Class 12 CBSE Sample Paper Solution – MATHEMATICS

1. Identify the function shown in the graph:

• (A) sin^-1 x
• (B) sin^-1(2x)
• (C) sin^-1(x/2)
• (D) 2 sin^-1 x

Answer: (B) sin^-1(2x)

Solution: The function represents an inverse trigonometric function. The domain is [-1/2, 1/2], which fits the graph of sin^-1(2x), as multiplying by 2 compresses the domain of the inverse sine function.

2. If for three matrices A = [a_ij]ₓ₄, B = [b_ij] and C = [c_ij], the products AB and AC are both defined and are square matrices of the same order, then the values of m, n, p, and q are:

• (A) m = q = 3 and n = p = 4
• (B) m = 2, q = 3 and n = p = 4
• (C) m = q = 4 and n = p = 3
• (D) m = 4, p = 2 and n = q = 3

Answer: (C) m = q = 4 and n = p = 3

Solution: For matrix multiplication AB and AC to be defined, the number of columns in A must match the rows in B and C respectively. Also, the resulting matrices must be square matrices, hence m = q, n = p, and the correct dimensions are m = 4, n = 3, p = 3, q = 4.

3. If the matrix A =
[ 0 r -2 ]
[ 3 p t ]
[ q -4 0 ] is skew-symmetric, then the value of (q + t) / (p + r) is:

• (A) -2
• (B) 0
• (C) 1
• (D) 2

Answer: (A) -2

Solution: A skew-symmetric matrix satisfies \(A^T = -A\), leading to the relationships: \(r = -q\), \(p = -t\), and \(q = -r\). Therefore, \(\frac{q + t}{p + r} = -2\).

4. If A is a square matrix of order 4 and |adj(A)| = 27, then A × adj(A) is equal to:

• (A) 3
• (B) 9
• (C) 3I
• (D) 9I

Answer: (C) 3I

Solution: From the property of adjoint matrices, \(A \times \text{adj}(A) = |A| \times I\). Since \(|\text{adj}(A)| = 27\), we conclude that \(|A| = 3\), and thus \(A \times \text{adj}(A) = 3I\).

5. The inverse of the matrix
[3 0 0]
[0 2 0]
[0 0 5]

• (A) [0 0 3] [0 2 0] [5 0 0]
• (B) [1/3 0 0] [0 1/2 0] [0 0 1/5]
• (C) [-1/3 0 0] [0 -1/2 0] [0 0 -1/5]
• (D) [-3 0 0] [0 -2 0] [0 0 -5]

Answer: (B) [1/3 0 0] [0 1/2 0] [0 0 1/5]

Solution: The inverse of a diagonal matrix is obtained by taking the reciprocal of each diagonal element. Therefore, the inverse is
[1/3 0 0] [0 1/2 0] [0 0 1/5].

6. Value of the determinant:
|cos 67° sin 67°
sin 23° cos 23°| is:

• (A) 0
• (B) 1/2
• (C) √3/2
• (D) 1

Answer: (A) 0

Solution: The determinant of the given matrix simplifies as follows:
cos(67° + 23°) = cos(90°) = 0.
Hence, the determinant is 0.

7. If a function defined by f(x) = {k x + 1, x ≤ π; cos x, x > π} is continuous at x = π, then the value of k is:

• (A) π
• (B) −1/π
• (C) 0
• (D) −2/π

Answer: (D) −2/π

Solution: For continuity at x = π, we require the left-hand limit and right-hand limit to be equal at x = π.
From the equation \( \lim_{x \to \pi^-} (k x + 1) = \lim_{x \to \pi^+} \cos x \), solving for k gives \( k = -2/\pi \).

8. If f(x) = x tan^-1 x, then f'(1) is equal to:

• (A) π/4 − 1/2
• (B) π/4 + 1/2
• (C) − π/4 − 1/2
• (D) − π/4 + 1/2

Answer: (B) π/4 + 1/2

Solution: By using the product rule for derivatives, we compute:
f'(x) = tan^-1 x + x/(1+x²). Substituting x = 1 gives f'(1) = π/4 + 1/2.

9. A function f(x) = 10 − x − 2x² is increasing on the interval:

• (A) (−∞, −1/4]
• (B) (−∞, 1/4)
• (C) [−1/4, ∞)
• (D) [−1/4, 1/4]

Answer: (A) (−∞, −1/4]

Solution: For the function to be increasing, the derivative must be greater than or equal to zero. The first derivative is f'(x) = -1 − 4x. Solving for f'(x) ≥ 0 gives the interval (−∞, −1/4].

10. The solution of the differential equation x dx + y dy = 0 represents a family of:

• (A) straight lines
• (B) parabolas
• (C) circles
• (D) ellipses

Answer: (C) circles

Solution: The equation x² + y² = 2k represents a family of circles.

11. If f(a + b − x) = f(x), then ∫_a^b f(x) dx is equal to:

• (A) (a+b)/2 ∫_a^b f(b − x) dx
• (B) (a+b)/2 ∫_a^b f(a − x) dx
• (C) (b−a)/2 ∫_a^b f(x) dx
• (D) (a+b)/2 ∫_a^b f(x) dx

Answer: (C) (b−a)/2 ∫_a^b f(x) dx

Solution: By applying the given condition, the result simplifies to (b−a)/2 ∫_a^b f(x) dx.

12. If ∫ x³ sin⁴(x⁴) cos(x⁴) dx = a sin⁵(x⁴) + C, then a is equal to:

• (A) −1/10
• (B) 1/20
• (C) 1/4
• (D) 1/5

Answer: (D) 1/5

Solution: By applying the substitution method, we find the constant \( a \) as \( \frac{1}{5} \).

13. A bird flies through a distance in a straight line given by the vector î + 2ĵ + k̂. A man standing beside a straight metro rail track given by r = (3 + λ)î + (2λ − 1)ĵ + 3λk̂ is observing the bird. The projected length of its flight on the metro track is:

• (A) 6√14 units
• (B) 14√6 units
• (C) 8√14 units
• (D) 5√6 units

Answer: (A) 6√14 units

Solution: The projection of the flight vector on the metro track is calculated using the scalar projection formula, resulting in a length of 6√14 units.

14. The distance of the point with position vector 3î + 4ĵ + 5k̂ from the y-axis is:

• (A) 4 units
• (B) √34 units
• (C) 5 units
• (D) 5√2 units

Answer: (B) √34 units

Solution: The distance of a point from the y-axis is given by the formula \( \sqrt{x^2 + z^2} \), where the x and z components are 3 and 5, respectively. Thus, the distance is \( \sqrt{9 + 25} = \sqrt{34} \).

15. If ⃗a = 3î + 2ĵ + 4k̂, ⃗b = î + ĵ − 3k̂ and c = 6î − ĵ + 2k̂ are three given vectors, then (2 ⃗a . î) î − (⃗b . ĵ) ĵ + (c . k̂) k̂ is same as the vector:

• (A) ⃗a
• (B) ⃗b + c
• (C) ⃗a − ⃗b
• (D) c

Answer: (D) c

Solution: The result of the vector expression simplifies to the vector c as shown in the calculation steps.

16. A student of class XII studying Mathematics comes across an incomplete question in a book. Maximize Z = 3x + 2y + 1, Subject to the constraints x ≥ 0, y ≥ 0, 3x + 4y ≤ 12, He/She notices the below shown graph for the said LPP problem, and finds that a constraint is missing in it: Help him/her choose the required constraint from the graph. The missing constraint is:

• (A) x + 2y ≤ 2
• (B) 2x + y ≥ 2
• (C) 2x + y ≤ 2
• (D) x + 2y ≥ 2

Answer: (C) 2x + y ≤ 2

Solution: Based on the graph, the missing constraint is determined as \(2x + y \leq 2\) to match the feasible region.

17. If Z = ax + by + c, where a, b, c > 0, attains its maximum value at two of its corner points (4,0) and (0,3) of the feasible region determined by the system of linear inequalities, then:

• (A) 4a = 3b
• (B) 3a = 4b
• (C) 4a + c = 3b
• (D) 3a + c = 4b

Answer: (B) 3a = 4b

Solution: The condition for the maximum value of Z at the corner points leads to the relation 3a = 4b, based on the corner points and the nature of the linear programming problem.

18. The feasible region of a linear programming problem is bounded but the objective function attains its minimum value at more than one point. One of the points is (5,0). Then one of the other possible points at which the objective function attains its minimum value is:

• (A) (2,9)
• (B) (6,6)
• (C) (4,7)
• (D) (0,0)

Answer: (D) (0,0)

Solution: Since the objective function attains its minimum at more than one point, (0,0) is another point where the function can attain its minimum value.

19. Assertion (A): Value of the expression sin^-1(√3/2) + tan^-1(1) − sec^-1(√2) is π/4.

Reason (R): Principal value branch of sin^-1(x) is [−π/2, π/2] and that of sec^-1(x) is [0, π] − {π/2}.

Answer:

• (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
• (B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
• (C) (A) is true but (R) is false.
• (D) (A) is false but (R) is true.

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

Solution:
Assertion (A) is correct because:
sin^-1(√3/2) = π/3, tan^-1(1) = π/4, and sec^-1(√2) = π/3.
Therefore, sin^-1(√3/2) + tan^-1(1) − sec^-1(√2) = π/3 + π/4 − π/3 = π/4.
Reason (R) is also correct because the principal value branches for sin^-1(x) and sec^-1(x) are as given in the statement.

20. Assertion (A): Given two non-zero vectors ⃗a and ⃗b. If r is another non-zero vector such that r × (⃗a + ⃗b) = 0, then r is perpendicular to ⃗a × ⃗b.

Reason (R): The vector (⃗a + ⃗b) is perpendicular to the plane of ⃗a and ⃗b.

Answer:

• (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
• (B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
• (C) (A) is true but (R) is false.
• (D) (A) is false but (R) is true.

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

Solution:
Assertion (A) is true because if r × (⃗a + ⃗b) = 0, this means r is parallel to the vector (⃗a + ⃗b), i.e., r is in the same plane as ⃗a and ⃗b.
Since r is parallel to ⃗a + ⃗b and ⃗a × ⃗b is perpendicular to the plane of ⃗a and ⃗b, r is perpendicular to ⃗a × ⃗b.
Reason (R) is also true because the vector (⃗a + ⃗b) lies in the same plane as ⃗a and ⃗b, so it is perpendicular to the cross product ⃗a × ⃗b, which is normal to the plane of the vectors ⃗a and ⃗b.