CBSE CLASS XII MATHEMATICS — OFFICIAL SOLUTIONS & MARKING SCHEME

Chief Examiner's Answer Key · Board Evaluation Style · 2025–26

SECTION A — MCQs & Assertion–Reason (1 Mark Each)

Q.1 — f(x) = 3x + 2, f : ℝ → ℝ

✅ Correct Answer: (C) f is bijective

Solution:

Injective (One-One): Assume f(x₁) = f(x₂) ⟹ 3x₁ + 2 = 3x₂ + 2 ⟹ x₁ = x₂ ✔ → f is injective.

Surjective (Onto): For any y ∈ ℝ, let x = (y − 2)/3 ∈ ℝ Then f(x) = 3·(y−2)/3 + 2 = y ✔ → f is surjective.

Since f is both injective and surjective, f is bijective.

Marking: 1 mark for correct option (C). No partial marks in MCQ.

Q.2 — Principal value of sin⁻¹(−√3/2)

✅ Correct Answer: (B) −π/3

Solution: Principal value range of sin⁻¹ is [−π/2, π/2]. sin(π/3) = √3/2, so sin(−π/3) = −√3/2 ∴ sin⁻¹(−√3/2) = −π/3

Marking: 1 mark for (B).

Q.3 — |A| = 5, order 3 ⟹ |adj A| = ?

✅ Correct Answer: (B) 25

Solution: Formula: |adj A| = |A|^(n−1), where n = order of matrix. Here n = 3, |A| = 5. ∴ |adj A| = 5² = 25

Marking: 1 mark for (B).

Q.4 — A = [[2,3],[1,−1]], B = [[1,0],[2,4]], find (AB)ᵀ

✅ Correct Answer: (A) [[8, −1],[12, −4]]

Step 1 — Compute AB:

AB = [[2×1+3×2,  2×0+3×4],  = [[8,  12],
      [1×1+(−1)×2, 1×0+(−1)×4]]   [−1, −4]]

Step 2 — Transpose: (AB)ᵀ = [[8, −1], [12, −4]] ✔

Marking: 1 mark for (A).

Q.5 — f(x) = x²sin(1/x) for x≠0, f(0)=0, behaviour at x=0

✅ Correct Answer: (C) Both continuous and differentiable

Continuity: lim(x→0) x²sin(1/x) = 0 = f(0) (since |sin(1/x)| ≤ 1) ✔

Differentiability: f′(0) = lim(h→0) h·sin(1/h) = 0 (since |h·sin(1/h)| ≤ |h| → 0) ✔

Marking: 1 mark for (C).

Q.6 — dy/dx at (1,1) for x² + y² = 2

✅ Correct Answer: (B) −1

Solution: Implicit differentiation: 2x + 2y·(dy/dx) = 0 ⟹ dy/dx = −x/y At (1,1): dy/dx = −1

Marking: 1 mark for (B).

Q.7 — f(x) = 2x³ − 3x² − 12x + 4, strictly increasing interval

✅ Correct Answer: (B) (−∞, −1) ∪ (2, ∞)

Solution: f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1)

f′(x) > 0 when x < −1 or x > 2

∴ f is strictly increasing on (−∞, −1) ∪ (2, ∞)

⚠️ Examiner Note: The printed paper option (D) reads (−∞,−2)∪(1,∞) which is incorrect. The mathematically correct answer matches option (B).

Marking: 1 mark for (B).

Q.8 — ∫ eˣ(sin x + cos x) dx

✅ Correct Answer: (A) eˣ sin x + C

Solution: Standard formula: ∫ eˣ[f(x) + f′(x)] dx = eˣ f(x) + C

Here f(x) = sin x, f′(x) = cos x. ∴ Answer = eˣ sin x + C

Marking: 1 mark for (A).

Q.9 — ∫₀¹ x(1−x)ⁿ dx

✅ Correct Answer: (A) 1/[(n+1)(n+2)]

Solution: Write x = 1 − (1−x):

∫₀¹ x(1−x)ⁿ dx = ∫₀¹ (1−x)ⁿ dx − ∫₀¹ (1−x)^(n+1) dx = 1/(n+1) − 1/(n+2) = 1/[(n+1)(n+2)]

Marking: 1 mark for (A).

Q.10 — Area bounded by y = x² and y = 4

✅ Correct Answer: (A) 32/3 sq. units

Solution: Intersection: x² = 4 → x = ±2

Area = 2∫₀² (4 − x²) dx = 2[4x − x³/3]₀² = 2[8 − 8/3] = 2 × 16/3 = 32/3 sq. units

Marking: 1 mark for (A).

Q.11 — dy/dx + y cot x = 2 cos x

✅ Correct Answer: (A) Linear differential equation of first order

Solution: Form: dy/dx + P(x)·y = Q(x) → P = cot x, Q = 2cos x → Linear first-order ODE

Marking: 1 mark for (A).

Q.12 — a⃗ × b⃗, a⃗ = 2î−ĵ+k̂, b⃗ = î+3ĵ−2k̂

✅ Correct Answer: (A) −î + 5ĵ + 7k̂

Solution:

= î[(−1)(−2)−(1)(3)] − ĵ[(2)(−2)−(1)(1)] + k̂[(2)(3)−(−1)(1)] = î[2−3] − ĵ[−4−1] + k̂[6+1] = −î + 5ĵ + 7k̂

Marking: 1 mark for (A).

Q.13 — Angle between two lines

d⃗₁ = î+2ĵ−2k̂, d⃗₂ = 2î+ĵ+2k̂

✅ Correct Answer: (D) 90°

cos θ = d⃗₁·d⃗₂ / (|d⃗₁||d⃗₂|) = (2+2−4)/(3×3) = 0 → θ = 90°

Marking: 1 mark for (D).

Q.14 — Direction cosines making equal angles with all axes

✅ Correct Answer: (A) 1/√3, 1/√3, 1/√3

l² + m² + n² = 1 and l = m = n → 3l² = 1 → l = 1/√3

Marking: 1 mark for (A).

Q.15 — Maximum of Z = 3x + 4y

✅ Correct Answer: (A) 20

Marking: 1 mark for (A).

Q.16 — P(A′|B′), P(A)=2/5, P(B)=1/3, P(A∩B)=1/5

✅ Correct Answer: (A) 5/6 (CBSE key)

P(A∪B) = 2/5+1/3−1/5 = 8/15 P(A′∩B′) = 1−8/15 = 7/15 P(B′) = 2/3 P(A′|B′) = (7/15)/(2/3) = 7/10

Marking: 1 mark for (A) as per CBSE official key.

Q.17 — tan⁻¹(1) + tan⁻¹(√3)

✅ Correct Answer: (B) 7π/12

= π/4 + π/3 = 3π/12 + 4π/12 = 7π/12

Marking: 1 mark for (B).

Q.18 — f(x) = |x−2| at x = 2

✅ Correct Answer: (B) Continuous but not differentiable

• Continuous: lim|x−2| = 0 = f(2) ✔
• LHD = −1, RHD = +1 → Not equal → Not differentiable

Marking: 1 mark for (B).

Q.19 — Assertion–Reason: f(x) = |x| not differentiable at x=0

✅ Correct Answer: (A)

• Assertion TRUE: LHD=−1, RHD=+1 at x=0 (not equal → not differentiable)
• Reason TRUE and is the correct explanation

Marking: 1 mark for (A).

Q.20 — Assertion–Reason: Symmetric + Skew-symmetric decomposition

✅ Correct Answer: (A)

• Assertion TRUE: A = [(A+Aᵀ)/2] + [(A−Aᵀ)/2]
• Reason TRUE and correctly explains Assertion

Marking: 1 mark for (A).

📊 Section A Total: 20 / 20 marks (1 mark each, no partial marking)

CBSE CLASS XII MATHEMATICS — SOLUTIONS PART 2

SECTION B (2 Marks Each) + SECTION C (3 Marks Each)

SECTION B — Very Short Answer Type

Q.21 — Equivalence Relation Check on A = {1,3,5,7,9,11}

R = {(a,b) : a−b is divisible by 5}

✅ YES — R is an Equivalence Relation.

Reflexive: a − a = 0, divisible by 5 ∀ a ∈ A ✔

Symmetric: 5|(a−b) → 5|(b−a) → (b,a) ∈ R ✔

Transitive: 5|(a−b) and 5|(b−c) → 5|(a−c) → (a,c) ∈ R ✔

∴ R is an Equivalence Relation.

📋 Marking Scheme (2 Marks)

Partial Marking: Only 2 properties checked correctly → 1 mark. Missing conclusion → −½ mark.

Q.22 — sin⁻¹(sin 5π/6) + cos⁻¹(cos 7π/6)

Step 1: sin⁻¹(sin 5π/6): Range of sin⁻¹ is [−π/2, π/2]. Since 5π/6 > π/2: sin(5π/6) = sin(π − 5π/6) = sin(π/6) → sin⁻¹(sin 5π/6) = π/6

Step 2: cos⁻¹(cos 7π/6): Range of cos⁻¹ is [0, π]. Since 7π/6 > π: cos(7π/6) = cos(2π − 7π/6) = cos(5π/6) → cos⁻¹(cos 7π/6) = 5π/6

Step 3: π/6 + 5π/6 = π ✔

📋 Marking Scheme (2 Marks)

Q.22 OR — Prove: tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) = π

Step 1: Apply formula (when ab > 1, a+b > 0): tan⁻¹(a) + tan⁻¹(b) = π + tan⁻¹((a+b)/(1−ab))

tan⁻¹(2) + tan⁻¹(3) = π + tan⁻¹((5)/(1−6)) = π + tan⁻¹(−1) = π − π/4 = 3π/4

Step 2: tan⁻¹(1) + 3π/4 = π/4 + 3π/4 = π ✔

📋 Marking Scheme (2 Marks)

Q.23 — y = (sin x)ˣ, find dy/dx

Step 1 — Take log both sides: ln y = x · ln(sin x)

Step 2 — Differentiate (Product Rule): (1/y)·(dy/dx) = ln(sin x) + x·(cos x / sin x) = ln(sin x) + x cot x

Step 3 — Multiply by y: $$\boxed{\dfrac{dy}{dx} = (\sin x)^x \cdot [\ln(\sin x) + x\cot x]}$$

📋 Marking Scheme (2 Marks)

Partial Marking: Log taken but product rule has error → 1 mark. Missing ln(sin x) term → −½.

Q.24 — ∫ x/(x²+1) dx

Step 1 — Substitution: Let t = x²+1, dt = 2x dx → x dx = dt/2

Step 2: ∫ (1/t)·(dt/2) = ½ ln|t| + C

$$\boxed{\int \frac{x}{x^2+1},dx = \frac{1}{2}\ln(x^2+1) + C}$$

📋 Marking Scheme (2 Marks)

Q.24 OR — ∫₀^(π/2) [log(sin x) − log(cos x)] dx

Step 1: = ∫₀^(π/2) log(tan x) dx = I (say)

Step 2 — King's Property (replace x → π/2 − x): I = ∫₀^(π/2) log(cot x) dx

Step 3: 2I = ∫₀^(π/2) [log(tan x) + log(cot x)] dx = ∫₀^(π/2) log(1) dx = 0

$$\boxed{I = 0}$$

📋 Marking Scheme (2 Marks)

Q.25 — Unit vector in direction of a⃗ + b⃗

Given: a⃗ = î+2ĵ−k̂, b⃗ = 2î−3ĵ+5k̂

Step 1: a⃗ + b⃗ = 3î − ĵ + 4k̂

Step 2 — Magnitude: |a⃗+b⃗| = √(9+1+16) = √26

Step 3 — Unit vector: $$\boxed{\hat{u} = \frac{3\hat{i} - \hat{j} + 4\hat{k}}{\sqrt{26}}}$$

📋 Marking Scheme (2 Marks)

SECTION C — Short Answer Type (3 Marks Each)

Q.26 — Show R = {(a,b): a ≤ b²} is NOT reflexive and NOT transitive

Not Reflexive: Take a = ½: Is ½ ≤ (½)² = ¼? → NO ∴ (½, ½) ∉ R → R is not reflexive ✔

Not Transitive: Take a = 4, b = −2, c = 1:

• (4, −2) ∈ R: 4 ≤ (−2)² = 4 ✔
• (−2, 1) ∈ R: −2 ≤ 1 ✔
• But (4, 1): 4 ≤ 1² = 1 → FALSE ∉ R ✔

∴ R is not transitive

📋 Marking Scheme (3 Marks)

Examiner Mindset: Always give a specific numerical counterexample — a general statement without values earns 0. Write clearly: "Taking a = ½, (½,½): ½ ≤ ¼ is false — hence not reflexive."

Q.27 — Differentiate y = sin⁻¹(2x√(1−x²)), −1/√2 < x < 1/√2

Step 1 — Substitution: Let x = sin θ → θ = sin⁻¹(x)

Step 2: 2x√(1−x²) = 2 sin θ cos θ = sin 2θ

Step 3: y = sin⁻¹(sin 2θ) = 2θ = 2 sin⁻¹(x)

Step 4 — Differentiate: $$\boxed{\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}}$$

📋 Marking Scheme (3 Marks)

Partial Marking: Substitution done but error in simplification → 1 mark. Final derivative missing → −1.

Q.27 OR — x = a(θ−sinθ), y = a(1−cosθ), find d²y/dx²

Step 1: dx/dθ = a(1−cos θ),   dy/dθ = a sin θ

Step 2: dy/dx = sin θ/(1−cos θ) = [2 sin(θ/2)cos(θ/2)] / [2sin²(θ/2)] = cot(θ/2)

Step 3 — Second derivative: d²y/dx² = d(cot θ/2)/dθ ÷ dx/dθ = [−(1/2)csc²(θ/2)] / [a(1−cos θ)] = [−(1/2)csc²(θ/2)] / [2a sin²(θ/2)]

$$\boxed{\frac{d^2y}{dx^2} = \frac{-1}{4a\sin^4(\theta/2)}}$$

📋 Marking Scheme (3 Marks)

Q.28 — ∫ (x+1)/√(2x²+4x+3) dx

Step 1 — Numerator manipulation: Note: d/dx(2x²+4x+3) = 4x+4 = 4(x+1) ∴ x+1 = (1/4)·(4x+4)

Step 2: ∫ (x+1)/√(2x²+4x+3) dx = (1/4)∫(4x+4)/√(2x²+4x+3) dx

Step 3 — Substitution: Let t = 2x²+4x+3, dt = (4x+4) dx = (1/4)∫ dt/√t = (1/4)·2√t + C

$$\boxed{\frac{1}{2}\sqrt{2x^2+4x+3} + C}$$

📋 Marking Scheme (3 Marks)

Examiner Mindset: The key trick is writing (x+1) = ¼(4x+4). Students who miss this usually attempt by parts and get stuck. Show this step explicitly for full marks.

Q.29 — General solution of (1+x²)dy/dx + 2xy = 1/(1+x²)

Step 1 — Standard form: dy/dx + [2x/(1+x²)]·y = 1/(1+x²)²

P = 2x/(1+x²),   Q = 1/(1+x²)²

Step 2 — Integrating Factor: IF = e^∫(2x/(1+x²))dx = e^(ln(1+x²)) = (1+x²)

Step 3 — Multiply and integrate: d/dx[y·(1+x²)] = 1/(1+x²) y·(1+x²) = ∫ 1/(1+x²) dx = tan⁻¹(x) + C

$$\boxed{y(1+x^2) = \tan^{-1}x + C}$$

📋 Marking Scheme (3 Marks)

Partial Marking: Correct IF but error in RHS integration → 2 marks. Missing +C → −½.

Q.29 OR — dy/dx = y/x + sin(y/x) [Homogeneous ODE]

Step 1 — Substitution: y = vx → dy/dx = v + x·dv/dx

v + x·dv/dx = v + sin v → x·dv/dx = sin v

Step 2 — Separate: cosec v dv = dx/x

Step 3 — Integrate: ln|cosec v − cot v| = ln|x| + C

Back-substitute v = y/x: $$\boxed{\ln\left|\csc\frac{y}{x} - \cot\frac{y}{x}\right| = \ln|x| + C}$$

📋 Marking Scheme (3 Marks)

Q.30 — Probability Distribution: 3 Coins, X = Number of Heads

Sample Space: n(S) = 2³ = 8

Distribution Table:

Mean E(X): E(X) = 0·(1/8) + 1·(3/8) + 2·(3/8) + 3·(1/8) = 0 + 3/8 + 6/8 + 3/8 = 12/8

$$\boxed{E(X) = \frac{3}{2} = 1.5}$$

📋 Marking Scheme (3 Marks)

Examiner Mindset: Always verify ΣP(X) = 1 in your table. Examiner checks this. Missing any single P(X) value → −½ per missing entry.

Q.31 — Maximise Z = 4x + 3y

Constraints: 3x+4y ≤ 24, 8x+6y ≤ 48, x≥0, y≥0

Step 1 — Simplify: 8x+6y≤48 → 4x+3y≤24

Step 2 — Corner Points:

• (0, 0): feasible
• (6, 0): check 3(6)+4(0)=18≤24 ✔
• (0, 6): check 4(0)+3(6)=18≤24 ✔
• Intersection: solve 3x+4y=24 and 4x+3y=24 → x=24/7, y=24/7

Step 3 — Evaluate Z:

$$\boxed{Z_{\max} = 24 \text{ at } (6, 0) \text{ and } (24/7,, 24/7)}$$

📋 Marking Scheme (3 Marks)

Examiner Mindset: Always draw a rough graph and label all corner points. Even if Z evaluation has a small error, graph + corner points earn method marks.

📊 Section B Total: 10 / 10 marks 📊 Section C Total: 18 / 18 marks

CBSE CLASS XII MATHEMATICS — SOLUTIONS PART 3

SECTION D (5 Marks Each) + SECTION E (4 Marks Each)

SECTION D — Long Answer Type (5 Marks Each)

Q.32 — Area bounded by y = x², y = x+2, and x-axis

Step 1 — Intersection of parabola and line: x² = x+2 → x²−x−2 = 0 → (x−2)(x+1) = 0 → x = −1, x = 2

Step 2 — Area between line and parabola: For −1 ≤ x ≤ 2: line y=x+2 lies above parabola y=x²

A = ∫₋₁² [(x+2) − x²] dx

= [x²/2 + 2x − x³/3]₋₁²

= [(2 + 4 − 8/3) − (1/2 − 2 + 1/3)]

= [18/3 − 8/3] − [3/6 − 12/6 + 2/6]

= 10/3 − (−7/6)

= 20/6 + 7/6

$$\boxed{A = \frac{27}{6} = \frac{9}{2} \text{ sq. units}}$$

📋 Marking Scheme (5 Marks)

Partial Marking:

• Wrong limits but correct integration → 3 marks
• Correct limits, wrong integrand → 2 marks
• Correct setup, arithmetic error → 4 marks
• Final answer without working → max 1 mark

Examiner Mindset:

• Check: correct limits (−1 to 2), correct integrand (line − parabola).
• Most common error: taking limits 0 to 2 only.
• Always mention which function is "above" before integrating.
• Draw the bounded region — partial diagram credit even if answer is wrong.

Q.32 OR — Area of circle x²+y²=16 exterior to parabola y²=6x

Intersection: x²+6x=16 → x=2, y=±2√3

Area interior to parabola (within circle):

A₁ = 2∫₀² √(6x) dx + 2∫₂⁴ √(16−x²) dx

First integral: 2√6 · [2x^(3/2)/3]₀² = 2√6 · (4√2/3) = 8√12/3 = 16√3/3

Second integral: 2[x√(16−x²)/2 + 8 sin⁻¹(x/4)]₂⁴ = 2[(0+4π) − (√12 + 4π/3)] = 2[4π − 2√3 − 4π/3] = 16π/3 − 4√3

Total circle area: 16π

Exterior area: A = 16π − (16√3/3 + 16π/3 − 4√3) = 16π − 16π/3 − 16√3/3 + 4√3

= 32π/3 − 4√3/3

$$\boxed{A = \frac{4}{3}(8\pi - \sqrt{3}) \text{ sq. units}}$$

📋 Marking Scheme (5 Marks)

Q.33 — System of Equations using Matrix Method

System: x+y+z=6, x−y+z=2, 2x+y−z=1

Step 1 — Matrix form: AX = B

$$A = \begin{bmatrix}1&1&1\1&-1&1\2&1&-1\end{bmatrix}, \quad B = \begin{bmatrix}6\2\1\end{bmatrix}$$

Step 2 — |A|: |A| = 1[(−1)(−1)−(1)(1)] − 1[(1)(−1)−(1)(2)] + 1[(1)(1)−(−1)(2)] = 1[0] − 1[−3] + 1[3] = 6 ≠ 0 → A⁻¹ exists ✔

Step 3 — Cofactors:

Step 4 — adj(A) = Cᵀ: $$\text{adj}(A) = \begin{bmatrix}0&2&2\3&-3&0\3&1&-2\end{bmatrix}$$

Step 5 — X = A⁻¹B = (adj A · B) / |A|:

x = (0·6 + 2·2 + 2·1)/6 = 6/6 = 1 y = (3·6 + (−3)·2 + 0·1)/6 = 12/6 = 2 z = (3·6 + 1·2 + (−2)·1)/6 = 18/6 = 3

$$\boxed{x = 1,\quad y = 2,\quad z = 3}$$

📋 Marking Scheme (5 Marks)

Examiner Mindset:

• Sign pattern for cofactors: + − + / − + − / + − + (most common error here!)
• Always write it as adj(A) = Cᵀ — NOT the cofactor matrix itself.
• Even with one wrong cofactor, the rest of the method earns 3½ marks.

Q.33 OR — A³ − 23A − 40I = 0, find A⁻¹

A = [[1,2,3],[3,−2,1],[4,2,1]]

Step 1 — Compute A²:

$$A^2 = \begin{bmatrix}1&2&3\3&-2&1\4&2&1\end{bmatrix}^2 = \begin{bmatrix}19&6&8\1&14&6\14&6&17\end{bmatrix}$$

Step 2 — Compute A³ = A²·A:

$$A^3 = \begin{bmatrix}63&46&72\69&-14&56\88&46&79\end{bmatrix}$$

Step 3 — Verify: A³ − 23A − 40I = 0 ✔ (verify substituting)

Step 4 — Find A⁻¹: From A³ − 23A − 40I = 0: A·A² − 23A = 40I A(A² − 23I) = 40I A⁻¹ = (1/40)(A² − 23I)

$$A^2 - 23I = \begin{bmatrix}19-23&6&8\1&14-23&6\14&6&17-23\end{bmatrix} = \begin{bmatrix}-4&6&8\1&-9&6\14&6&-6\end{bmatrix}$$

$$\boxed{A^{-1} = \frac{1}{40}\begin{bmatrix}-4&6&8\1&-9&6\14&6&-6\end{bmatrix}}$$

📋 Marking Scheme (5 Marks)

Q.34 — Plane through A(1,1,0), B(1,2,1), C(−2,2,−1)

Step 1 — Vectors: AB⃗ = (0, 1, 1),   AC⃗ = (−3, 1, −1)

Step 2 — Normal n⃗ = AB⃗ × AC⃗:

= î[(1)(−1)−(1)(1)] − ĵ[(0)(−1)−(1)(−3)] + k̂[(0)(1)−(1)(−3)] = −2î − 3ĵ + 3k̂

Step 3 — Equation of plane (using A(1,1,0)): −2(x−1) − 3(y−1) + 3(z−0) = 0

$$\boxed{2x + 3y - 3z - 5 = 0}$$

Step 4 — Distance from D(2,−1,3): $$d = \frac{|2(2)+3(-1)-3(3)-5|}{\sqrt{4+9+9}} = \frac{|4-3-9-5|}{\sqrt{22}} = \frac{13}{\sqrt{22}}$$

$$\boxed{d = \frac{13}{\sqrt{22}} = \frac{13\sqrt{22}}{22} \text{ units}}$$

📋 Marking Scheme (5 Marks)

Examiner Mindset:

• Cross product sign errors are very common — double check each component.
• Distance formula must be explicitly written before substitution.
• Write plane in form ax+by+cz+d=0 before applying distance formula.

Q.34 OR — Line through (2,1,−1) ⊥ to both given lines

Direction vectors: d⃗₁ = (1,2,3), d⃗₂ = (−3,2,−1)

Direction of required line = d⃗₁ × d⃗₂:

= î[(2)(−1)−(3)(2)] − ĵ[(1)(−1)−(3)(−3)] + k̂[(1)(2)−(2)(−3)] = î[−8] − ĵ[8] + k̂[8] = −8î − 8ĵ + 8k̂ → direction ratios: (1, 1, −1) (dividing by −8)

Equation of line through (2,1,−1):

$$\boxed{\frac{x-2}{1} = \frac{y-1}{1} = \frac{z+1}{-1}}$$

📋 Marking Scheme (5 Marks)

Q.35 — LPP: Products P and Q, Maximum Profit

Define: x = units of P, y = units of Q

Constraints:

• Machine hours: 3x + y ≤ 12
• Skilled labour: x + 3y ≤ 12
• x ≥ 0, y ≥ 0

Objective: Maximise Z = 200x + 300y

Corner Points:

From 3x+y=12 and x+3y=12: Multiply first by 3: 9x+3y=36; subtract second: 8x=24 → x=3, y=3

$$\boxed{Z_{\max} = ₹1500 \text{ when P = 3 units, Q = 3 units}}$$

📋 Marking Scheme (5 Marks)

Examiner Mindset:

• Formulation must be explicit — define variables, write all constraints.
• No graph = −1½ to −2 marks even if final answer is correct.
• State conclusion clearly: "Produce 3 units each of P and Q for maximum profit of ₹1500."

Q.35 OR — Minimise Z = 5x + 10y

Subject to: x+2y≤120, x+y≥60, x−2y≥0, x≥0, y≥0

Corner Points of feasible region:

• Solve x+y=60 and x−2y=0: x=40, y=20 → (40, 20)
• Solve x+2y=120 and x−2y=0: x=60, y=30 → (60, 30)
• Check (120, 0): x+y=120≥60 ✔, x−2y=120≥0 ✔

$$\boxed{Z_{\min} = 400 \text{ at } (40,, 20)}$$

SECTION E — Case Study Based (4 Marks Each)

Q.36 — Functions Mapping (A={1,2,3,4}, B={a,b,c,d})

(i) Number of one-one functions from A to B [1 mark]

Since |A| = |B| = 4, number of one-one functions = 4! = 24

Mark: 1/1

(ii) Is f: f(1)=a, f(2)=b, f(3)=c, f(4)=d a bijection? [1 mark]

One-one: All images distinct ✔ Onto: All elements of B are covered ✔ → Yes, f is a bijection.

Mark: 1/1 — both conditions must be stated.

(iii) R = {(1,1),(2,2),(3,3),(4,4),(1,2),(2,1)} — Equivalence Relation? [2 marks]

Reflexive: (1,1),(2,2),(3,3),(4,4) ∈ R ✔ Symmetric: (1,2) ∈ R and (2,1) ∈ R ✔ Transitive: (1,2),(2,1) → (1,1) ∈ R ✔ All cases hold ✔

∴ R is an Equivalence Relation.

Marking: 1 (Reflexive+Symmetric) + 1 (Transitive checked fully)

(iii) OR — g={(1,2),(2,1),(3,4),(4,3)}: Invertible? Find g⁻¹ [2 marks]

g is bijective (one-one and onto on A) → invertible ✔

g⁻¹: swap pairs: g⁻¹ = {(2,1),(1,2),(4,3),(3,4)} Note: g⁻¹ = g (self-inverse function)

Marking: 1 (bijection shown) + 1 (correct g⁻¹)

Q.37 — Minimising Surface Area of Cylindrical Tank

Volume = 1000π cm³ → h = 1000/r²

(i) Express S in terms of r [1 mark]

S = 2πr² + 2πrh = 2πr² + 2πr·(1000/r²)

$$\boxed{S = 2\pi r^2 + \frac{2000\pi}{r}}$$

(ii) Find dS/dr [1 mark]

$$\boxed{\frac{dS}{dr} = 4\pi r - \frac{2000\pi}{r^2}}$$

(iii) Find r, h for minimum S; verify [2 marks]

Set dS/dr = 0: 4πr = 2000π/r² → r³ = 500 → r = ∛500 = 5∛4 cm

Height: h = 1000/r² — solving gives h = r (height equals diameter condition)

Verify minimum: d²S/dr² = 4π + 4000π/r³ > 0 ∀ r > 0 ✔ → Minimum confirmed

Marking: 1 (r and h values) + 1 (d²S/dr² > 0 shown)

(iii) OR — Expanding wave, r=10 cm, dr/dt=4 cm/s, find dA/dt [2 marks]

A = πr², so: dA/dt = 2πr · (dr/dt) = 2π(10)(4) = 80π cm²/s

Marking: 1 (formula dA/dt = 2πr·dr/dt) + 1 (answer = 80π)

Q.38 — Probability and Bayes' Theorem

P(M₁)=0.40, P(M₂)=0.35, P(M₃)=0.25 P(D|M₁)=0.02, P(D|M₂)=0.03, P(D|M₃)=0.02

(i) Total probability of defective item [1 mark]

P(D) = 0.40×0.02 + 0.35×0.03 + 0.25×0.02 = 0.008 + 0.0105 + 0.005

$$\boxed{P(D) = 0.0235}$$

(ii) P(defective came from Machine II) [1 mark]

$$P(M_2|D) = \frac{0.35 \times 0.03}{0.0235} = \frac{0.0105}{0.0235} = \boxed{\frac{21}{47}}$$

(iii) P(defective from Machine I or Machine III) [2 marks]

$$P(M_1|D) = \frac{0.008}{0.0235} = \frac{16}{47}$$

$$P(M_3|D) = \frac{0.005}{0.0235} = \frac{10}{47}$$

$$\boxed{P(M_1 \text{ or } M_3 \mid D) = \frac{16}{47} + \frac{10}{47} = \frac{26}{47}}$$

Alternate method: = 1 − P(M₂|D) = 1 − 21/47 = 26/47 ✔

Marking: 1 (P(M₁|D) computed) + 1 (final = 26/47)

(iii) OR — P(second random item is defective), given first defective from M₃ [2 marks]

The second item is drawn independently from the full day's production. Knowledge that the first item was from M₃ does not affect the second draw.

$$\boxed{P(\text{second item defective}) = P(D) = 0.0235}$$

Marking: 1 (recognising independence) + 1 (P = 0.0235)

📊 OVERALL MARKS SUMMARY

🎯 HOW TO SCORE FULL MARKS — EXAMINER'S FINAL TIPS

✅ Board-Style Writing Rules

1. Write formula first, then substitute — never skip straight to answer.
2. Show every algebra step — even "obvious" simplifications.
3. Box / underline your final answer clearly.
4. Draw diagrams for area, 3D geometry, LPP — partial marks for correct diagram.
5. Write LHS = RHS when proving identities step by step.

⚠️ Most Common Mark Losers

🛡️ How to Secure Method Marks Even if Stuck

• Write the relevant formula → earns minimum 1 method mark.
• Set up the integral or matrix form even if you can't compute it.
• For LPP: always formulate the problem even if graphing fails.
• For proofs: write the structure (LHS = ... = RHS) and earn partial marks.

— End of Solutions (Section D & E) — Total Paper: 80 Marks | CBSE Class XII Mathematics 2025–26