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Applications of Derivatives practice questions — A-Level Maths

13 free multiple-choice problems on applications of derivatives, ordered to match A-Level Maths difficulty, each with a full worked solution. No account needed — and there's a fresh problem set every day.

Problems & worked solutions

Problem #0220 International

Problem 1Applications of Derivatives

Among rectangles with perimeter 2020, the one with maximum area has dimensions:

Show answer & worked solution
  1. A. 4×64 \times 6
  2. B. 3×73 \times 7
  3. C. 5×55 \times 5✓ correct
  4. D. 2×82 \times 8

A(x)=10xx2A(x) = 10x - x^2. A(x)=102x=0x=5A'(x) = 10 - 2x = 0 \Rightarrow x = 5. The maximum-area rectangle is the square 5×55 \times 5, with area 2525.

Problem #0219 International

Problem 2Applications of Derivatives

The function f(x)=x33x2f(x) = x^3 - 3x^2 is decreasing on:

Show answer & worked solution
  1. A. R\mathbb{R}
  2. B. (,0)(-\infty, 0)
  3. C. (0,2)(0, 2)✓ correct
  4. D. (2,+)(2, +\infty)

f(x)<00<x<2f'(x) < 0 \Leftrightarrow 0 < x < 2. So ff is decreasing on (0,2)(0, 2). (Increasing on (,0)(-\infty, 0) and (2,+)(2, +\infty).)

Problem #0215 International

Problem 3Applications of Derivatives

For the same position x(t)=t36t2+9tx(t) = t^3 - 6t^2 + 9t, the acceleration at t=1t = 1 is:

Show answer & worked solution
  1. A. 6-6 m/s²✓ correct
  2. B. 00 m/s²
  3. C. 66 m/s²
  4. D. 99 m/s²

a(t)=x(t)=6t12a(t) = x''(t) = 6t - 12. At t=1t = 1: a=6a = -6 m/s².

Problem #0217 International

Problem 4Applications of Derivatives

The point of inflection of f(x)=x33xf(x) = x^3 - 3x is at:

Show answer & worked solution
  1. A. x=1x = -1
  2. B. x=0x = 0✓ correct
  3. C. x=1x = 1
  4. D. ff has no inflection

f(x)=3x23f'(x) = 3x^2 - 3, f(x)=6xf''(x) = 6x. ff'' changes sign at x=0x = 0, so the inflection is at x=0x = 0.

Problem #0216 International

Problem 5Applications of Derivatives

The function f(x)=x3f(x) = x^3 is concave on:

Show answer & worked solution
  1. A. R\mathbb{R}
  2. B. (,0)(-\infty, 0)✓ correct
  3. C. (0,+)(0, +\infty)
  4. D. nowhere

f(x)=6xf''(x) = 6x. f<0x<0f'' < 0 \Leftrightarrow x < 0. So ff is concave on (,0)(-\infty, 0).

Problem #0881 US AP

Problem 6Applications of Derivatives

The function f(x)=x2f(x) = x^{2} is:

Show answer & worked solution
  1. A. convex on R\mathbb{R} (f(x)>0f''(x) > 0)✓ correct
  2. B. concave on R\mathbb{R}
  3. C. convex only on (0,)(0,\infty)
  4. D. has an inflection point at x=0x = 0

f(x)=2xf'(x) = 2x and f(x)=2>0f''(x) = 2 > 0 for all xRx \in \mathbb{R}.

Since f(x)>0f''(x) > 0 everywhere, ff is convex (concave up) on all of R\mathbb{R} — no inflection point.

Problem #0214 International

Problem 7Applications of Derivatives

A particle's position is x(t)=t36t2+9tx(t) = t^3 - 6t^2 + 9t (in metres, tt in seconds). The velocity at t=2t = 2 is:

Show answer & worked solution
  1. A. 3-3 m/s✓ correct
  2. B. 00 m/s
  3. C. 33 m/s
  4. D. 99 m/s

v(t)=3t212t+9v(t) = 3t^2 - 12t + 9. At t=2t = 2: v=1224+9=3v = 12 - 24 + 9 = -3 m/s (the particle is moving backward).

Problem #0218 International

Problem 8Applications of Derivatives

The function f(x)=x2+4x1f(x) = -x^2 + 4x - 1 has a local maximum at:

Show answer & worked solution
  1. A. x=2x = -2
  2. B. x=0x = 0
  3. C. x=2x = 2✓ correct
  4. D. x=4x = 4

f(x)=2x+4=0x=2f'(x) = -2x + 4 = 0 \Rightarrow x = 2. f(x)=2<0f''(x) = -2 < 0, confirming a local maximum.

Problem #0880 US AP

Problem 9Applications of Derivatives

The function f(x)=x33xf(x) = x^{3} - 3x is decreasing on the interval:

Show answer & worked solution
  1. A. (,1)(-\infty, -1)
  2. B. (1,1)(-1, 1)✓ correct
  3. C. (1,)(1, \infty)
  4. D. R\mathbb{R}

f(x)=3x23=3(x1)(x+1)f'(x) = 3x^{2} - 3 = 3(x - 1)(x + 1).

f(x)<0f'(x) < 0 when (x1)(x+1)<0(x-1)(x+1) < 0, i.e. on (1,1)(-1, 1).

So ff is decreasing on (1,1)(-1, 1).

Problem #0017 International

Problem 10Applications of Derivatives

For which value of aa does f(x)=x33ax+1f(x) = x^3 - 3ax + 1 have a local minimum at x=2x = 2?

Show answer & worked solution
  1. A. a=1a = 1
  2. B. a=2a = 2
  3. C. a=3a = 3
  4. D. a=4a = 4✓ correct

f(x)=3x23af'(x) = 3x^2 - 3a, so f(2)=123a=0a=4f'(2) = 12 - 3a = 0 \Rightarrow a = 4. Check: f(x)=6xf''(x) = 6x, so f(2)=12>0f''(2) = 12 > 0, confirming a local minimum.

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