Lesson F2
F2 Exponential model and the gradient of e to the kx Quiz: AQA Maths, Unit 6
20 questions
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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
-
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)
-
What is the gradient of y = e^(3x) at x = 0?
- 0
- e^3
- 1
- 3
-
What is the rate of change of y = e^(kx) proportional to?
- The product kx
- Its current value
- The value x
- The constant k
-
Find dy/dx for y = 5e^(-2x).
- 10e^(-2x)
- -2e^(-2x)
- 5e^(-2x)
- -10e^(-2x)
-
What happens to y = A e^(kx) as x increases, where A > 0 and k > 0?
- It approaches zero
- It approaches infinity
- It approaches A
- It remains constant
-
The curve y = e^(2x) passes through the point where y = 4. What is the gradient there?
- 8
- 2
- 4
- 16
-
For y = e^(kx), the gradient at x = ln 2 / k is which value?
- k
- 2
- ln 2
- 2k
-
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
-
A quantity satisfies N = 50 e^(0.2t). What is the rate of change of N when N = 100?
- 20
- 0.2
- 10
- 50
-
A temperature model is T = 20 + 60 e^(-0.1t). What is dT/dt at t = 0?
- 6
- -0.1
- -60
- -6
-
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
-
What is the sign of the gradient of y = e^(-kt) for k > 0?
- Always positive
- Always zero
- Always negative
- Positively increasing
-
For y = 3e^(2x), find the gradient at the point where y = 6.
- 3
- 12
- 6
- 24
-
Which function satisfies dy/dx = 2y with y = 5 when x = 0?
- y = 5x^2
- y = 2e^(5x)
- y = 10e^x
- y = 5e^(2x)
-
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
-
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
-
Why does an exponential model overestimate population growth over long time periods?
- Birth rates increase
- Gradient is zero
- Resources are limited
- Death rates decrease
-
The gradient of y = e^(kx) at the point (0, 1) is 4. What is k?
- e^4
- 1/4
- ln 4
- 4
-
For y = e^(3x), find d^2y/dx^2.
- e^(9x)
- 6e^(3x)
- 9e^(3x)
- 3e^(3x)
-
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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