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0606/12 Feb/Mar 2026, Question 2: Graphs of Modulus Functions

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 6 marks

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The question, in short

Sketch y = |x² − 2x − 3| on the given axes, giving axis intercepts, then find an expression for the inverse of g(x) = 4e^(x−2) − 1.

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

Part (a) · 4 marks

Draw the modulus graph of y = |x² − 2x − 3| on the axes and state where it crosses the x-axis and y-axis.

  1. 1x² − 2x − 3 = (x + 1)(x − 3): roots x = −1 and x = 3
  2. 2At x = 0: y = |−3| = 3
  3. 3The part of the parabola below the x-axis (−1 < x < 3) is reflected above it
  4. 4This gives cusps at (−1, 0) and (3, 0) and a maximum at (1, 4)

AnswerW-shaped curve: cusps on the x-axis at x = −1 and x = 3, y-intercept 3, the middle section reflected above the x-axis with its maximum in the first quadrant, outer branches curving upwards

Part (b) · 2 marks

For g(x) = 4e^(x−2) − 1, find an expression for g⁻¹(x).

  1. 1y = 4e^(x−2) − 1
  2. 2e^(x−2) = (y + 1)/4
  3. 3x = 2 + ln((y + 1)/4)
  4. 4g⁻¹(x) = 2 + ln((x + 1)/4)

Answerg⁻¹(x) = 2 + ln((x + 1)/4)

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Also searched as: 0606/12/F/M/26 · 0606_m26_qp_12 · 0606 m26 12 · question 2.
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