CBSE Class 12 Mathematics Board Paper 1 — MCQs with Answers

8 solved questions from this paper

Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.

1. Q1Easy

   On the set A = {1, 2, 3}, the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is:

   A.reflexive but neither symmetric nor transitive✓ Correct

   B.reflexive and symmetric but not transitive

   C.reflexive and transitive but not symmetric

   D.an equivalence relation

   Answer: A. reflexive but neither symmetric nor transitive

   Explanation: Reflexivity holds because (1, 1), (2, 2) and (3, 3) are all present. Symmetry fails since (1, 2) belongs to R while (2, 1) does not, and transitivity fails since (1, 2) and (2, 3) are in R but (1, 3) is not. Assuming a reflexive relation automatically carries symmetry is the usual slip.

2. Q2Easy

   If A is a square matrix such that A² = A, then (I + A)³ − 7A equals:

   A.A

   B.I✓ Correct

   C.I + A

   D.2A

   Answer: B. I

   Explanation: Since I commutes with A, (I + A)³ = I + 3A + 3A² + A³. Using A² = A gives A³ = A·A = A, so the expansion collapses to I + 7A, and subtracting 7A leaves I. Treating A² = A as though it made A itself an identity is what leads to the answer A.

3. Q3Easy

   If f(x) = (x² − 9)/(x − 3) for x ≠ 3 and f(3) = k, then f is continuous at x = 3 when k equals:

   A.3

   B.6✓ Correct

   C.0

   D.9

   Answer: B. 6

   Explanation: Factorising, (x² − 9)/(x − 3) = x + 3 for every x ≠ 3, so the limit as x → 3 is 6. Continuity requires f(3) to equal that limit, hence k = 6. Substituting x = 3 into the uncancelled fraction gives the indeterminate 0/0 and tempts the answer 0.

4. Q4Medium

   The value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2) is:

   A.13π/12

   B.5π/12

   C.3π/4✓ Correct

   D.11π/12

   Answer: C. 3π/4

   Explanation: Read each inverse on its own principal branch: tan⁻¹1 = π/4, cos⁻¹(−1/2) = 2π/3 because cos⁻¹ has range [0, π], and sin⁻¹(−1/2) = −π/6 because sin⁻¹ has range [−π/2, π/2]. Adding, 3π/12 + 8π/12 − 2π/12 = 3π/4. Taking sin⁻¹(−1/2) as +π/6 gives 13π/12, and omitting that term altogether gives 11π/12.

5. Q5Medium

   If A = [[0, −1], [1, 0]], then A²⁰²⁶ equals:

   A.I

   B.A

   C.−A

   D.−I✓ Correct

   Answer: D. −I

   Explanation: Direct multiplication gives A² = [[−1, 0], [0, −1]] = −I, so A⁴ = (−I)² = I and the powers repeat with period 4. Since 2026 = 4 × 506 + 2, A²⁰²⁶ = A² = −I. Forgetting the minus sign that A² carries leaves the identity.

6. Q6Medium

   If A is a square matrix of order 3 with |A| = 5, then |adj (2A)| equals:

   A.1600✓ Correct

   B.200

   C.100

   D.25

   Answer: A. 1600

   Explanation: For order 3, |kA| = k³|A| and |adj B| = |B|². Here |2A| = 2³ × 5 = 40, so |adj (2A)| = 40² = 1600. Scaling only once, as |2A| = 2 × 5 = 10, produces 100, and dropping the scalar entirely gives |adj A| = 25.

7. Q7Hard

   If xʸ = e^(x − y), then dy/dx equals:

   A.(x − y)/x

   B.1/(1 + ln x)

   C.ln x/(1 + ln x)

   D.ln x/(1 + ln x)²✓ Correct

   Answer: D. ln x/(1 + ln x)²

   Explanation: Taking logarithms gives y ln x = x − y, so y(1 + ln x) = x and y = x/(1 + ln x). Differentiating the log form, y′ ln x + y/x = 1 − y′, so y′ = (x − y)/(x(1 + ln x)); substituting x − y = x ln x/(1 + ln x) gives ln x/(1 + ln x)². Stopping before the second power of (1 + ln x) appears leaves ln x/(1 + ln x).

8. Q8Hard

   The height of the right circular cylinder of maximum volume that can be inscribed in a sphere of radius 6 cm is:

   A.4√3 cm✓ Correct

   B.6√2 cm

   C.12 cm

   D.2√3 cm

   Answer: A. 4√3 cm

   Explanation: If the cylinder has height h, its radius satisfies r² = 36 − h²/4, so V = πh(36 − h²/4). Setting dV/dh = π(36 − 3h²/4) = 0 gives h² = 48, so h = 4√3 cm. The value 6√2 solves the different problem of maximising curved surface area, where h = R√2.

More Class 12 Mathematics board papers

• Board Paper 1 — you are here
• Board Paper 2 — 20 questions
• Board Paper 3 — 20 questions
• Board Paper 4 — 20 questions
• Board Paper 5 — 20 questions

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