Class 12 Maths Chapter 6: Application of Derivatives MCQs with answers

Answer: (c) 20π cm³/cm

V = πr²(5) = 5πr², so dV/dr = 10πr = 10π(2) = 20π cm³ per cm of radius.

Answer: (d) 10 units/s

dy/dt = (2x + 3) dx/dt = (2 × 1 + 3) × 2 = 10 units per second. 5 forgets the factor dx/dt.

Answer: (c) 20

MC = dC/dx = 4x + 15. Setting 4x + 15 = 95 gives x = 20. 40 comes from writing MC = 2x + 15.

Answer: (d) −8 cm²/s

A = x², dA/dt = 2x dx/dt = 2(8)(−0.5) = −8 cm²/s; the negative sign says the area is decreasing.

Answer: (b) strictly decreasing

If x₁ < x₂, then x₁³ < x₂³, so −2x₁³ > −2x₂³ and f(x₁) > f(x₂): f is strictly decreasing on R. f′(x) = −6x² is 0 at x = 0 only, which does not spoil strictness.

Answer: (b) x³ + 5x

For x³ + 5x, f′(x) = 3x² + 5 > 0 for every real x. e^−x has derivative −e^−x < 0; x² decreases for x < 0; cos x changes direction.

Answer: (b) x = −3 and x = 3

f′(x) = 3x² − 27 = 3(x − 3)(x + 3) = 0 at x = ±3; f is differentiable everywhere, so these are the only critical points.

Answer: (d) 5

f′(x) = 3x² − 3 = 0 at x = 1 in [0, 2]. f(0) = k, f(1) = k − 2, f(2) = k + 2, so the minimum is k − 2 = 3 and k = 5. 3 takes the minimum at x = 0; 1 takes it at x = 2.

Answer: (c) has minimum value 0 but no maximum value

f(x) = x² ≥ 0 = f(0), and 0 is in (−1, 2), so the minimum value is 0. As x approaches 2, f(x) approaches 4, but x = 2 is not in the interval; for any x₀ in [0, 2), a point between x₀ and 2 gives a larger value, so there is no maximum.

Answer: (a) attains its absolute maximum and its absolute minimum value, each at least once, in [a, b]

Theorem 5: a continuous function on a closed interval attains its absolute maximum and absolute minimum, each at least once. f need not be differentiable, so (c) need not hold, and the maximum may be at an interior point.