Mathematics 9709 · syllabus 1.2
Cambridge A Level Mathematics 9709 — Functions topical past papers
Domain, range, composite and inverse functions, and how translations, stretches and reflections change the graph of y = f(x).
- Questions
- 173
- Years
- 2009–2026
- With mark scheme
- 98%
- Questions 2022–2026
- 66
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Functions covers
A function maps each value in its domain to exactly one value in its range. Paper 1 asks you to find ranges, often by completing the square, to form composite functions such as fg(x), where g acts first, and to decide whether a function is one-one. Only a one-one function has an inverse, so questions often restrict a domain, for example to one side of a vertex, before asking for f⁻¹(x).
The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x, and the domain of f⁻¹ is the range of f. A composite gf exists only when the range of f lies inside the domain of g. The second half of the topic is transformations: y = f(x) + a, y = f(x + a), y = af(x), y = f(ax) and simple combinations of them, described in words as translations, stretches and reflections.
Functions is section 1.2 of Paper 1. It leans on quadratics for ranges and inverses, and its transformations come back in trigonometry, where graphs such as y = 2 − 3 cos 2x are built from y = cos x, and in the exponential and logarithmic graphs of Papers 2 and 3.
In the syllabus
The headings this topic has in the Cambridge International 2026–2027 syllabus, as the board prints them.
- 1.2 Functions
What you should be able to do
- Find the range of a function on a given domain, using a sketch or a completed square
- Form composite functions in the right order and say when one exists
- Decide whether a function is one-one, and restrict a domain so that it becomes one
- Find an inverse function algebraically and state its domain
- Sketch a function and its inverse on the same axes, showing the mirror line y = x
- Describe and apply translations, stretches and reflections, including combinations
Key ideas
- fg(x) = f(g(x)): apply g first, then f
- f⁻¹ exists only if f is one-one; the domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f
- y = f(x) + a is a translation of a units parallel to the y-axis; y = f(x + a) is a translation of −a units parallel to the x-axis
- y = af(x) is a stretch parallel to the y-axis with factor a; y = f(ax) is a stretch parallel to the x-axis with factor 1/a
- y = −f(x) is a reflection in the x-axis; y = f(−x) is a reflection in the y-axis
How Functions has been examined
From 2022 to 2026, Functions was on Paper 1 · Pure Mathematics 1 in 11 of 11 exam series, worth about 19% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Pure Mathematics 1 | 66 | 19% | 11 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 7.
Filed in Momentum under: Functions.
A worked example
The function f is defined by f(x) = 2x² − 8x + 5 for x ≥ 2. Find the range of f, then find f⁻¹(x) and state its domain.
- Complete the square: 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
- For x ≥ 2 the squared bracket is never negative and equals 0 at x = 2, so the range of f is f(x) ≥ −3.
- Write y = 2(x − 2)² − 3, so (x − 2)² = (y + 3)/2.
- Take the positive square root, because x ≥ 2: x = 2 + √((y + 3)/2).
- Swap x and y. The domain of f⁻¹ is the range of f. Check: f(3) = −1 and 2 + √((−1 + 3)/2) = 3.
Answer: Range f(x) ≥ −3; f⁻¹(x) = 2 + √((x + 3)/2), with domain x ≥ −3
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Working out fg(x) as g(f(x)): the function written next to x acts first, and the two orders usually give different answers.
- Giving f⁻¹(x) without its domain, or quoting the domain of f: the inverse's domain is the range of f, and stating it is often a separate mark.
- Describing y = f(x − 3) as a translation 3 units to the left: replacing x by x − 3 moves the graph 3 units to the right.
- Describing y = f(2x) as a stretch with factor 2: it is a stretch parallel to the x-axis with factor ½.
- Applying a combined transformation in the wrong order: for y = 2f(x) + 1 the stretch comes before the translation, because 1 is added after the doubling.
Exam technique
- To find a range, sketch the graph over the given domain and check the end points and any vertex, saying whether each end is included (≤) or not (<).
- To find f⁻¹(x), write y = f(x), make x the subject, then swap x and y; after completing the square, choose the sign of the square root that matches the restricted domain.
- Name each transformation in full: its type, its direction or axis, and its factor or distance. 'Stretch' on its own earns nothing.
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.