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0606/23 May/Jun 2025, Question 4: Graphs of Modulus Functions

Cambridge IGCSE Additional Mathematics 0606 · Paper 23 (Calculator) · Calculator · 8 marks

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The question, in short

For f(x) = 2x − 1 on ℝ, justify composing f with itself, sketch the modulus of that composite, and find a and b so that |f²(x)| ≤ ax + b exactly on −1 ≤ x ≤ 3.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 1 mark

Give a reason why the composite f² is valid for this function.

  1. 1The range of f is ℝ
  2. 2The domain of f is ℝ, so every output of f is a valid input of f

AnswerThe range of f is equal to (contained in) the domain of f

Part (b) · 4 marks

Draw y = |f²(x)| on the axes and give where it crosses the axes.

  1. 1f²(x) = f(2x − 1) = 2(2x − 1) − 1 = 4x − 3
  2. 2y = |4x − 3| meets the x-axis at x = 3/4
  3. 3When x = 0, y = |−3| = 3
  4. 4The part below the x-axis is reflected in the x-axis, giving a V shape

AnswerV-shaped graph of y = |4x − 3| with vertex on the x-axis at (3/4, 0) and y-intercept 3

Part (c) · 3 marks

Given |f²(x)| ≤ ax + b holds for −1 ≤ x ≤ 3 only, determine a and b.

  1. 1The line y = ax + b meets y = |4x − 3| at x = −1 and x = 3
  2. 2When x = −1, y = |−7| = 7; when x = 3, y = 9
  3. 3a = (9 − 7)/(3 − (−1)) = 1/2
  4. 47 = −1/2 + b, so b = 15/2

Answera = 1/2, b = 15/2

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Also searched as: 0606/23/M/J/25 · 0606_s25_qp_23 · 0606 s25 23 · question 4.
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