Calculus: rate of change — KCSE Mathematics

KCSE Mathematics · 113 practice questions · 3 syllabus objectives

38 easy38 medium37 hard

What You'll Learn

Key learning outcomes for this topic, aligned to the KNEC KCSE syllabus.

Define the gradient of a curve at a point as the limit of the gradient of a chord as its length tends to zero

Interpret the derivative as a rate of change; identify when a function is increasing (f'(x) > 0) or decreasing (f'(x) < 0)

Calculus: rate of change

Sample Questions

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1
easySHORT ANSWER2 marks

State when the function g(t) = -2t^2 + 8t - 3 is decreasing. (2 marks)

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Part (a) — 2 marks
The derivative g'(t) must be less than 0 (1 mk)
Identify the interval where g'(t) < 0 (1 mk)
2
easySHORT ANSWER3 marks

State the conditions under which the function f(x) = 3x^2 - 12x + 5 is increasing. (3 marks)

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Part (a) — 3 marks
The derivative f'(x) must be greater than 0 (1 mk)
Identify the interval where f'(x) > 0 (1 mk)
For this function, x > 2 is the interval where f'(x) > 0 (1 mk)
3
easySHORT ANSWER4 marks

Consider the function g(t) = t^3 - 3t^2 + 2. (a) State the definition of the gradient of a curve at a point. (1 mark) (b) Calculate the gradient of the curve at t = 2. (2 marks) (c) Discuss the behaviour of the gradient as t approaches 2 from the left. (1 mark)

View Marking Scheme
Part (a) — 1 mark
The gradient at a point is the limit of the gradient of the chord as the length of the chord tends to zero (1 mk)
Part (b) — 2 marks
Differentiate g(t) to get g'(t) = 3t^2 - 6t (1 mk)
Substitute t = 2 into g'(t) to get g'(2) = 3(2^2) - 6(2) = 0 (1 mk)
Part (c) — 1 mark
As t approaches 2 from the left, the gradient g'(t) decreases and approaches 0 (1 mk)
4

Given the function f(x) = 3x^2 + 2x, (a) define the gradient of the curve at a point. (1 mark) (b) Find the gradient of the curve at the point where x = 1. (2 marks) (c) Explain how the gradient changes as x increases from 1. (1 mark)

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