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

Q11 mark

Range of y when 2 cos⁻¹x = y

Answer: (C) 0 ≤ y ≤ 2π

1. The principal value of cos⁻¹x lies in [0, π], so 2 cos⁻¹x lies in [0, 2π].1 mark

Q21 mark

Which order cannot belong to a row matrix

Answer: (A) 2 × 1

1. A row matrix has exactly one row (order 1 × n); 2 × 1 has two rows.1 mark

Q31 mark

Which transpose properties are true

Answer: (D) (i) and (iv)

1. (A + B)′ = A′ + B′ and (kAB)′ = kB′A′ are true. (A − B)′ = A′ − B′, not B′ − A′, and (AB)′ = B′A′, not A′B′.1 mark

Q41 mark

Compare two 3 × 3 determinants Δ₁ and Δ₂

Answer: (B) Δ₂ = −2Δ₁

1. Δ₁ = 1 × 2 × 3 = 6 (diagonal). Expanding Δ₂ along the third column: 6 × (0 × 0 − 2 × 1) = −12 = −2Δ₁.1 mark

Q51 mark

A value of x making a 2 × 2 trig determinant equal 1

Answer: (B) π/4

1. The determinant is cos x sin x + sin x cos x = sin 2x = 1, so 2x = π/2 and x = π/4.1 mark

Q61 mark

Which combination of skew symmetric A and B is skew symmetric

Answer: (D) A − B

1. (A − B)′ = A′ − B′ = −A + B = −(A − B).1 mark

Q71 mark

Least value of x³ − 12x on [0, 3]

Answer: (A) −16

1. f′(x) = 3x² − 12 = 0 at x = 2. f(0) = 0, f(2) = −16, f(3) = −9, so the least value is −16.1 mark

Q81 mark

Find A in a log-form integral of 3ax/(b² + c²x²)

Answer: (D) 3a/(2c²)

1. The derivative of b² + c²x² is 2c²x, so $\int \frac{3ax}{b^2+c^2{x}^2}dx=\frac{3a}{2c^2}\log |b^2+c^2{x}^2|+K$.1 mark

Q91 mark

${\int }_{-1}^1\frac{x^3}{x^2+2|x|+1}dx$

Answer: (A) 0

1. The integrand is odd (the numerator is odd, the denominator even), so the integral over [−1, 1] is 0.1 mark

Q101 mark

Area between y = x|x|, the x-axis, x = −1 and x = 1

Answer: (C) 2/3

1. The curve is symmetric about the origin, so area = $2{\int }_0^1{x}^{2}dx=\frac{2}{3}$.1 mark

Q111 mark

Integrating factor of R dx/dy + Px = Q (P, Q, R functions of y)

Answer: (C) $e^{\int \frac{P}{R}dy}$

1. Divide by R: $\frac{dx}{dy}+\frac{P}{R}x=\frac{Q}{R}$, so IF = $e^{\int \frac{P}{R}dy}$.1 mark

Q121 mark

Order and degree of d/dx(eʸ) = 0

Answer: (B) 1, 1

1. d/dx(eʸ) = eʸ dy/dx = 0: the highest derivative is dy/dx, to power 1.1 mark

Q131 mark

p making î + 2ĵ + 3k̂ and 2î − pĵ + k̂ perpendicular

Answer: (C) 5/2

1. Dot product = 2 − 2p + 3 = 0, so p = 5/2.1 mark

Q141 mark

m making three position vectors collinear

Answer: (A) 8

1. AB = 3î + (m + 1)ĵ + 3k̂ and AC = 4î + 12ĵ + 4k̂. Parallel: 3/4 = (m + 1)/12, so m = 8.1 mark

Q151 mark

|a⃗·b⃗| when |a⃗| = 8, |b⃗| = 3, |a⃗ × b⃗| = 12

Answer: (C) 12√3

1. |a⃗·b⃗|² + |a⃗ × b⃗|² = |a⃗|²|b⃗|² = 576, so |a⃗·b⃗|² = 432 and |a⃗·b⃗| = 12√3.1 mark

Q161 mark

Perpendicular distance from (2, 5, 7) to the x-axis line

Answer: (C) √74

1. The line x/1 = y/0 = z/0 is the x-axis; the foot is (2, 0, 0), so the distance = √(5² + 7²) = √74.1 mark

Q171 mark

LPP from a graph: maximum − minimum of Z = 5x + 7y

Answer: (D) 43

1. Corners (0, 0), (0, 2), (3, 4), (7, 0) give Z = 0, 14, 43, 35. Max − min = 43 − 0 = 43.1 mark

Q181 mark

Degree of an LPP objective function

Answer: (B) 1

1. The objective function of a linear programming problem is linear.1 mark

Q191 mark

Assertion-reason: P(prime | odd) on a die, and the formula for P(A|B)

Answer: (C) A is true, R is false

1. Odd: {1, 3, 5}; prime and odd: {3, 5}, so P = 2/3 (A true). P(A|B) = P(A ∩ B)/P(B), not P(A ∪ B)/P(B) (R false).1 mark

Q201 mark

Assertion-reason: perpendicular lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′

Answer: (D) A is false, R is true

1. The direction ratios are (p, 1, r) and (p′, 1, r′). Perpendicular means pp′ + 1 + rr′ = 0, i.e. pp′ + rr′ = −1, not 1 (A false). R is the true condition b⃗₁·b⃗₂ = 0.1 mark

Q212 marks

(a) Continuity of a piecewise |x − 3| function at x = 3. OR (b) dy/dx for √3(x² + y²) = 4xy at (½, √3/2)

(a)

Answer: Continuous at x = 3

1. For x < 3, |x − 3| = −(x − 3), so f(x) = −½; f(3) = (3 − 6)/6 = −½. LHL = −½.1 mark
2. RHL = lim (x − 6)/6 = −½.½ marks
3. LHL = RHL = f(3), so f is continuous at x = 3.½ marks

(b)

Answer: dy/dx = √3

1. Differentiate: √3(2x + 2y dy/dx) = 4(x dy/dx + y).1 mark
2. dy/dx = (2y − √3x)/(√3y − 2x).½ marks
3. At (½, √3/2): (√3 − √3/2)/(3/2 − 1) = √3.½ marks

Q222 marks

Inverted cone, semi-vertical angle π/6, volume falls 1 mm³/min: rate the level drops at 10 mm

Answer: 3/(100π) mm/min

1. r/h = tan(π/6), so r = h/√3.½ marks
2. v = (1/3)πr²h = πh³/9.½ marks
3. dv/dt = (π/9) × 3h² × dh/dt.½ marks
4. dh/dt = −3/(πh²) = −3/(100π) at h = 10: the level drops at 3/(100π) mm/min.½ marks

Q232 marks

Vector of magnitude 14 along QP, P(1, 3, 2), Q(−1, 0, 8)

Answer: 4î + 6ĵ − 12k̂

1. QP⃗ = 2î + 3ĵ − 6k̂.½ marks
2. |QP⃗| = √(4 + 9 + 36) = 7.½ marks
3. Required vector = 14 × QP⃗/7 = 4î + 6ĵ − 12k̂.1 mark

Q242 marks

Diagonals of the parallelogram on a⃗ = 3î − 2ĵ + 2k̂, b⃗ = î + 2k̂, and their lengths

Answer: 4î − 2ĵ + 4k̂ (length 6) and 2î − 2ĵ (length 2√2)

1. One diagonal: a⃗ + b⃗ = 4î − 2ĵ + 4k̂.½ marks
2. The other: a⃗ − b⃗ = 2î − 2ĵ.½ marks
3. Lengths: √(16 + 4 + 16) = 6 and √(4 + 4) = 2√2.1 mark

Q252 marks

(a) Simplify tan⁻¹[(cos 2x − sin 2x)/(cos 2x + sin 2x)], 0 < x < π/4. OR (b) tan(sin⁻¹1 − cos⁻¹(−½))

(a)

Answer: π/4 − 2x

1. Divide top and bottom by cos 2x: tan⁻¹[(1 − tan 2x)/(1 + tan 2x)].1 mark
2. = tan⁻¹[tan(π/4 − 2x)] = π/4 − 2x, since π/4 − 2x lies in (−π/4, π/4).1 mark

(b)

Answer: −1/√3

1. sin⁻¹1 = π/2 and cos⁻¹(−½) = 2π/3, so the value is tan(π/2 − 2π/3).1 mark
2. = tan(−π/6) = −1/√3.1 mark

Q263 marks

${\int }_0^{1}x{\tan }^{-1}xdx$

Answer: π/4 − ½

1. By parts: ${[\frac{x^2}{2}{\tan }^{-1}x]}_0^1-{\int }_0^1\frac{x^2}{2(1+x^2)}dx$.1 mark
2. $=\frac{\pi }{8}-\frac{1}{2}{\int }_0^1(1-\frac{1}{1+x^2})dx=\frac{\pi }{8}-\frac{1}{2}{[x-{\tan }^{-1}x]}_0^1$1 mark
3. $=\frac{\pi }{8}-\frac{1}{2}+\frac{\pi }{8}=\frac{\pi }{4}-\frac{1}{2}$1 mark

Q273 marks

(a) $\int \sqrt{\frac{x+2}{x-2}}dx$. OR (b) $\int \frac{x^2}{(x^2+9)(x^2+16)}dx$

(a)

Answer: $\sqrt{x^2-4}+2\log |x+\sqrt{x^2-4}|+C$

1. Multiply inside the root by (x + 2)/(x + 2): $I=\int \frac{x+2}{\sqrt{x^2-4}}dx$.1 mark
2. $=\frac{1}{2}\int \frac{2x}{\sqrt{x^2-4}}dx+2\int \frac{dx}{\sqrt{x^2-4}}$1 mark
3. With t = x² − 4, the first part is $\sqrt{x^2-4}$.½ marks
4. The second is $2\log |x+\sqrt{x^2-4}|$, so $I=\sqrt{x^2-4}+2\log |x+\sqrt{x^2-4}|+C$.½ marks

(b)

Answer: $-\frac{3}{7}{\tan }^{-1}\frac{x}{3}+\frac{4}{7}{\tan }^{-1}\frac{x}{4}+C$

1. Put x² = t: $\frac{t}{(t+9)(t+16)}=\frac{A}{t+9}+\frac{B}{t+16}$.½ marks
2. A = −9/7 and B = 16/7.1 mark
3. Integral = $-\frac{9}{7}\int \frac{dx}{x^2+9}+\frac{16}{7}\int \frac{dx}{x^2+16}$½ marks
4. $=-\frac{3}{7}{\tan }^{-1}\frac{x}{3}+\frac{4}{7}{\tan }^{-1}\frac{x}{4}+C$1 mark

Q283 marks

Show I₁ − 4I₂ = 0 for $I_1={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1+\cos 2x}$, $I_2={\int }_{-1/2}^{1/2}|x|dx$

Answer: I₁ = 1, I₂ = ¼

1. The integrand is even: $I_1=2{\int }_0^{\pi /4}\frac{dx}{2{\cos }^{2}x}={\int }_0^{\pi /4}{\sec }^{2}xdx=[\tan x{]}_0^{\pi /4}=1$.1½ marks
2. |x| is even: $I_2=2{\int }_0^{1/2}xdx$.½ marks
3. $=[x^2{]}_0^{1/2}=\frac{1}{4}$.½ marks
4. I₁ − 4I₂ = 1 − 1 = 0.½ marks

Q293 marks

(a) General solution of x² dy/dx = x² + xy + y². OR (b) particular solution of xy dy/dx = (x + 2)(y + 2), y(1) = −1

(a)

Answer: tan⁻¹(y/x) = log|x| + C

1. dy/dx = 1 + y/x + (y/x)²: homogeneous.½ marks
2. Put y = vx, dy/dx = v + x dv/dx: x dv/dx = 1 + v².½ marks
3. Separate: dv/(1 + v²) = dx/x.½ marks
4. Integrate: tan⁻¹v = log|x| + C.1 mark
5. tan⁻¹(y/x) = log|x| + C.½ marks

(b)

Answer: y − 2 log|y + 2| = x + 2 log|x| − 2

1. Separate: y/(y + 2) dy = (x + 2)/x dx.1 mark
2. ∫(1 − 2/(y + 2)) dy = ∫(1 + 2/x) dx gives y − 2 log|y + 2| = x + 2 log|x| + C.1½ marks
3. x = 1, y = −1: −1 − 0 = 1 + 0 + C, so C = −2. Solution: y − 2 log|y + 2| = x + 2 log|x| − 2.½ marks

Q303 marks

LPP: minimise Z = 13x − 15y with x + y ≤ 7, 2x − 3y + 6 ≥ 0, x, y ≥ 0

Answer: Minimum Z = −30 at (0, 2)

1. Draw x + y = 7 and 2x − 3y + 6 = 0 and shade the feasible region (bounded, with corners (0, 0), (7, 0), (3, 4), (0, 2)).1½ marks
2. Z at the corners: 0, 91, −21, −30.1 mark
3. Minimum Z = −30 at x = 0, y = 2.½ marks

Q313 marks

(a) Die chooses a bag (I: 3R 4W, II: 8R 6W): P(red). OR (b) prove P(X′) + P(Y′) = 2 − 2a + b

(a)

Answer: 11/21

1. E₁: die shows less than 3; E₂: 3 or more; A: a red ball is drawn.½ marks
2. P(E₁) = 2/6, P(A|E₁) = 3/7, P(E₂) = 4/6, P(A|E₂) = 8/14.1 mark
3. P(A) = (2/6)(3/7) + (4/6)(8/14)1 mark
4. = 1/7 + 8/21 = 11/21.½ marks

(b)

Answer: Shown

1. At least one occurs: P(X ∪ Y) = a.½ marks
2. Exactly one occurs: P(X ∪ Y) − P(X ∩ Y) = b, so P(X ∩ Y) = a − b.1 mark
3. P(X′) + P(Y′) = 2 − [P(X) + P(Y)].1 mark
4. = 2 − [P(X ∪ Y) + P(X ∩ Y)] = 2 − [a + (a − b)] = 2 − 2a + b.½ marks

Q325 marks

(a) Is R = {(x, y): |x − y| divisible by a prime p} on Z an equivalence relation? OR (b) show f(x) = (3x + 2)/(5x − 3) is one-one and onto

(a)

Answer: Yes, an equivalence relation

1. Reflexive: |x − x| = 0 is divisible by p, so (x, x) ∈ R.1½ marks
2. Symmetric: |y − x| = |x − y|, so (x, y) ∈ R gives (y, x) ∈ R.1½ marks
3. Transitive: x − y = kp and y − z = mp give x − z = (k + m)p, so (x, z) ∈ R.1½ marks
4. Reflexive, symmetric and transitive: R is an equivalence relation.½ marks

(b)

Answer: One-one and onto

1. One-one: f(x₁) = f(x₂) gives (3x₁ + 2)(5x₂ − 3) = (3x₂ + 2)(5x₁ − 3), i.e. 15x₁x₂ − 9x₁ + 10x₂ − 6 = 15x₁x₂ + 10x₁ − 9x₂ − 6, so 19x₁ = 19x₂ and x₁ = x₂.2½ marks
2. Onto: for y ≠ 3/5 put x = (3y + 2)/(5y − 3). This x ≠ 3/5 (that would need 15y + 10 = 15y − 9) and f(x) = y, so every y in the codomain has a pre-image.2½ marks

Q335 marks

(a) Find A⁻¹ for a 3 × 3 matrix and use it to solve a system. OR (b) x in [0, π/2] making a matrix singular

(a)

Answer: x = 0, y = −5, z = −3

1. |A| = −1 ≠ 0 (expanding: 0 − 2(0 + 2) + 1(2 + 1) = −1).1 mark
2. $\text{adj}A=\left[\begin{array}{ccc}-2& -1& -3\\ -2& -1& -2\\ 3& 2& 4\end{array}\right]$1½ marks
3. $A^{-1}=\frac{1}{|A|}\text{adj}A=\left[\begin{array}{ccc}2& 1& 3\\ 2& 1& 2\\ -3& -2& -4\end{array}\right]$½ marks
4. The system's coefficient matrix is A′: A′X = B with B = [7, 8, 10]ᵀ.½ marks
5. X = (A′)⁻¹B = (A⁻¹)′B, giving x = 0, y = −5, z = −3.1½ marks

(b)

Answer: x = 0 or x = π/6

1. Singular: the determinant is 0.1 mark
2. Expanding: 10 − 10 cos 2x − 5 sin 3x = 0, i.e. 2 cos 2x + sin 3x = 2.1 mark
3. Use cos 2x = 1 − 2 sin²x and sin 3x = 3 sin x − 4 sin³x: 4 sin³x + 4 sin²x − 3 sin x = 0.1 mark
4. sin x (2 sin x − 1)(2 sin x + 3) = 0.1 mark
5. sin x = 0 or ½ (sin x = −3/2 is impossible), so x = 0 or π/6.1 mark

Q345 marks

x = cos t, y = cos mt: prove (1 − x²)y″ − xy′ + m²y = 0

Answer: Shown

1. dx/dt = −sin t, dy/dt = −m sin mt.1 mark
2. dy/dx = m sin mt/sin t, i.e. sin t · dy/dx = m sin mt.1 mark
3. Differentiate w.r.t. x: sin t · y″ + cos t · (dt/dx) · y′ = m² cos mt · (dt/dx), with dt/dx = −1/sin t. Multiply by sin t: sin²t · y″ − cos t · y′ = −m² cos mt.2 marks
4. Replace sin²t = 1 − x², cos t = x, cos mt = y: (1 − x²)y″ − xy′ + m²y = 0.1 mark

Q355 marks

Are two given lines parallel? Distance if parallel, else point of intersection

Answer: Not parallel; they meet at (−1, −1, −1)

1. Direction ratios 〈2, 3, 4〉 and 〈5, 2, 1〉.1 mark
2. Not proportional, so the lines are not parallel.1 mark
3. General points: (2λ + 1, 3λ + 2, 4λ + 3) and (5μ + 4, 2μ + 1, μ).1 mark
4. 2λ + 1 = 5μ + 4 and 3λ + 2 = 2μ + 1 give λ = −1, μ = −1, which also satisfy 4λ + 3 = μ: the lines intersect.1 mark
5. Point of intersection: (−1, −1, −1).1 mark

Q364 marks

Case study: delivery subscriptions (5000 at ₹300; 10 leave per ₹1 rise)

(i)

Answer: 10x subscribers

1. 10 leave for each ₹1, so 10x leave.1 mark

(ii)

Answer: R(x) = (5000 − 10x)(300 + x)

1. Revenue = subscribers × fee = (5000 − 10x)(300 + x).1 mark

(iii)(a)

Answer: x = 100

1. R′(x) = 2000 − 20x.½ marks
2. R′(x) = 0 at x = 100.½ marks
3. R″(x) = −20 < 0.½ marks
4. So R is maximum at x = 100 (a fee of ₹400).½ marks

(iii)(b)

Answer: Increasing on (0, 100), decreasing on (100, 5000)

1. R′(x) = 2000 − 20x.½ marks
2. R′(x) = 0 at x = 100.½ marks
3. On (0, 100), R′ > 0: R is increasing.½ marks
4. On (100, 5000), R′ < 0: R is decreasing.½ marks

Q374 marks

Case study: online jackpot (1 × ₹3,00,000, 2 × ₹2,00,000, 3 × ₹50,000; 1,00,000 tickets)

(i)

Answer: ₹3,00,000, ₹2,00,000 or ₹50,000

1. The prize amounts on offer (or nothing).1 mark

(ii)(a)

Answer: 3/100000

1. At least ₹2,00,000 means the first or a second prize: 1 + 2 = 3 tickets of 1,00,000, so P = 3/100000.2 marks

(ii)(b)

Answer: 49997/50000

1. 6 winning tickets: P(no prize) = 1 − 6/100000 = 99994/100000 = 49997/50000.2 marks

(iii)

Answer: 699988/10¹⁰ = 174997/2500000000

1. Win on the first ticket (any of its 6 prizes) and not the second, or the other way round: (6/100000)(99999/100000) + (99994/100000)(1/100000) = 699988/10¹⁰. CBSE's scheme reads 'wins' on the first ticket as any prize; if you count only its first prize the answer is 2 × (1/100000)(99999/100000).1 mark

Q384 marks

Case study: roundabout x² + y² = 64 and pond x² + y² = 4

(i)

Answer: Two circles centred at the origin, radii 8 and 2

1. Sketch both circles about O: radius 8 (roundabout) and radius 2 (pond).1 mark

(ii)

Answer: y = ±√(64 − x²) and y = ±√(4 − x²)

1. C₁: y = √(64 − x²) or y = −√(64 − x²).½ marks
2. C₂: y = √(4 − x²) or y = −√(4 − x²).½ marks

(iii)(a)

Answer: 64π square units

1. Area = $4{\int }_0^8\sqrt{64-x^2}dx$1 mark
2. $=2{[x\sqrt{64-x^2}+64{\sin }^{-1}\frac{x}{8}]}_0^8=64\pi$1 mark

(iii)(b)

Answer: 4π square units

1. Area = $4{\int }_0^2\sqrt{4-x^2}dx$1 mark
2. $=2{[x\sqrt{4-x^2}+4{\sin }^{-1}\frac{x}{2}]}_0^2=4\pi$1 mark