CBSE Class 12 Mathematics Board Paper 5 — 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, 2), (2, 1)} is:

   A.reflexive only

   B.transitive only

   C.symmetric only✓ Correct

   D.an equivalence relation

   Answer: C. symmetric only

   Explanation: R contains none of (1, 1), (2, 2), (3, 3), so it is not reflexive. It is symmetric because (1, 2) and (2, 1) both belong, but transitivity fails since those two pairs demand (1, 1), which is absent. A relation must satisfy all three properties to be an equivalence relation.

2. Q2Easy

   If [[x + y, 2], [5, xy]] = [[6, 2], [5, 8]], then x² + y² equals:

   A.36

   B.16

   C.52

   D.20✓ Correct

   Answer: D. 20

   Explanation: Equating corresponding entries gives x + y = 6 and xy = 8. Then x² + y² = (x + y)² − 2xy = 36 − 16 = 20. Reporting 36 stops at (x + y)² and forgets to remove the 2xy cross term.

3. Q3Easy

   The maximum value of f(x) = sin x + cos x is:

   A.√2✓ Correct

   B.2

   C.1

   D.1/√2

   Answer: A. √2

   Explanation: Rewriting, sin x + cos x = √2 sin(x + π/4), whose amplitude is √2, so the greatest value is √2, attained at x = π/4. Adding the individual maxima 1 and 1 to get 2 is wrong because sin x and cos x never reach 1 at the same value of x.

4. Q4Medium

   The value of sin⁻¹(3/5) + sin⁻¹(4/5) is:

   A.π/2✓ Correct

   B.π

   C.sin⁻¹(24/25)

   D.π/4

   Answer: A. π/2

   Explanation: Let α = sin⁻¹(3/5) and β = sin⁻¹(4/5), so sin α = 3/5, cos α = 4/5, sin β = 4/5 and cos β = 3/5. Then sin(α + β) = (3/5)(3/5) + (4/5)(4/5) = 1, and since both angles are acute their sum is π/2. Applying the sin 2θ pattern instead produces 24/25, which is the sine of a different angle altogether.

5. Q5Medium

   If A is a square matrix of order 3 with |A| = 4, then |adj A| equals:

   A.4

   B.64

   C.16✓ Correct

   D.12

   Answer: C. 16

   Explanation: For a matrix of order n, |adj A| = |A|^(n−1); with n = 3 this is |A|² = 4² = 16. Using the exponent n instead gives 64, and quoting 4 forgets the adjoint step altogether.

6. Q6Medium

   If f(x) = kx + 1 for x ≤ π and f(x) = cos x for x > π is continuous at x = π, then k equals:

   A.2/π

   B.−2/π✓ Correct

   C.−1/π

   D.π/2

   Answer: B. −2/π

   Explanation: Continuity at x = π requires the two pieces to agree there, so kπ + 1 = cos π = −1. Solving, kπ = −2 and k = −2/π. Using cos π = 1 rather than −1 flips the sign and gives 2/π.

7. Q7Hard

   If A = [[cos α, −sin α], [sin α, cos α]] and A + A′ = I, where I is the identity matrix of order 2, then α equals:

   A.π/6

   B.π/3✓ Correct

   C.π

   D.3π/2

   Answer: B. π/3

   Explanation: Transposing swaps the off-diagonal entries, so A + A′ = [[2 cos α, 0], [0, 2 cos α]], which is 2 cos α · I. Setting that equal to I forces cos α = 1/2, so α = π/3. Choosing π/6 makes cos α = √3/2 and yields √3 I rather than I.

8. Q8Hard

   If x, y and z are all different and the determinant of [[x, x², 1 + x³], [y, y², 1 + y³], [z, z², 1 + z³]] is 0, then xyz equals:

   A.−1✓ Correct

   B.1

   C.0

   D.3

   Answer: A. −1

   Explanation: Split the third column: the determinant becomes the one with third column of 1s plus the one with third column x³, y³, z³. The first is a cyclic shift of the Vandermonde determinant V formed by columns 1, x, x² and so equals V, while the second factors as xyz·V. Hence V(1 + xyz) = 0, and since x, y, z are distinct V ≠ 0, forcing xyz = −1; concluding xyz = 0 would instead require two of the values to coincide.

More Class 12 Mathematics board papers

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

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