Class 12 Maths Chapter 6: Application of Derivatives worksheet with answer key

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) 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: (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.

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: (c) 20π cm³/cm

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

For |x| > 2, f(x) = x² − 4 and f′(x) = 2x; for |x| < 2, f(x) = 4 − x² and f′(x) = −2x, which is 0 at x = 0. At x = 2 the left derivative is −4 and the right derivative 4; at x = −2 they are likewise −4 and 4. So f is not differentiable at x = ±2. Critical points: x = 0, x = −2, x = 2.

• x = 0, where f′(x) = −2x = 0 (1)
• x = ±2, where f is not differentiable (left and right derivatives differ) (1)

f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 at x = 1, 3. f(0) = 0, f(1) = 4, f(3) = 0, f(4) = 4. Absolute maximum 4, at x = 1 and x = 4; absolute minimum 0, at x = 0 and x = 3.

• critical points x = 1, 3 and values at 0, 1, 3, 4 (1)
• absolute maximum 4 (x = 1, 4), absolute minimum 0 (x = 0, 3) (1)

f′(x) = 3x² − 12, f′(2) = 0. f″(x) = 6x, f″(2) = 12 > 0, so x = 2 is a point of local minima, and the local minimum value is f(2) = 8 − 24 + 1 = −15.

• f′(2) = 0 and f″(2) = 12 > 0, so local minima (1)
• local minimum value f(2) = −15 (1)

f′(x) = 3x² − 6x − 24 = 3(x − 4)(x + 2), zero at x = −2 and 4. Through x = −2, f′ changes from positive to negative: local maxima, value f(−2) = −8 − 12 + 48 + 5 = 33. Through x = 4, f′ changes from negative to positive: local minima, value f(4) = 64 − 48 − 96 + 5 = −75.

• f′(x) = 3(x − 4)(x + 2), critical points −2 and 4 (1)
• sign changes: local maxima at −2, local minima at 4 (1)
• local maximum value 33, local minimum value −75 (1)

f′(x) = 1 − 2 cos x = 0 when cos x = 1/2, that is x = π/3, 5π/3. On [0, π/3), cos x > 1/2, so f′ < 0; on (π/3, 5π/3), cos x < 1/2, so f′ > 0; on (5π/3, 2π], f′ < 0. So f is decreasing in [0, π/3) and (5π/3, 2π], and increasing in (π/3, 5π/3).

• f′(x) = 1 − 2 cos x, zero at π/3 and 5π/3 (1)
• sign of f′ on the three intervals (1)
• increasing in (π/3, 5π/3); decreasing in [0, π/3) and (5π/3, 2π] (1)

Let each side perpendicular to the canal be w m; the side along the canal is 800/w m. Fencing L(w) = 2w + 800/w. L′(w) = 2 − 800/w² = 0 at w = 20 (w > 0). L″(w) = 1600/w³ > 0, a minimum. Dimensions 20 m by 40 m (40 m along the canal); least fencing 40 + 40 = 80 m.

• L(w) = 2w + 800/w (1)
• L′(w) = 0 at w = 20 with L″ > 0 (1)
• 20 m × 40 m, fencing 80 m (1)