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0606/12 Feb/Mar 2025, Question 1: Modulus Equations and Inequalities

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

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The question, in short

Sketch two modulus graphs on given axes, marking axis intercepts, then solve the inequality comparing 4|x − 1| with |3x + 2|.

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

Part (a) · 3 marks

On the axes, draw y = 4|x − 1| and y = |3x + 2|, and state where each graph crosses the axes.

  1. 1y = 4|x − 1|: vertex where x − 1 = 0, so (1, 0)
  2. 2At x = 0: y = 4|−1| = 4
  3. 3y = |3x + 2|: vertex where 3x + 2 = 0, so (−2/3, 0)
  4. 4At x = 0: y = |2| = 2
  5. 5Draw both as V-shapes reflected up from the x-axis

Answery = 4|x − 1|: V-shape with vertex on the x-axis at x = 1 and y-intercept 4; y = |3x + 2|: V-shape with vertex on the x-axis at x = −2/3 and y-intercept 2; both drawn fully with all intercepts marked

Part (b) · 4 marks

Find the values of x satisfying 4|x − 1| ≤ |3x + 2|.

  1. 1Square both sides: 16(x − 1)² ≤ (3x + 2)²
  2. 216x² − 32x + 16 ≤ 9x² + 12x + 4
  3. 37x² − 44x + 12 ≤ 0
  4. 4(7x − 2)(x − 6) ≤ 0, critical values 2/7 and 6
  5. 52/7 ≤ x ≤ 6

Answer2/7 ≤ x ≤ 6

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Also searched as: 0606/12/F/M/25 · 0606_m25_qp_12 · 0606 m25 12 · question 1.
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