CBSE Class 12 Maths Board Paper 2026 (65/1/1): Solutions

If y = 2 cos⁻¹x, what range of values can y take?

1. The principal range of the inverse cosine function cos⁻¹x is [0, π].
2. Since y = 2 cos⁻¹x, the minimum value is 2 × 0 = 0 and the maximum value is 2 × π = 2π.
3. Therefore, the range of y is [0, 2π], which corresponds to option (C).

Answer: cos⁻¹x lies in [0, π], so 0 ≤ y ≤ 2π — option (C).

Which order is impossible for a row matrix: 2 × 1, 1 × 2, 1 × 1 or 1 × n?

1. A row matrix is defined as a matrix with exactly one row and any number of columns.
2. The order 2 × 1 represents a matrix with 2 rows and 1 column, which is a column matrix, not a row matrix.
3. Therefore, 2 × 1 is the impossible order for a row matrix.

Answer: A row matrix has exactly one row, so 2 × 1 is impossible — option (A).

For matrices of suitable order, which transpose rules are true: (i) sum rule (A+B)' = A'+B'; (ii) (A−B)' equals B'−A'; (iii) (AB)' equals A'B'; (iv) (kAB)' equals kB'A' for a scalar k?

1. Recall the standard transpose properties: (A+B)' = A'+B', (A−B)' = A'−B', (AB)' = B'A', and (kAB)' = kB'A' for any scalar k.
2. Compare these with the given statements: (i) matches the sum rule, (ii) incorrectly reverses the order, (iii) incorrectly keeps the order, and (iv) matches the scalar product rule.
3. Thus, only statements (i) and (iv) are true, which corresponds to option (D).

Answer: Only (i) and (iv) are true; (A−B)' = A'−B' and (AB)' = B'A' — option (D).

Δ₁ has rows (1,0,0), (0,2,0), (0,0,3) and Δ₂ has rows (0,2,0), (1,0,0), (0,0,6). How are Δ₁ and Δ₂ related?

1. Calculate the determinant of Δ₁: Δ₁ = 1(2×3 - 0×0) - 0 + 0 = 6.
2. Calculate the determinant of Δ₂: Δ₂ = 0 - 2(1×6 - 0×0) + 0 = -12.
3. Compare the values: Since Δ₂ = -12 and Δ₁ = 6, we have Δ₂ = -2Δ₁, which corresponds to option (B).

Answer: Δ₁ = 6 and Δ₂ = -12, so Δ₂ = -2Δ₁ — option (B).

Find one value of x for which the determinant with rows (cos x, sin x) and (−cos x, sin x) equals 1.

1. The determinant is calculated as (cos x)(sin x) - (sin x)(-cos x) = 2 sin x cos x.
2. Using the identity 2 sin x cos x = sin 2x, we set sin 2x = 1.
3. Solving for x gives 2x = π/2, so x = π/4, which corresponds to option (B).

Answer: The determinant is 2 sin x cos x = sin 2x = 1, so x = π/4 — option (B).

A and B are skew symmetric of the same order. Which of AB, AB + BA, (A + B)², A − B is skew symmetric?

1. Since A and B are skew symmetric, their transposes are A' = -A and B' = -B.
2. Taking the transpose of (A - B) gives (A - B)' = A' - B' = -A - (-B) = -A + B.
3. This simplifies to -(A - B), which satisfies the condition for a skew symmetric matrix, so option (D) is correct.

Answer: (A − B)' = A' − B' = −A + B = −(A − B), so A − B is skew symmetric — option (D).

Least value of f(x) = x³ − 12x on [0, 3].

1. Differentiate the function to find critical points: f'(x) = 3x² − 12.
2. Set f'(x) = 0 to solve for x: 3x² − 12 = 0, which gives x² = 4, so x = 2 (since x is in [0, 3]).
3. Evaluate f(x) at the critical point and endpoints: f(0) = 0, f(2) = 2³ − 12(2) = 8 − 24 = -16, and f(3) = 3³ − 12(3) = 27 − 36 = -9.
4. Compare the values 0, -16, and -9 to determine the least value is -16, corresponding to option (A).

Answer: f'(x) = 0 at x = 2; f(0) = 0, f(2) = -16, f(3) = -9, so the least value is -16 — option (A).

If ∫ 3ax/(b² + c²x²) dx = A log|b² + c²x²| + K, find A.

1. Let t = b² + c²x², then dt = 2c²x dx, which implies x dx = dt/(2c²).
2. Substitute into the integral to get ∫ 3a/(2c²) (dt/t) = (3a/(2c²)) log|t| + K.
3. Comparing with A log|b² + c²x²| + K, we find A = 3a/(2c²).

Answer: Put t = b² + c²x², dt = 2c²x dx, so A = 3a/(2c²) — option (D).

Evaluate the integral of x³/(x² + 2|x| + 1) from x = −1 to 1.

1. Let f(x) = x³ / (x² + 2|x| + 1).
2. Check if f(x) is odd: f(-x) = (-x)³ / ((-x)² + 2|-x| + 1) = -x³ / (x² + 2|x| + 1) = -f(x).
3. Since f(x) is an odd function and the interval [−1, 1] is symmetric about 0, the integral is 0.

Answer: The integrand is odd, so the integral over [−1, 1] is 0 — option (A).

Area enclosed by y = x|x|, the x-axis and the lines x = −1 and x = 1.

1. Since y = x|x| is an odd function, the area enclosed by the curve, the x-axis, and the lines x = -1 and x = 1 is twice the area in the first quadrant, so Area = 2 × ∫₀¹ x² dx.
2. Evaluating the integral gives 2 × [x³/3]₀¹ = 2 × (1/3 - 0) = 2/3 square units, which corresponds to option (C).

Answer: Area = 2 × ∫₀¹ x² dx = 2/3 square units — option (C).

Integrating factor of R dx/dy + Px = Q, where P, Q, R are functions of y.

1. Divide the entire equation by R to obtain the standard linear form dx/dy + (P/R)x = Q/R.
2. Identify the coefficient of x as P/R, so the integrating factor is e^(∫(P/R) dy), which corresponds to option (C).

Answer: Dividing by R gives dx/dy + (P/R)x = Q/R, so IF = e^(∫(P/R) dy) — option (C).

Find the order and the degree of d/dx (eʸ) = 0.

1. Differentiate eʸ with respect to x using the chain rule to get eʸ dy/dx = 0.
2. The highest derivative present is dy/dx, so the order is 1.
3. The power of the highest derivative is 1, so the degree is 1.

Answer: It is eʸ dy/dx = 0, so order 1 and degree 1 — option (B).

Find p so that the vectors (1, 2, 3) and (2, −p, 1) are perpendicular.

1. For two vectors to be perpendicular, their dot product must be zero.
2. Calculate the dot product: (1)(2) + (2)(−p) + (3)(1) = 2 − 2p + 3.
3. Set the dot product to zero and solve: 5 − 2p = 0, so p = 5/2.

Answer: Dot product 2 − 2p + 3 = 0 gives p = 5/2 — option (C).

Find m so that the points (−1, −1, 2), (2, m, 5) and (3, 11, 6) are collinear.

1. Take A(−1, −1, 2), B(2, m, 5) and C(3, 11, 6).
2. AB = (2 + 1, m + 1, 5 − 2) = (3, m + 1, 3) and AC = (3 + 1, 11 + 1, 6 − 2) = (4, 12, 4).
3. Collinear points need AB parallel to AC, so the component ratios must match: 3/4 (x), (m + 1)/12 (y), 3/4 (z).
4. From (m + 1)/12 = 3/4, m + 1 = 9, so m = 8 — option (A).

Answer: AB = (3, m+1, 3) must be parallel to AC = (4, 12, 4), so m + 1 = 9 and m = 8 — option (A).

Vectors a, b have lengths 8 and 3, and the cross product has length 12. Find the absolute value of their dot product.

1. Use the identity |a·b|² + |a×b|² = |a|²|b|² to relate the dot and cross products.
2. Substitute the given values: |a·b|² + 12² = 8² × 3², which simplifies to |a·b|² + 144 = 576.
3. Solve for the dot product magnitude: |a·b|² = 576 − 144 = 432, so |a·b| = √432 = 12√3.

Answer: |a·b|² + |a×b|² = |a|²|b|², so |a·b| = √(576 − 144) = 12√3 — option (C).

Length of the perpendicular from (2, 5, 7) to the line x/1 = y/0 = z/0.

1. The given line x/1 = y/0 = z/0 represents the x-axis, which passes through the origin with direction ratios 1, 0, 0.
2. The length of the perpendicular from point (2, 5, 7) to the x-axis is the distance from the point to its projection (2, 0, 0), calculated as √((5-0)² + (7-0)²) = √(25 + 49) = √74.

Answer: The line is the x-axis, so the distance is √(5² + 7²) = √74 — option (C).

For Z = 5x + 7y over the feasible region shown, find (maximum of Z) − (minimum of Z).

Uses a figure printed in CBSE's paper — what the figure shows: The shaded feasible region is the quadrilateral with corners (0, 0), (0, 2), (3, 4) and (7, 0).

1. Evaluate Z = 5x + 7y at each corner of the feasible region.
2. At (0, 0), Z = 5(0) + 7(0) = 0.
3. At (0, 2), Z = 5(0) + 7(2) = 14.
4. At (3, 4), Z = 5(3) + 7(4) = 15 + 28 = 43.
5. At (7, 0), Z = 5(7) + 7(0) = 35.
6. The maximum value is 43 and the minimum value is 0, so the difference is 43 − 0 = 43.

Answer: Z at (0,0), (0,2), (3,4), (7,0) is 0, 14, 43, 35, so max − min = 43 — option (D).

What is the degree of the objective function in a linear programming problem?

1. A linear programming problem requires the objective function to be a linear expression.
2. Since the highest power of any variable in a linear expression is 1, the degree of the objective function is 1.

Answer: The objective function is linear, so its degree is 1 — option (B).

Assertion: for a fair die, the probability of a prime given an odd number is 2/3. Reason: for any events A, B, the conditional probability of A given B equals P(A ∪ B)/P(B).

1. Identify the odd outcomes on a die as 1, 3, and 5, which gives a total of 3 possible results.
2. Count the prime numbers within these odd outcomes, which are 3 and 5, resulting in 2 favorable cases.
3. Calculate the conditional probability as 2 divided by 3, confirming that the Assertion is true.
4. State that the correct formula for conditional probability involves the intersection of events, not the union, so the Reason is false.
5. Conclude that since the Assertion is true and the Reason is false, the correct option is C.

Answer: Odd outcomes {1, 3, 5} contain primes 3 and 5, so the Assertion is true; the correct formula uses A ∩ B, so the Reason is false — option (C).

Assertion: line 1 (x = py + q, z = ry + s) is perpendicular to line 2 (z = r'y + s', x = p'y + q') when pp' + rr' = 1. Reason: lines r = a₁ + λb₁ and r = a₂ + μb₂ are perpendicular if b₁·b₂ = 0.

1. The direction ratios of line 1 are (p, 1, r) and of line 2 are (p', 1, r').
2. For the lines to be perpendicular, the dot product of their direction vectors must be zero, so pp' + 1(1) + rr' = 0, which implies pp' + rr' = -1.
3. Since the assertion claims pp' + rr' = 1, it is false, while the reason correctly states the condition for perpendicularity, making option (D) the correct choice.

Answer: Directions are (p, 1, r) and (p', 1, r'), so perpendicular needs pp' + rr' = −1; the Assertion is false and the Reason is true — option (D).

Is f continuous at x = 3, where f(x) = |x − 3|/(2(x − 3)) for x < 3 and f(x) = (x − 6)/6 for x ≥ 3?

1. For x < 3, f(x) = |x − 3| / (2(x − 3)) = −(x − 3) / (2(x − 3)) = −1/2.
2. Left Hand Limit (LHL) as x → 3⁻ is −1/2.
3. Right Hand Limit (RHL) as x → 3⁺ is (3 − 6)/6 = −3/6 = −1/2.
4. Value of function at x = 3 is f(3) = (3 − 6)/6 = −1/2.
5. Since LHL = RHL = f(3), f is continuous at x = 3.

Answer: For x < 3, f(x) = −1/2; LHL = −1/2, RHL = (3 − 6)/6 = −1/2 and f(3) = −1/2, so f is continuous at x = 3.

Given √3(x² + y²) = 4xy, find dy/dx at the point (1/2, √3/2).

1. Differentiate the given equation √3(x² + y²) = 4xy implicitly with respect to x to get 2√3x + 2√3y(dy/dx) = 4y + 4x(dy/dx).
2. Rearrange the terms to isolate dy/dx, resulting in dy/dx(2√3y − 4x) = 4y − 2√3x, which simplifies to dy/dx = (2y − √3x)/(√3y − 2x).
3. Substitute the coordinates x = 1/2 and y = √3/2 into the derivative expression to obtain dy/dx = (√3 − √3/2)/(3/2 − 1).
4. Simplify the numerator and denominator to find dy/dx = (√3/2)/(1/2) = √3.

Answer: dy/dx = (2y − √3x)/(√3y − 2x); at (1/2, √3/2) this is (√3/2)/(1/2) = √3.

Perfume in an inverted conical bottle (semi-vertical angle π/6) decreases at 1 mm³/min. How fast is the perfume level falling when it is 10 mm deep?

Uses a figure printed in CBSE's paper — what the figure shows: Figure shows the inverted cone with perfume depth 10 mm and 30° semi-vertical angle (same data as the text).

1. Let h be the depth of the perfume and r be the radius of the surface. From the geometry of the inverted cone with semi-vertical angle π/6, we have tan(π/6) = r/h, so r = h/√3.
2. The volume of the perfume is V = (1/3)πr²h. Substituting r = h/√3 gives V = (1/3)π(h/√3)²h = (1/3)π(h²/3)h = πh³/9.
3. Differentiating with respect to time t, we get dV/dt = (π/9)(3h²)(dh/dt) = (πh²/3)(dh/dt).
4. Given dV/dt = −1 mm³/min and h = 10 mm, substitute these values: −1 = (π(10)²/3)(dh/dt) = (100π/3)(dh/dt).
5. Solving for dh/dt gives dh/dt = −3/(100π) mm/min. The rate at which the level is falling is 3/(100π) mm/min.

Answer: r = h tan(π/6) = h/√3, so V = πh³/9 and dV/dt = (πh²/3) dh/dt = −1. At h = 10, dh/dt = −3/(100π); the level falls at (3/100)/π = 3/(100π) mm/min.

With P(1, 3, 2) and Q(−1, 0, 8), find the vector along QP whose magnitude is 14.

1. Calculate the vector QP by subtracting the coordinates of Q from P: QP = (1 - (-1))i + (3 - 0)j + (2 - 8)k = 2i + 3j - 6k.
2. Find the magnitude of QP: |QP| = √(2² + 3² + (-6)²) = √(4 + 9 + 36) = √49 = 7.
3. Scale the vector QP to have a magnitude of 14 by multiplying it by the ratio 14/7 = 2.
4. The required vector is 2 × (2i + 3j - 6k) = 4i + 6j - 12k.

Answer: QP = 2i + 3j − 6k with |QP| = 7, so the required vector is 2 × QP = 4i + 6j - 12k.

A parallelogram has adjacent sides given by the vectors (3, −2, 2) and (1, 0, 2). Give its diagonal vectors and their lengths.

1. Let a = 3i − 2j + 2k and b = i + 2k be the adjacent sides.
2. One diagonal is a + b = 4i − 2j + 4k, with length √(16 + 4 + 16) = √36 = 6.
3. The other diagonal is a − b = 2i − 2j + 0k, with length √(4 + 4) = √8 = 2√2.
4. So the diagonals are 4i − 2j + 4k (length 6) and 2i − 2j (length 2√2); their negatives are also accepted.

Answer: Diagonals are a + b = 4i − 2j + 4k and a − b = 2i − 2j (or their negatives); lengths 6 and √8 = 2√2.

Simplify tan⁻¹[(cos 2x − sin 2x)/(cos 2x + sin 2x)] for 0 < x < π/4.

1. Divide the numerator and denominator by cos 2x to get tan⁻¹[(1 − tan 2x)/(1 + tan 2x)].
2. Apply the formula tan(A − B) = (tan A − tan B)/(1 + tan A tan B) with A = π/4 and B = 2x to obtain tan⁻¹[tan(π/4 − 2x)].
3. Since 0 < x < π/4, the angle π/4 − 2x lies in (−π/4, π/4), so the expression simplifies to π/4 − 2x.

Answer: Divide by cos 2x: tan⁻¹[(1 − tan 2x)/(1 + tan 2x)] = tan⁻¹[tan(π/4 − 2x)] = π/4 − 2x (valid as π/4 − 2x lies in (−π/4, π/4)).

Evaluate tan(sin⁻¹1 − cos⁻¹(−1/2)).

1. Evaluate the inverse trigonometric terms: sin⁻¹1 = π/2 and cos⁻¹(−1/2) = 2π/3.
2. Substitute these values into the expression to get tan(π/2 − 2π/3).
3. Simplify the angle inside the tangent function: π/2 − 2π/3 = −π/6.
4. Evaluate tan(−π/6) to obtain the final answer −1/√3 ≈ -0.5774.

Answer: sin⁻¹1 = π/2 and cos⁻¹(−1/2) = 2π/3, so the value is tan(−π/6) = −1/√3 ≈ -0.5774.

Evaluate the integral of x tan⁻¹x from 0 to 1.

1. Let I = ∫₀¹ x tan⁻¹x dx. Use integration by parts with u = tan⁻¹x and dv = x dx, so du = dx/(1 + x²) and v = x²/2.
2. Apply the formula ∫ u dv = uv − ∫ v du to get I = [(x²/2) tan⁻¹x]₀¹ − ½∫₀¹ x²/(1 + x²) dx.
3. Evaluate the boundary term: at x = 1, (1/2) tan⁻¹1 = (1/2)(π/4) = π/8; at x = 0, the term is 0. Thus the boundary part is π/8.
4. Simplify the remaining integral: ∫₀¹ x²/(1 + x²) dx = ∫₀¹ (1 − 1/(1 + x²)) dx = [x − tan⁻¹x]₀¹ = 1 − π/4.
5. Combine the results: I = π/8 − ½(1 − π/4) = π/8 − 1/2 + π/8 = π/4 − 1/2 ≈ 0.2854.

Answer: By parts: [(x²/2) tan⁻¹x]₀¹ − ½∫₀¹ x²/(1 + x²) dx = π/8 − ½(1 − π/4) = π/4 − 1/2 ≈ 0.2854.

Find ∫ √((x + 2)/(x − 2)) dx.

1. For the integrand to be real we need x > 2 or x ≤ −2; multiply top and bottom inside the root by (x + 2).
2. √((x + 2)/(x − 2)) = (x + 2)/√((x + 2)(x − 2)) = (x + 2)/√(x² − 4).
3. Split: ∫ x/√(x² − 4) dx + 2∫ 1/√(x² − 4) dx.
4. With u = x² − 4, du = 2x dx, the first part is ∫ du/(2√u) = √u = √(x² − 4).
5. Using ∫ dx/√(x² − a²) = log|x + √(x² − a²)| with a = 2, the second part is 2 log|x + √(x² − 4)|.
6. So the integral is √(x² − 4) + 2 log|x + √(x² − 4)| + C.

Answer: Write it as ∫ (x + 2)/√(x² − 4) dx = √(x² − 4) + 2 log|x + √(x² − 4)| + C.

Find ∫ x²/((x² + 9)(x² + 16)) dx.

1. Let I = ∫ x²/((x² + 9)(x² + 16)) dx.
2. Decompose the integrand: x²/((x² + 9)(x² + 16)) = A/(x² + 9) + B/(x² + 16).
3. Equate numerators: x² = A(x² + 16) + B(x² + 9).
4. Set x² = -9 to find A: -9 = 7A, so A = -9/7.
5. Set x² = -16 to find B: -16 = -7B, so B = 16/7.
6. Substitute A and B into the integral: I = (-9/7)∫ dx/(x² + 9) + (16/7)∫ dx/(x² + 16).
7. Integrate using ∫ dx/(x² + a²) = (1/a)tan⁻¹(x/a): I = (-9/7)(1/3)tan⁻¹(x/3) + (16/7)(1/4)tan⁻¹(x/4) + C = -(3/7)tan⁻¹(x/3) + (4/7)tan⁻¹(x/4) + C.

Answer: With t = x², x²/((x²+9)(x²+16)) = (−9/7)/(x² + 9) + (16/7)/(x² + 16), so the integral is −(3/7) tan⁻¹(x/3) + (4/7) tan⁻¹(x/4) + C.

Let I₁ = ∫ dx/(1 + cos 2x) over [−π/4, π/4] and I₂ = ∫ |x| dx over [−1/2, 1/2]. Prove I₁ − 4I₂ = 0.

1. Use the identity 1 + cos 2x = 2cos²x to rewrite the integrand of I₁ as 1/(2cos²x) = (1/2)sec²x.
2. Since sec²x is an even function, I₁ = 2 × (1/2) ∫₀^(π/4) sec²x dx = ∫₀^(π/4) sec²x dx.
3. Evaluate the integral to get I₁ = [tan x]₀^(π/4) = tan(π/4) − tan(0) = 1 − 0 = 1.
4. Since |x| is an even function, I₂ = 2 ∫₀^(1/2) x dx.
5. Evaluate the integral to get I₂ = 2 [x²/2]₀^(1/2) = [(1/2)² − 0] = 1/4.
6. Substitute the values into the expression: I₁ − 4I₂ = 1 − 4(1/4) = 1 − 1 = 0.
7. Thus, I₁ − 4I₂ = 0 is proved.

Answer: 1 + cos 2x = 2cos²x, so I₁ = 2 × ½∫₀^(π/4) sec²x dx = [tan x]₀^(π/4) = 1. I₂ = 2∫₀^(1/2) x dx = 1/4. Hence I₁ − 4I₂ = 1 − 1 = 0.

General solution of x² dy/dx = x² + xy + y².

1. Identify the differential equation as homogeneous since all terms are of degree 2.
2. Substitute y = vx, which implies dy/dx = v + x(dv/dx).
3. Substitute into the equation to get x²(v + x dv/dx) = x² + x²v + v²x².
4. Simplify by dividing by x² and cancelling v to obtain x dv/dx = 1 + v².
5. Separate variables to write dv/(1 + v²) = dx/x.
6. Integrate both sides to get tan⁻¹(v) = log|x| + C.
7. Replace v with y/x to obtain the general solution tan⁻¹(y/x) = log|x| + C.

Answer: Put y = vx: x dv/dx = 1 + v², so tan⁻¹(y/x) = log|x| + C.

Particular solution of xy dy/dx = (x + 2)(y + 2) with y(1) = −1.

1. Separate the variables: xy dy = (x + 2)(y + 2) dx gives y/(y + 2) dy = (x + 2)/x dx.
2. Write y/(y + 2) = 1 − 2/(y + 2) and (x + 2)/x = 1 + 2/x, so (1 − 2/(y + 2)) dy = (1 + 2/x) dx.
3. Integrate both sides: y − 2 log|y + 2| = x + 2 log|x| + C.
4. Put x = 1, y = −1: −1 − 2 log 1 = 1 + 2 log 1 + C, so −1 = 1 + C and C = −2.
5. Particular solution: y − 2 log|y + 2| = x + 2 log|x| − 2.

Answer: Separate: (1 − 2/(y + 2)) dy = (1 + 2/x) dx, so y − 2 log|y + 2| = x + 2 log|x| + C. y(1) = −1 gives C = −2: y − 2 log|y + 2| = x + 2 log|x| − 2.

Minimise Z = 13x − 15y graphically subject to: x + y at most 7; 2x − 3y + 6 non-negative; x and y non-negative.

1. Draw x + y = 7 (through (7, 0) and (0, 7)) and 2x − 3y + 6 = 0 (through (0, 2) and (−3, 0)); the region is below the first line, on the origin side of the second, and in the first quadrant.
2. Corners: (0, 0), (7, 0), (0, 2), and (3, 4) where x + y = 7 meets 2x − 3y = −6 (substituting y = 7 − x gives 5x = 15, so x = 3, y = 4).
3. Z = 13x − 15y: at (0, 0) Z = 0; at (7, 0) Z = 91; at (3, 4) Z = 39 − 60 = −21; at (0, 2) Z = −30.
4. The region is bounded, so the minimum is Z = −30 at x = 0, y = 2.

Answer: Corners (0, 0), (7, 0), (3, 4), (0, 2) give Z = 0, 91, −21, −30. Minimum Z = -30 at x = 0, y = 2.

Bag I: 3 red, 4 white; bag II: 8 red, 6 white. A die decides: below 3 → bag I, else bag II. One ball is drawn. P(red)?

1. Let E1 be the event that a number less than 3 is obtained, so Bag I is chosen, and E2 be the event that a number 3 or more is obtained, so Bag II is chosen.
2. The probability of choosing Bag I is P(E1) = 2/6, and the probability of choosing Bag II is P(E2) = 4/6.
3. The probability of drawing a red ball from Bag I is P(R|E1) = 3/7, and from Bag II is P(R|E2) = 8/14.
4. Using the law of total probability, P(R) = P(E1)P(R|E1) + P(E2)P(R|E2) = (2/6)(3/7) + (4/6)(8/14).
5. Simplifying the expression gives P(R) = 1/7 + 16/42 = 6/42 + 16/42 = 22/42 = 11/21.

Answer: P(red) = (2/6)(3/7) + (4/6)(8/14) = 11/21.

P(X or Y happens) = a and P(exactly one of X, Y happens) = b. Prove P(X') + P(Y') = 2 − 2a + b.

1. Given P(X ∪ Y) = a and P(exactly one of X, Y happens) = b.
2. Using the formula for exactly one event, P(X ∪ Y) − P(X ∩ Y) = b.
3. Substitute P(X ∪ Y) = a to find P(X ∩ Y) = a − b.
4. Express the required sum as P(X') + P(Y') = (1 − P(X)) + (1 − P(Y)) = 2 − [P(X) + P(Y)].
5. Use the addition rule P(X) + P(Y) = P(X ∪ Y) + P(X ∩ Y) to substitute a + (a − b).
6. Simplify the expression to get 2 − [2a − b] = 2 − 2a + b.

Answer: P(X ∪ Y) = a and P(X ∪ Y) − P(X ∩ Y) = b, so P(X ∩ Y) = a − b. Then P(X') + P(Y') = 2 − [P(X) + P(Y)] = 2 − [P(X ∪ Y) + P(X ∩ Y)] = 2 − [a + a − b] = 2 − 2a + b.

On the integers Z define R by: (x, y) ∈ R when a fixed prime p divides |x − y|. Is R an equivalence relation?

1. To prove R is an equivalence relation, we must verify reflexivity, symmetry, and transitivity.
2. For reflexivity, |x - x| = 0, and since any prime p divides 0, (x, x) is in R.
3. For symmetry, if (x, y) is in R, then p divides |x - y|, which implies p divides |y - x| because |y - x| = |x - y|.
4. For transitivity, if (x, y) and (y, z) are in R, then p divides x - y and y - z.
5. Since p divides both differences, it divides their sum (x - y) + (y - z) = x - z.
6. Therefore, p divides |x - z|, so (x, z) is in R.
7. As R satisfies all three properties, it is an equivalence relation.

Answer: Yes. Reflexive: |x − x| = 0 is divisible by p. Symmetric: |y − x| = |x − y|. Transitive: if p divides x − y and y − z, it divides their sum x − z. So R is an equivalence relation.

Show f(x) = (3x + 2)/(5x − 3), from R − {3/5} to R − {3/5}, is one-one and onto.

1. One-one: let f(x₁) = f(x₂), so (3x₁ + 2)/(5x₁ − 3) = (3x₂ + 2)/(5x₂ − 3).
2. Cross-multiply: 15x₁x₂ − 9x₁ + 10x₂ − 6 = 15x₁x₂ − 9x₂ + 10x₁ − 6, so −19x₁ = −19x₂ and x₁ = x₂. Hence f is one-one.
3. Onto: take any y ≠ 3/5 and solve y = (3x + 2)/(5x − 3): 5xy − 3y = 3x + 2, so x(5y − 3) = 3y + 2 and x = (3y + 2)/(5y − 3).
4. This x is a real number because y ≠ 3/5, and x ≠ 3/5, since x = 3/5 would give 3y + 2 = 3y − 9/5, i.e. 2 = −9/5, which is false.
5. Also f(x) = y for this x, so every y in R − {3/5} has a pre-image. Hence f is one-one and onto.

Answer: One-one: f(x₁) = f(x₂) cross-multiplies to −19x₁ = −19x₂, so x₁ = x₂. Onto: for y ≠ 3/5, x = (3y + 2)/(5y − 3) lies in R − {3/5} and f(x) = y. Hence f is one-one and onto.

A has rows (0, 2, 1), (−2, −1, −2), (1, −1, 0). Find A⁻¹ and use it to solve the system: z − 2y = 7; −y − z + 2x = 8; x − 2y = 10.

1. |A| = 0(0 − 2) − 2(0 + 2) + 1(2 + 1) = 0 − 4 + 3 = −1, so A⁻¹ exists.
2. Cofactors: C11 = −2, C12 = −2, C13 = 3; C21 = −1, C22 = −1, C23 = 2; C31 = −3, C32 = −2, C33 = 4.
3. adj A (transpose of the cofactor matrix) has rows (−2, −1, −3), (−2, −1, −2), (3, 2, 4), so A⁻¹ = (1/|A|) adj A = −adj A, with rows (2, 1, 3), (2, 1, 2), (−3, −2, −4).
4. Write the system as 0x − 2y + z = 7, 2x − y − z = 8, x − 2y + 0z = 10; its coefficient matrix has rows (0, −2, 1), (2, −1, −1), (1, −2, 0), which is A'.
5. So X = (A')⁻¹B = (A⁻¹)'B, where (A⁻¹)' has rows (2, 2, −3), (1, 1, −2), (3, 2, −4) and B = (7, 8, 10).
6. x = 14 + 16 − 30 = 0, y = 7 + 8 − 20 = −5, z = 21 + 16 − 40 = −3.
7. Check: z − 2y = −3 + 10 = 7, −y − z + 2x = 5 + 3 + 0 = 8, x − 2y = 0 + 10 = 10. So x = 0, y = −5, z = −3.

Answer: |A| = −1; A⁻¹ has rows (2, 1, 3), (2, 1, 2), (−3, −2, −4). The system's coefficient matrix is A', so X = (A⁻¹)'B, giving x, y, z = 0, -5, -3.

The matrix with rows (3, −1, sin 3x), (−7, 4, cos 2x), (−11, 7, 2) is singular. Find all x in [0, π/2].

1. Since the matrix is singular, its determinant is zero: 3(8 - 7cos 2x) + 1(-14 + 11cos 2x) + sin 3x(-49 + 44) = 0.
2. Simplify the determinant expression to get 10 - 10cos 2x - 5sin 3x = 0, which reduces to 2 - 2cos 2x = sin 3x.
3. Apply the identity 1 - cos 2x = 2sin²x to the left side to obtain 4sin²x = sin 3x.
4. Use the triple angle formula sin 3x = 3sin x - 4sin³x to rewrite the equation as 4sin²x = 3sin x - 4sin³x.
5. Rearrange and factor the equation: 4sin³x + 4sin²x - 3sin x = 0 becomes sin x(2sin x - 1)(2sin x + 3) = 0.
6. The possible values are sin x = 0, sin x = 1/2, or sin x = -3/2; since x is in [0, π/2], sin x = -3/2 is rejected.
7. Solving for x gives x = 0 for sin x = 0 and x = π/6 for sin x = 1/2, so the solutions are x = 0 and x = π/6.

Answer: Determinant = 10 − 10 cos 2x − 5 sin 3x = 0 ⇒ 2 − 2cos 2x = sin 3x ⇒ 4sin²x = 3 sin x − 4 sin³x ⇒ sin x (2 sin x − 1)(2 sin x + 3) = 0. So sin x = 0 or 1/2: x = 0 or π/6.

Given x = cos t and y = cos mt, prove (1 − x²) y'' − x y' + m²y = 0.

1. dx/dt = −sin t and dy/dt = −m sin mt, so y' = dy/dx = m sin mt / sin t.
2. So sin t · y' = m sin mt; since sin t = √(1 − x²), this is √(1 − x²) y' = m sin mt.
3. Differentiate w.r.t. x. Left side: √(1 − x²) y'' − x y'/√(1 − x²).
4. Right side: m cos mt × m × dt/dx = m² cos mt × (−1/sin t) = −m² cos mt / √(1 − x²).
5. Multiply both sides by √(1 − x²): (1 − x²)y'' − x y' = −m² cos mt.
6. Since cos mt = y, (1 − x²)y'' − x y' = −m²y, i.e. (1 − x²)y'' − x y' + m²y = 0.

Answer: dy/dx = m sin mt / sin t, so sin t · y' = m sin mt, i.e. √(1 − x²) y' = m sin mt. Differentiating w.r.t. x and multiplying by √(1 − x²): (1 − x²)y'' − x y' = −m² cos mt = −m²y. Hence (1 − x²)y'' − xy' + m²y = 0.

Are the lines (x − 1)/2 = (y − 2)/3 = (z − 3)/4 and (x − 4)/5 = (y − 1)/2 = z parallel? If yes, find the gap between them; if not, find where they meet (if they do).

1. Direction ratios are 2, 3, 4 and 5, 2, 1; since 2/5 ≠ 3/2 they are not proportional, so the lines are not parallel.
2. General points: (2λ + 1, 3λ + 2, 4λ + 3) on the first line and (5μ + 4, 2μ + 1, μ) on the second.
3. Equate x and y: 2λ − 5μ = 3 and 3λ − 2μ = −1.
4. Multiply the first by 2 and the second by 5 and subtract: 4λ − 15λ = 6 + 5, so −11λ = 11, λ = −1; then 2(−1) − 5μ = 3 gives μ = −1.
5. Check z: 4(−1) + 3 = −1 and μ = −1, which agree, so the lines intersect.
6. Putting λ = −1 gives the point of intersection (−1, −1, −1).

Answer: Direction ratios 2, 3, 4 and 5, 2, 1 are not proportional, so not parallel. Equating (2λ+1, 3λ+2, 4λ+3) and (5μ+4, 2μ+1, μ) gives λ = μ = −1, which satisfies all three, so the lines meet at (-1, -1, -1).

A delivery firm has 5000 subscribers at ₹300 a year; each ₹1 rise loses 10 subscribers. After a rise of ₹x, how many leave?

1. The problem states that for every ₹1 rise in subscription fee, 10 subscribers are lost.
2. Therefore, for a rise of ₹x, the number of subscribers who discontinue is 10 × x = 10x.

Answer: 10x subscribers discontinue.

Same firm (5000 subscribers, ₹300 fee, 10 lost per ₹1 rise): write revenue R(x) after a fee rise of ₹x.

1. The number of subscribers after a fee rise of ₹x is 5000 − 10x.
2. The new fee per subscriber is 300 + x.
3. Revenue R(x) is the product of subscribers and fee, so R(x) = (5000 − 10x)(300 + x).

Answer: R(x) = (5000 − 10x)(300 + x).

With R(x) = (5000 − 10x)(300 + x), find x giving maximum revenue.

1. Expand R(x) = (5000 − 10x)(300 + x) to get R(x) = 1500000 + 2000x − 10x².
2. Find the first derivative R'(x) = 2000 − 20x.
3. Set R'(x) = 0 to find the critical point: 2000 − 20x = 0 gives x = 100.
4. Find the second derivative R''(x) = −20.
5. Since R''(x) = −20 < 0, R is maximum at x = 100.

Answer: R'(x) = 2000 − 20x = 0 gives x = 100, and R''(x) = −20 < 0, so R is maximum at x = 100.

With R(x) = (5000 − 10x)(300 + x), find where R increases and decreases in (0, 5000).

1. Expand R(x) to get R(x) = 1500000 + 2000x - 10x².
2. Find the derivative R'(x) = 2000 - 20x.
3. Set R'(x) = 0 to find the critical point: 2000 - 20x = 0, so x = 100.
4. Test intervals: for x < 100, R'(x) > 0 (increasing); for x > 100, R'(x) < 0 (decreasing).
5. Conclusion: R is increasing on (0, 100) and decreasing on (100, 5000).

Answer: R'(x) = 2000 − 20x: R is increasing on (0, 100) and decreasing on (100, 5000).

A jackpot sells 1,00,000 tickets with prizes: one ₹3,00,000, two ₹2,00,000 and three ₹50,000. What amounts can a ticket holder win?

1. The possible winning amounts are the specific prize values listed in the problem statement.
2. These amounts are ₹3,00,000, ₹2,00,000, and ₹50,000, with the possibility of winning nothing if the ticket is not a winner.

Answer: ₹3,00,000, ₹2,00,000 or ₹50,000 (or nothing).

Same jackpot (1,00,000 tickets; 1 prize of ₹3,00,000, 2 of ₹2,00,000, 3 of ₹50,000): P(one ticket wins ₹2,00,000 or more)?

1. Identify the tickets that win ₹2,00,000 or more: 1 ticket of ₹3,00,000 and 2 tickets of ₹2,00,000.
2. Calculate the total number of qualifying tickets: 1 + 2 = 3.
3. Determine the probability by dividing the number of qualifying tickets by the total number of tickets: P = 3/100000.

Answer: 3 tickets qualify, so P = 3/100000.

Same jackpot (1,00,000 tickets, 6 prizes in all): P(one ticket wins nothing)?

1. Total number of tickets is 100000 and the number of winning tickets is 6.
2. The probability that a single ticket wins a prize is 6/100000.
3. The probability that the ticket wins nothing is 1 - 6/100000, which simplifies to 99994/100000 or 49997/50000.

Answer: P = 1 − 6/100000 = 99994/100000 = 49997/50000.

Rohan holds one ticket of the above jackpot (6 winning tickets out of 1,00,000) and one ticket of another jackpot with win chance 1 in 1,00,000. P(he wins on exactly one ticket)?

1. Let A be the event of winning the first jackpot and B be the event of winning the second jackpot.
2. P(A) = 6/100000 and P(B) = 1/100000, so P(A') = 99994/100000 and P(B') = 99999/100000.
3. P(exactly one win) = P(A ∩ B') + P(A' ∩ B) = (6/100000)(99999/100000) + (99994/100000)(1/100000).
4. = 599994/10¹⁰ + 99994/10¹⁰ = 699988/10¹⁰ = 174997/2500000000.

Answer: (6/100000)(99999/100000) + (99994/100000)(1/100000) = 699988/10¹⁰ = 174997/2500000000.

Note: CBSE's marking scheme counts any of the 6 prizes as a win on the first ticket, so P(win) = 6/100000 there.

A roundabout's boundary is C₁: x² + y² = 64 and its central pond is C₂: x² + y² = 4. Sketch both.

1. Identify the equations as circles centered at the origin (0, 0) in the standard form x² + y² = r².
2. Calculate the radius of C₁ as √64 = 8 and the radius of C₂ as √4 = 2, resulting in two concentric circles.

Answer: Two concentric circles centred at the origin: C₁ of radius 8 and C₂ of radius 2.

Write y as a function of x for C₁: x² + y² = 64 and C₂: x² + y² = 4.

1. For C₁, rearrange x² + y² = 64 to get y² = 64 − x², so y = ±√(64 − x²).
2. For C₂, rearrange x² + y² = 4 to get y² = 4 − x², so y = ±√(4 − x²).
3. The positive sign represents the upper halves and the negative sign represents the lower halves.

Answer: C₁: y = ±√(64 − x²); C₂: y = ±√(4 − x²) (upper/lower halves).

By integration, find the area inside the roundabout boundary x² + y² = 64.

1. The equation x² + y² = 64 represents a circle with radius r = 8 centered at the origin.
2. The area of the circle is given by the integral 4∫₀⁸ √(64 − x²) dx, where the factor 4 accounts for symmetry in all four quadrants.
3. Using the standard integral formula ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2)sin⁻¹(x/a), we substitute a = 8 to get 4[(x/2)√(64 − x²) + 32sin⁻¹(x/8)] evaluated from 0 to 8.
4. Substituting the limits: at x = 8, the term (8/2)√(64 − 64) is 0 and 32sin⁻¹(1) is 32(π/2) = 16π; at x = 0, both terms are 0.
5. Multiplying the result by 4 gives 4(16π) = 64π square units.

Answer: Area = 4∫₀⁸ √(64 − x²) dx = 4[(x/2)√(64 − x²) + 32 sin⁻¹(x/8)]₀⁸ = 64π square units.

Note: The paper asks for the area covered by the roundabout; CBSE's marking scheme takes the whole disc of C₁ (64π square units). If the pond is left out, the area is 64π − 4π = 60π square units.

By integration, find the area of the pond x² + y² = 4.

1. The equation x² + y² = 4 represents a circle with radius r = 2.
2. The area of the circle is given by 4 times the area in the first quadrant, which is 4∫₀² √(4 − x²) dx.
3. Using the standard integral formula ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2)sin⁻¹(x/a), we evaluate the definite integral.
4. Substituting the limits, we get 4[(x/2)√(4 − x²) + 2sin⁻¹(x/2)]₀² = 4[(0 + 2(π/2)) − (0 + 0)] = 4π square units.

Answer: Area = 4∫₀² √(4 − x²) dx = 4[(x/2)√(4 − x²) + 2 sin⁻¹(x/2)]₀² = 4π square units.