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0606/11 May/Jun 2025, Question 5: Modulus Equations and Inequalities

Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 5 marks

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The question, in short

Solve an absolute-value equation in x, then sketch a transformed sine curve over 0° to 360° on given axes.

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

Part (a) · 3 marks

Solve 5|2x − 1| + 8 = 23 for x.

  1. 15|2x − 1| = 15, so |2x − 1| = 3
  2. 22x − 1 = 3 or 2x − 1 = −3
  3. 3x = 2 or x = −1

Answerx = 2, x = −1

Part (b) · 2 marks

Sketch y = 5sin x − 2 for 0° ≤ x ≤ 360° on the axes, showing key features.

  1. 1Midline y = −2 (the −2 shifts y = 5sin x down)
  2. 2Amplitude 5: maximum 5 − 2 = 3 at x = 90°
  3. 3Minimum −5 − 2 = −7 at x = 270°
  4. 4y = −2 at x = 0°, 180°, 360°

AnswerOne full cycle of a sine curve for 0° ≤ x ≤ 360°: starts at (0, −2), maximum 3 at x = 90°, crosses y = −2 at 180°, minimum −7 at x = 270°, back to −2 at 360°

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Also searched as: 0606/11/M/J/25 · 0606_s25_qp_11 · 0606 s25 11 · question 5.
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