0606/21 May/Jun 2025, Question 7: Functions: One-One, Inverses and Restricted Domains
Cambridge IGCSE Additional Mathematics 0606 · Paper 21 (Calculator) · Calculator · 8 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
Sketch y = 2e^(−x) + 3 and its inverse on the same axes, marking axis crossings and asymptotes, then invert g(x) = 2 − 3/(eˣ + 2) for x ≥ 0 and give the domain.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Part (a) · 4 marks
Draw both curves on the given axes, indicating where each crosses the coordinate axes and any asymptotes.
- 1f(0) = 2 + 3 = 5, so y-intercept (0, 5)
- 2As x → ∞, e^(−x) → 0, so asymptote y = 3; f is decreasing
- 3Reflect in y = x: f⁻¹ crosses the x-axis at (5, 0)
- 4f⁻¹ has vertical asymptote x = 3 and exists for x > 3
Answery = f(x): decreasing exponential curve crossing the y-axis at (0, 5) and tending to the asymptote y = 3 as x → ∞. y = f⁻¹(x): its reflection in y = x, crossing the x-axis at (5, 0) with vertical asymptote x = 3
Part (b) · 4 marks
Find a formula for the inverse of g and give the set of values it accepts.
- 1y = 2 − 3/(eˣ + 2), so 3/(eˣ + 2) = 2 − y
- 2eˣ + 2 = 3/(2 − y), so eˣ = 3/(2 − y) − 2
- 3g⁻¹(x) = ln(3/(2 − x) − 2)
- 4g(0) = 2 − 1 = 1 and g(x) → 2 as x → ∞, so the range of g is 1 ≤ g(x) < 2
- 5Domain of g⁻¹: 1 ≤ x < 2
Answerg⁻¹(x) = ln(3/(2 − x) − 2), domain 1 ≤ x < 2
Stuck on a step? Read the free lesson: Functions: One-One, Inverses and Restricted Domains
Practise more functions: one-one, inverses and restricted domains questions
Real 0606 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on functions: one-one, inverses and restricted domains
Also searched as: 0606/21/M/J/25 · 0606_s25_qp_21 · 0606 s25 21 · question 7.
Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education. The worked solutions are Lemmara’s own.