the denominator to obtain [tan(x − b) − tan(x − a)] / sin(a − b).
4. Integrate term by term: ∫ tan(x − b) dx = −log|cos(x − b)| and ∫ tan(x − a) dx = −log|cos(x − a)|.
5. Combine the results to get (1/sin(a − b)) [−log|cos(x − b)| + log|cos(x − a)|] + C, which simplifies to (1/sin(a − b)) log|cos(x − a)/cos(x − b)| + C.

Answer: Write 1 = sin[(x − b) − (x − a)]/sin(a − b); the integrand becomes [tan(x − b) − tan(x − a)]/sin(a − b), giving (1/sin(a − b)) log|cos(x − a)/cos(x − b)| + C.

On R, (x, y) ∈ R when the product xy is irrational. Test R for reflexive, symmetric, transitive.

1. To test for reflexivity, consider the element 1 in R.
2. The product 1 × 1 = 1 is rational, so (1, 1) is not in R, meaning R is not reflexive.
3. To test for symmetry, assume (x, y) is in R, which implies xy is irrational.
4. Since multiplication is commutative, yx = xy is also irrational, so (y, x) is in R, meaning R is symmetric.
5. To test for transitivity, consider the elements 1, √2, and 2.
6. The products 1 × √2 = √2 and √2 × 2 = 2√2 are both irrational, so (1, √2) and (√2, 2) are in R.
7. However, 1 × 2 = 2 is rational, so (1, 2) is not in R, meaning R is not transitive.

Answer: Not reflexive: 1·1 = 1 is rational, so (1, 1) ∉ R. Symmetric: xy irrational means yx irrational. Not transitive: (1, √2) and (√2, 2) are in R but 1·2 = 2 is rational, so (1, 2) ∉ R.

A = [1 2 −2; −1 3 0; 0 −2 1] and B⁻¹ = [3 −1 1; −15 6 −5; 5 −2 2]. Find (AB)⁻¹.

1. |A| = 1(3 × 1 − 0) − 2(−1 × 1 − 0) + (−2)((−1)(−2) − 0) = 3 + 2 − 4 = 1.
2. Cofactors: C11 = 3, C12 = 1, C13 = 2; C21 = 2, C22 = 1, C23 = 2; C31 = 6, C32 = 2, C33 = 5.
3. adj A (transpose of the cofactor matrix) = [3, 2, 6; 1, 1, 2; 2, 2, 5], and since |A| = 1, A⁻¹ = adj A.
4. (AB)⁻¹ = B⁻¹A⁻¹ = [3, −1, 1; −15, 6, −5; 5, −2, 2] × [3, 2, 6; 1, 1, 2; 2, 2, 5].
5. Row 1: 9 − 1 + 2 = 10, 6 − 1 + 2 = 7, 18 − 2 + 5 = 21. Row 2: −45 + 6 − 10 = −49, −30 + 6 − 10 = −34, −90 + 12 − 25 = −103.
6. Row 3: 15 − 2 + 4 = 17, 10 − 2 + 4 = 12, 30 − 4 + 10 = 36. So (AB)⁻¹ = [10, 7, 21; −49, −34, −103; 17, 12, 36].

Answer: |A| = 1 and A⁻¹ = adj A = [3, 2, 6; 1, 1, 2; 2, 2, 5]. (AB)⁻¹ = B⁻¹A⁻¹ = [10, 7, 21; -49, -34, -103; 17, 12, 36].

Using matrices, solve x + 2y + 3z = 6, 2x + z − y = 2, 3x + 2y − 2z = 3.

1. Write AX = B with A = [1, 2, 3; 2, −1, 1; 3, 2, −2], X = [x; y; z], B = [6; 2; 3].
2. |A| = 1(2 − 2) − 2(−4 − 3) + 3(4 + 3) = 0 + 14 + 21 = 35 ≠ 0, so A⁻¹ exists.
3. Cofactors: C11 = 0, C12 = 7, C13 = 7; C21 = 10, C22 = −11, C23 = 4; C31 = 5, C32 = 5, C33 = −5.
4. adj A = [0, 10, 5; 7, −11, 5; 7, 4, −5], so A⁻¹ = (1/35)[0, 10, 5; 7, −11, 5; 7, 4, −5].
5. X = A⁻¹B = (1/35)[0 + 20 + 15; 42 − 22 + 15; 42 + 8 − 15] = (1/35)[35; 35; 35].
6. So x = 1, y = 1, z = 1.

Answer: |A| = 35 ≠ 0; A⁻¹ = (1/35)[0, 10, 5; 7, −11, 5; 7, 4, −5], and X = A⁻¹B gives (x, y, z) = 1, 1, 1.

A line goes through (1, 2, −4) parallel to AB, where A(3, 3, −5), B(1, 0, −11). Give its vector and Cartesian forms, then the distance between it and line AB.

1. AB has direction (1 − 3, 0 − 3, −11 + 5) = (−2, −3, −6), so take b = 2î + 3ĵ + 6k̂ (same direction line), with |b| = √(4 + 9 + 36) = 7.
2. Vector form: r = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂). Cartesian form: (x − 1)/2 = (y − 2)/3 = (z + 4)/6.
3. Line AB passes through a₂ = 3î + 3ĵ − 5k̂ with the same direction, so the lines are parallel. With a₁ = î + 2ĵ − 4k̂, a₂ − a₁ = 2î + ĵ − k̂.
4. (a₂ − a₁) × b = î(1 × 6 − (−1) × 3) − ĵ(2 × 6 − (−1) × 2) + k̂(2 × 3 − 1 × 2) = 9î − 14ĵ + 4k̂.
5. Its length is √(81 + 196 + 16) = √293.
6. Distance between the parallel lines = |(a₂ − a₁) × b|/|b| = √293/7 units.

Answer: r = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂); (x − 1)/2 = (y − 2)/3 = (z + 4)/6. With a₂ − a₁ = 2î + ĵ − k̂, (a₂ − a₁) × b = 9î − 14ĵ + 4k̂, so the distance is √293/7 units.

Write the line joining A(1, 2, 3) to B(3, 5, 9); then find its points lying 14 units from B.

1. Direction ratios of AB: 3 − 1 = 2, 5 − 2 = 3, 9 − 3 = 6, so the line is (x − 1)/2 = (y − 2)/3 = (z − 3)/6.
2. Setting each ratio equal to λ, a general point is P(2λ + 1, 3λ + 2, 6λ + 3).
3. PB² = (2λ − 2)² + (3λ − 3)² + (6λ − 6)² = (4 + 9 + 36)(λ − 1)² = 49(λ − 1)², so PB = 7|λ − 1|.
4. PB = 14 gives |λ − 1| = 2, so λ = 3 or λ = −1.
5. λ = 3 gives (7, 11, 21); λ = −1 gives (−1, −1, −3).

Answer: (x − 1)/2 = (y − 2)/3 = (z − 3)/6. A general point (2λ + 1, 3λ + 2, 6λ + 3) is 7|λ − 1| from B, so λ = 3 or −1: the points are (7, 11, 21) and (-1, -1, -3).

By integration, find the area enclosed by y = x², y = x + 2 and the x-axis.

1. x² = x + 2 gives (x − 2)(x + 1) = 0, so the curves meet at x = −1 and x = 2. The line meets the x-axis at x = −2; the parabola touches it at x = 0.
2. The region bounded by all three lies above the x-axis from x = −2 to x = 0: under the line from −2 to −1, and under the parabola from −1 to 0.
3. ∫ from −2 to −1 of (x + 2) dx = [x²/2 + 2x] = (1/2 − 2) − (2 − 4) = −3/2 + 2 = 1/2.
4. ∫ from −1 to 0 of x² dx = [x³/3] = 0 − (−1/3) = 1/3.
5. Area = 1/2 + 1/3 = 5/6 ≈ 0.8333 square units.

Answer: The curves meet at x = −1 and 2; the line cuts the x-axis at −2. Area = ∫₋₂⁻¹ (x + 2) dx + ∫₋₁⁰ x² dx = 1/2 + 1/3 = 5/6 ≈ 0.8333 square units.

Venn diagram of Yoga types A, B, C (see figure); P(C) = 0.44. Find x.

Uses a figure printed in CBSE's paper — what the figure shows: Region values: A only 0.32; A and B only 0.09; B only y; B and C only x; C only 0.21; outside all 0.11. A and C do not overlap.

1. From the Venn diagram, the set C consists of the region 'B and C only' (x) and the region 'C only' (0.21).
2. Since P(C) = 0.44, we have x + 0.21 = 0.44.
3. Solving for x gives x = 0.44 − 0.21 = 0.23.

Answer: C is made of x and 0.21, so x = 0.44 − 0.21 = 0.23.

Same Venn diagram, P(C) = 0.44. Find y.

Uses a figure printed in CBSE's paper — what the figure shows: Region values: A only 0.32; A and B only 0.09; B only y; B and C only x; C only 0.21; outside all 0.11. A and C do not overlap.

1. All regions together have probability 1. The two regions of C add to P(C) = x + 0.21 = 0.44.
2. So 0.32 + 0.09 + y + 0.44 + 0.11 = 1, i.e. 0.96 + y = 1.
3. y = 1 − 0.96 = 0.04.

Answer: All regions add to 1: 0.32 + 0.09 + y + 0.44 + 0.11 = 1, so y = 0.04.

Same Venn diagram. Find P(C | B).

Uses a figure printed in CBSE's paper — what the figure shows: Region values: A only 0.32; A and B only 0.09; B only y = 0.04; B and C only x = 0.23; C only 0.21; outside all 0.11. A and C do not overlap.

1. Identify the probability of the intersection of B and C from the Venn diagram, which is P(B ∩ C) = 0.23.
2. Calculate the total probability of event B by summing the probabilities of all regions within B: P(B) = 0.09 + 0.04 + 0.23 = 0.36.
3. Apply the conditional probability formula P(C | B) = P(B ∩ C) / P(B) to get 0.23 / 0.36.
4. Simplify the fraction to obtain the final answer 23/36.

Answer: P(B ∩ C) = x = 0.23 and P(B) = 0.09 + 0.04 + 0.23 = 0.36, so P(C | B) = 23/36.

Same Venn diagram. Probability that a person practises A or B and not C.

Uses a figure printed in CBSE's paper — what the figure shows: Region values: A only 0.32; A and B only 0.09; B only y = 0.04; B and C only x = 0.23; C only 0.21; outside all 0.11. A and C do not overlap.

1. The event 'A or B and not C' corresponds to the regions containing A or B that do not include C.
2. These regions are A only (0.32), A and B only (0.09), and B only (0.04).
3. Sum these values: 0.32 + 0.09 + 0.04 = 0.45.

Answer: Add the A-only, A-and-B-only and B-only regions: 0.32 + 0.09 + 0.04 = 0.45.

A conical tank part (semi-vertical angle 45°) is full of water draining at 2 cm³/s. Express the water volume through the radius r.

1. Since the semi-vertical angle is 45°, the height h equals the radius r (h = r cot 45° = r).
2. Substitute h = r into the volume formula V = (1/3)πr²h to get V = (1/3)πr²(r) = πr³/3.

Answer: With a 45° half-angle, h = r, so V = (1/3)πr²h = πr³/3.

Same cone (V decreasing at 2 cm³/s). How fast does r change when r = 2√2 cm?

1. From (i), V = πr³/3 (h = r, so h is not constant). Differentiating, dV/dt = πr² dr/dt.
2. Water drains at 2 cm³/s, so dV/dt = −2. At r = 2√2, r² = 8: −2 = 8π dr/dt.
3. dr/dt = −2/(8π) = −1/(4π) cm/s (the radius decreases at 1/(4π) cm/s).

Answer: dV/dt = πr² dr/dt = −2, so dr/dt = −2/(8π) = −1/(4π) cm/s.

Same cone. At r = 2√2 cm, at what rate does the wet curved surface shrink?

1. Given the relationship between slant height and radius, l = √2 r.
2. The wet curved surface area is S = πrl = πr(√2 r) = √2πr².
3. Differentiating with respect to time, dS/dt = 2√2πr (dr/dt).
4. Substituting r = 2√2 and dr/dt = -1/(4π), we get dS/dt = 2√2π(2√2)(-1/(4π)) = -2 cm²/s.

Answer: Slant height l = √2 r, so wet surface S = πrl = √2πr² and dS/dt = 2√2πr dr/dt = 2√2π(2√2)(−1/(4π)) = -2 cm²/s, i.e. it decreases at 2 cm²/s.

Same cone. Find dh/dt at the moment the slant height is 4 cm.

1. With semi-vertical angle 45°, h = r, so l² = h² + r² = 2h². For l = 4: 2h² = 16, h² = 8, h = 2√2 cm.
2. In terms of h, V = πh³/3, so dV/dt = πh² dh/dt.
3. With dV/dt = −2 and h² = 8: −2 = 8π dh/dt, so dh/dt = −1/(4π) cm/s.

Answer: l² = h² + r² with h = r gives r = h = 2√2 cm; dh/dt = dr/dt = −2/(πr²) = −1/(4π) cm/s.

A roller-coaster path is f(x) = a(x − 3)(x + 1)(x + 9) and it meets the y-axis at (0, −1). Find a.

1. The path meets the y-axis at (0, -1), so f(0) = -1.
2. Substitute x = 0 into f(x) = a(x - 3)(x + 1)(x + 9) to get f(0) = a(0 - 3)(0 + 1)(0 + 9).
3. Simplify the expression: f(0) = a(-3)(1)(9) = -27a.
4. Equate to the y-intercept: -27a = -1, so a = 1/27.

Answer: f(0) = a(9)(1)(−3) = −27a = −1, so a = 1/27.

With the same f and the value of a found, evaluate f″(1).

1. With a = 1/27: (x − 3)(x + 1) = x² − 2x − 3, and (x² − 2x − 3)(x + 9) = x³ + 7x² − 21x − 27, so f(x) = (x³ + 7x² − 21x − 27)/27.
2. f′(x) = (3x² + 14x − 21)/27 and f″(x) = (6x + 14)/27.
3. f″(1) = (6 + 14)/27 = 20/27.

Answer: f(x) = (x³ + 7x² − 21x − 27)/27, f″(x) = (6x + 14)/27, so f″(1) = 20/27.