0606/12 Oct/Nov 2025, Question 4: Functions: One-One, Inverses and Restricted Domains
Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 10 marks
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The question, in short
Complete the square on 2x² + 5x + 3, locate the stationary point, then restrict the domain so the inverse exists and sketch both graphs with axis intercepts.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 2 marks
Rewrite 2x² + 5x + 3 as 2(x + a)² + b and find the constants a and b.
- 12x² + 5x + 3 = 2(x² + (5/2)x) + 3
- 2= 2[(x + 5/4)² − 25/16] + 3
- 3= 2(x + 5/4)² − 25/8 + 3
- 4= 2(x + 5/4)² − 1/8
Answer2(x + 5/4)² − 1/8 (a = 5/4, b = −1/8)
Part (b) · 2 marks
Use your completed-square form to state the coordinates of the stationary point of y = 2x² + 5x + 3.
- 1Minimum of 2(x + 5/4)² − 1/8 is when x + 5/4 = 0
- 2x = −5/4, y = −1/8
Answer(−5/4, −1/8)
Part (ci) · 1 mark
Given f(x) = 2x² + 5x + 3 for x ≥ p and that f⁻¹ exists, state the smallest possible p.
- 1f must be one-one, so its domain starts at the vertex
- 2p = −5/4
Answerp = −5/4
Part (cii) · 5 marks
With that p, draw y = f(x) and y = f⁻¹(x) on the blank axes, label each curve, and give where each meets the axes.
- 1f(x) = (2x + 3)(x + 1) = 0 ⇒ x = −1 (x = −3/2 is outside the domain)
- 2f(0) = 3, so f meets the y-axis at 3
- 3f starts at the vertex (−5/4, −1/8) and increases
- 4f⁻¹ is the reflection of f in y = x
- 5f⁻¹ meets the x-axis at 3 and the y-axis at −1
Answery = f(x): right half of the parabola from its vertex (−5/4, −1/8), crossing the x-axis at x = −1 and the y-axis at y = 3, increasing; y = f⁻¹(x): its reflection in y = x, starting at (−1/8, −5/4), crossing the x-axis at x = 3 and the y-axis at y = −1; each graph labelled
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