CBSE Class 12 Mathematics Board Paper 3 — 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

   The domain of the function f(x) = sin⁻¹(2x − 1) is:

   A.[−1, 1]

   B.[−1/2, 1/2]

   C.[0, 2]

   D.[0, 1]✓ Correct

   Answer: D. [0, 1]

   Explanation: sin⁻¹t is defined only for −1 ≤ t ≤ 1, so the requirement is −1 ≤ 2x − 1 ≤ 1. Adding 1 gives 0 ≤ 2x ≤ 2 and halving gives 0 ≤ x ≤ 1. Quoting [−1, 1] copies the domain of sin⁻¹ itself without undoing the substitution.

2. Q2Easy

   If the points (2, −3), (λ, −1) and (0, 4) are collinear, then λ equals:

   A.7/10

   B.10/7✓ Correct

   C.−10/7

   D.5/7

   Answer: B. 10/7

   Explanation: Three points are collinear when the determinant giving twice the triangle's area vanishes: 2(−1 − 4) + λ(4 + 3) + 0(−3 + 1) = −10 + 7λ = 0. Solving gives λ = 10/7. Inverting the fraction to 7/10 comes from dividing 7 by 10 rather than 10 by 7.

3. Q3Easy

   The function f(x) = x² − 4x + 6 is strictly increasing on the interval:

   A.(−∞, 2)

   B.(−2, ∞)

   C.(2, ∞)✓ Correct

   D.(−∞, −2)

   Answer: C. (2, ∞)

   Explanation: Here f′(x) = 2x − 4, which is positive exactly when x > 2, so f increases strictly on (2, ∞). The vertex sits at x = 2 and the parabola falls to its left. Reading the inequality the other way gives (−∞, 2), which is the interval of decrease.

4. Q4Medium

   On the set Z of integers, the relation R defined by a R b if and only if a − b is divisible by 5 is an equivalence relation. The number of distinct equivalence classes it produces is:

   A.4

   B.5✓ Correct

   C.infinitely many

   D.1

   Answer: B. 5

   Explanation: The condition says a and b leave the same remainder on division by 5, so the relation is congruence modulo 5. That partitions Z according to remainder, giving the classes of 0, 1, 2, 3 and 4 — exactly five of them. Answering 'infinitely many' confuses the number of classes with the number of integers inside each class.

5. Q5Medium

   If A = [[1, 2], [3, 4]] and A (adj A) = kI, then k equals:

   A.−2✓ Correct

   B.2

   C.10

   D.−10

   Answer: A. −2

   Explanation: For every square matrix, A (adj A) = |A| I, so k is simply the determinant. Here |A| = (1)(4) − (2)(3) = 4 − 6 = −2. Adding the two products instead of subtracting them gives 10, which is the classic sign error in a 2 × 2 determinant.

6. Q6Medium

   If A is an invertible matrix of order 3 with |A| = 2, then |A⁻¹ (adj A)| equals:

   A.1/2

   B.4

   C.8

   D.2✓ Correct

   Answer: D. 2

   Explanation: For order 3, |adj A| = |A|² = 4, and |A⁻¹| = 1/|A| = 1/2. Since the determinant of a product is the product of the determinants, |A⁻¹ (adj A)| = (1/2)(4) = 2. Quoting 4 alone reports |adj A| and ignores the A⁻¹ factor entirely.

7. Q7Hard

   If A = [[3, −4], [1, −1]], then Aⁿ, for every positive integer n, equals:

   A.[[1 + 2n, 4n], [n, 1 − 2n]]

   B.[[1 + 2n, −4n], [−n, 1 − 2n]]

   C.[[1 + 2n, −4n], [n, 1 − 2n]]✓ Correct

   D.[[3n, −4n], [n, −n]]

   Answer: C. [[1 + 2n, −4n], [n, 1 − 2n]]

   Explanation: Check the pattern at n = 1, where it reproduces A itself, and at n = 2, where direct multiplication gives A² = [[5, −8], [2, −3]], matching [[1 + 4, −8], [2, 1 − 4]]. Induction completes it: multiplying the pattern by A once more raises n by 1. Simply scaling every entry of A by n, as in [[3n, −4n], [n, −n]], fails already at n = 2.

8. Q8Hard

   If y = (tan⁻¹x)², then (1 + x²)² y₂ + 2x(1 + x²) y₁ equals:

   A.2✓ Correct

   B.0

   C.1

   D.2/(1 + x²)

   Answer: A. 2

   Explanation: From y = (tan⁻¹x)², y₁ = 2 tan⁻¹x/(1 + x²), so (1 + x²)y₁ = 2 tan⁻¹x. Differentiating again gives (1 + x²)y₂ + 2x y₁ = 2/(1 + x²), and multiplying through by (1 + x²) yields (1 + x²)²y₂ + 2x(1 + x²)y₁ = 2. Leaving that final multiplication undone is what produces the 2/(1 + x²) form.

More Class 12 Mathematics board papers

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