Lesson F2

F2 Exponential model and the gradient of e to the kx Quiz: AQA Maths, Unit 6

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson F2, Exponential model and the gradient of e to the kx: 20 multiple choice questions for the AQA Maths (7357), Unit 6: Exponentials and logarithms, written with Revision Ninja.

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The 20 questions

  1. What is the derivative of e^(kx) with respect to x?

    • e^(kx)/k
    • k x e^(kx - 1)
    • e^x / k
    • k e^(kx)

  2. What is the gradient of y = e^(3x) at x = 0?

    • 0
    • e^3
    • 1
    • 3

  3. Which property makes e^(kx) suitable for modelling many growth processes?

    • Its values repeat with a fixed period
    • Its rate of change is proportional to its current value
    • Its rate of change is constant at every value of x
    • Its gradient is always zero at x = 0

  4. Find dy/dx for y = 5e^(-2x).

    • 10e^(-2x)
    • -2e^(-2x)
    • 5e^(-2x)
    • -10e^(-2x)

  5. If y = A e^(kx) with k > 0 and A > 0, what happens to y as x increases?

    • It decays towards zero
    • It grows without limit
    • It oscillates between positive and negative values
    • It approaches A from above and stays bounded

  6. The curve y = e^(2x) passes through the point where y = 4. What is the gradient there?

    • 8
    • 2
    • 4
    • 16

  7. For y = e^(kx), the gradient at x = ln 2 / k is which value?

    • k
    • 2
    • ln 2
    • 2k

  8. A population is modelled by P = 100 e^(0.05t), with t in years. What is dP/dt when t = 0?

    • 0.05
    • 100
    • 105
    • 5

  9. A quantity satisfies N = 50 e^(0.2t). What is the rate of change of N when N = 100?

    • 20
    • 0.2
    • 10
    • 50

  10. A temperature model is T = 20 + 60 e^(-0.1t). What is dT/dt at t = 0?

    • 6
    • -0.1
    • -60
    • -6

  11. A car value is V = 12000 e^(-0.15t) pounds, with t in years. What is the initial rate of change of V?

    • -12000 pounds per year
    • -0.15 pounds per year
    • 1800 pounds per year
    • -1800 pounds per year

  12. Which statement about y = e^(-kt) with k > 0 is correct?

    • It increases and approaches 1 as t tends to infinity
    • It reaches zero at a finite time t
    • It decreases towards zero as t increases, and its gradient is always negative
    • Its gradient is positive for all values of t

  13. For y = 3e^(2x), find the gradient at the point where y = 6.

    • 3
    • 12
    • 6
    • 24

  14. Which function satisfies dy/dx = 2y with y = 5 when x = 0?

    • y = 5x^2
    • y = 2e^(5x)
    • y = 10e^x
    • y = 5e^(2x)

  15. A quantity Q = Q0 e^(-kt) halves in time T. Which expression gives T?

    • T = 2k
    • T = k ln(1/2)
    • T = ln 2 / k
    • T = k / ln 2

  16. For y = e^(kx), the gradient at x = 2 equals 3 times the value of y there. What is k?

    • 2
    • 3
    • 6
    • 1/3

  17. Which limitation applies to an exponential growth model of a population over a long period?

    • Exponential models always give negative population values
    • The model gives a constant population for all times
    • Resources are limited, so growth slows and the model overestimates the population
    • Population cannot grow at all in an exponential model

  18. The gradient of y = e^(kx) at the point (0, 1) is 4. What is k?

    • e^4
    • 1/4
    • ln 4
    • 4

  19. For y = e^(3x), find d^2y/dx^2.

    • e^(9x)
    • 6e^(3x)
    • 9e^(3x)
    • 3e^(3x)

  20. A quantity doubles every 5 years and follows Q = Q0 e^(kt). What is k to 3 significant figures?

    • 0.322
    • 5 ln 2
    • 0.400
    • 0.139

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