Cambridge International AS & A Level Mathematics 9709: Functions – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 1.2 Functions of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 1.2 is part of Pure Mathematics 1, which is examined on Paper 1 and is compulsory for both AS Level and A Level. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Give exact answers where the question asks for them.

Related: the Functions study guide, the Functions revision notes, the Quadratics practice questions, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

1. The function f is defined by f(x) = 5 − 2x for −1 ≤ x < 3. Find the range of f. [2]

2. For each function, determine whether it is one-one. Give a reason for each answer.

(a) p(x) = x² + 4x for x ≥ −1 [2] (b) q(x) = sin x for 0 ≤ x ≤ π [2]

3. The point P(4, −6) lies on the curve y = f(x). State the coordinates of the image of P on each of these curves.

(a) y = f(x − 3) [1] (b) y = −½f(x) [1] (c) y = f(2x) + 5 [2]

4. The functions f and g are defined for x ∈ ℝ by f(x) = 3x − 2 and g(x) = x² + 1.

(a) Find expressions for fg(x) and gf(x). [2] (b) Solve the equation fg(x) = gf(x), giving your answers in exact form. [3]

5. The function f is defined by f(x) = √(2x + 5) − 1 for x ≥ −5/2.

(a) State the range of f. [1] (b) Find an expression for f⁻¹(x) and state the domain of f⁻¹. [3]

6. The functions f and g are defined by f(x) = 4 − x² for x ∈ ℝ, and g(x) = √x for x ≥ 0.

(a) Explain why the composite function gf cannot be formed. [2] (b) Find an expression for fg(x) and state the range of fg. [2]

7. (a) Describe fully a sequence of two transformations that maps the graph of y = cos x onto the graph of y = cos 2x + 1. [4] (b) State the range of y = cos 2x + 1 for 0 ≤ x ≤ π. [1]

8. The curve y = x² − 6x + 4 is translated 2 units in the positive x-direction. The result is then stretched parallel to the y-axis with scale factor 3.

(a) Find the equation of the final curve in the form y = ax² + bx + c. [3] (b) Find the coordinates of the minimum point of the final curve. [2]

9. The function f is defined by f(x) = 3 − 2 cos x for 0 ≤ x ≤ π.

(a) Find the range of f. [2] (b) Explain why f has an inverse. [1] (c) Find an expression for f⁻¹(x) and state its domain. [2]

10. The function f is defined by f(x) = (x² + 6)/5 for x ≥ 0.

(a) Find an expression for f⁻¹(x) and state the domain of f⁻¹. [3] (b) Find the coordinates of the points where the graphs of y = f(x) and y = f⁻¹(x) meet. [3]

11. The function f is defined by f(x) = 2x² + 12x + 11 for x ≤ k.

(a) Express 2x² + 12x + 11 in the form a(x + b)² + c, where a, b and c are constants. [3] (b) State the largest value of k for which f is one-one. [1] (c) For this value of k, find an expression for f⁻¹(x) and state the domain of f⁻¹. [4] (d) On the same diagram, sketch the graphs of y = f(x) and y = f⁻¹(x), making clear the relationship between them. [3]

12. The functions f and g are defined by f(x) = 6/(x + 2) for x ≥ 0, and g(x) = 2x − 1 for x ∈ ℝ.

(a) Find the range of f. [2] (b) Find an expression for f⁻¹(x) and state the domain of f⁻¹. [3] (c) Find gf(x), and solve the equation gf(x) = x. [4] (d) Determine whether the composite function fg can be formed. Give a reason. [2]

Answers

1. f is decreasing. f(−1) = 7 and f(3) = −1 [1]. x = 3 is not included, so −1 < f(x) ≤ 7 [1]. [2] Examiner insight: the accuracy mark needs both inequality signs right; a range written in terms of x instead of f(x) usually loses it.

2. (a) p(x) = (x + 2)² − 4, so the vertex is at x = −2, outside the domain, and p is increasing for x ≥ −1 [1]. p is one-one [1]. (b) sin(π/6) = sin(5π/6) = ½, so two inputs give the same output [1]. q is not one-one [1]. [4] Examiner insight: a bare “yes” or “no” earns nothing here; the mark for the conclusion depends on a correct reason, such as a vertex position or a specific pair of equal outputs.

3. (a) (7, −6) [1]. (b) (4, 3) [1]. (c) x-coordinate 4 ÷ 2 = 2 [1]; y-coordinate −6 + 5 = −1, giving (2, −1) [1]. [4] Examiner insight: each coordinate is often marked separately, so a correct y-coordinate still scores in (c) even if you multiply the x-coordinate by 2 by mistake.

4. (a) fg(x) = 3(x² + 1) − 2 = 3x² + 1 [1]; gf(x) = (3x − 2)² + 1 = 9x² − 12x + 5 [1]. (b) 3x² + 1 = 9x² − 12x + 5 gives 6x² − 12x + 4 = 0, so 3x² − 6x + 2 = 0 [1]. x = (6 ± √12)/6 [1], so x = 1 ± (√3)/3 [1]. [5] Examiner insight: the final mark asks for exact form, so decimals such as 1.58 and 0.423 lose the accuracy mark even though the method marks are kept.

5. (a) f(x) ≥ −1 [1]. (b) y + 1 = √(2x + 5), so (y + 1)² = 2x + 5 [1]. f⁻¹(x) = ((x + 1)² − 5)/2 [1], with domain x ≥ −1 [1]. [4] Examiner insight: the domain mark is usually independent of the expression, so state x ≥ −1 even if your algebra has gone wrong; it is simply the range from (a).

6. (a) The range of f is f(x) ≤ 4 [1]. This includes negative values, which are not in the domain of g (x ≥ 0), so gf cannot be formed [1]. (b) fg(x) = 4 − (√x)² = 4 − x [1] for x ≥ 0, so the range is fg(x) ≤ 4 [1]. [4] Examiner insight: in (a) you must compare the range of f with the domain of g explicitly; “you cannot square-root a negative” without naming the range often earns only the first mark.

7. (a) Stretch parallel to the x-axis [1], scale factor ½ [1]. Translation [1] by the vector (0, 1), written as a column vector [1]. (Either order is correct.) (b) 0 ≤ y ≤ 2 [1]. [5] Examiner insight: “shift”, “move” or “squash” are not accepted for the name of a transformation, and a stretch with no direction or factor loses its detail mark.

8. (a) Replace x by x − 2: y = (x − 2)² − 6(x − 2) + 4 = x² − 10x + 20 [1]. Multiply by 3 [1]: y = 3x² − 30x + 60 [1]. (b) The original curve is y = (x − 3)² − 5, with minimum (3, −5) [1]. The image is (5, −15) [1]. [5] Examiner insight: (b) is usually marked with follow-through from your equation in (a), so completing the square on your own answer still earns credit if (a) was wrong.

9. (a) f(0) = 3 − 2 = 1 and f(π) = 3 + 2 = 5 [1], so 1 ≤ f(x) ≤ 5 [1]. (b) cos x is decreasing for 0 ≤ x ≤ π, so f is increasing and therefore one-one, so it has an inverse [1]. (c) cos x = (3 − x)/2, so f⁻¹(x) = cos⁻¹((3 − x)/2) [1], for 1 ≤ x ≤ 5 [1]. [5] Examiner insight: in (b), “because it is one-one” alone rarely earns the mark; you need a reason why f is one-one on this domain.

10. (a) 5y = x² + 6, so x² = 5y − 6 [1]. Since x ≥ 0, take the positive root: f⁻¹(x) = √(5x − 6) [1], domain x ≥ 6/5 [1]. (b) f is increasing for x ≥ 0, so the graphs meet on y = x: (x² + 6)/5 = x gives x² − 5x + 6 = 0 [1]. So x = 2 or x = 3 [1], and the points are (2, 2) and (3, 3) [1]. [6] Examiner insight: both x = 2 and x = 3 are in the domain and must be kept; rejecting one without a reason loses the final accuracy mark.

11. (a) a = 2 [1], b = 3 [1], c = −7 [1], so 2x² + 12x + 11 = 2(x + 3)² − 7. (b) The vertex is at x = −3, so k = −3 [1]. (c) y = 2(x + 3)² − 7, so (x + 3)² = (y + 7)/2 [1]. Since x ≤ −3, x + 3 ≤ 0, so take the negative root: x + 3 = −√((y + 7)/2) [1]. f⁻¹(x) = −3 − √((x + 7)/2) [1], domain x ≥ −7 [1]. (d) y = f(x): the left half of a parabola, decreasing, ending at (−3, −7) [1]. y = f⁻¹(x): its mirror image, starting at (−7, −3) and decreasing, for example through (1, −5) [1]. The line y = x is drawn, and the two curves are clearly symmetrical about it [1]. [11] Examiner insight: in (c) the mark for the sign of the root needs a reason linked to x ≤ −3; a “±” left in the final answer loses the accuracy mark.

12. (a) f is decreasing and f(0) = 3 [1]. f(x) never reaches 0, so 0 < f(x) ≤ 3 [1]. (b) y(x + 2) = 6, so x = 6/y − 2 [1]. f⁻¹(x) = 6/x − 2 [1], for 0 < x ≤ 3 [1]. (c) gf(x) = 2 × 6/(x + 2) − 1 = 12/(x + 2) − 1 [1]. Setting this equal to x: 12 − (x + 2) = x(x + 2), so x² + 3x − 10 = 0 [1]. (x + 5)(x − 2) = 0 gives x = 2 [1]; x = −5 is rejected because the domain of f is x ≥ 0 [1]. (d) The range of g is all real numbers [1]. This is not within the domain of f (x ≥ 0); for example, g(0) = −1. So fg cannot be formed [1]. [11] Examiner insight: in (c), the last mark is for rejecting x = −5 with a reason; keeping both roots, or dropping −5 silently, loses it.

Where marks are usually lost

• Using ≤ instead of < at an end point that is not in the domain.
• Giving the inverse’s domain as the domain of f instead of the range of f.
• Leaving “±” in an inverse, or choosing the positive root when the domain of f is to the left of the vertex.
• Working out gf when fg was asked for.
• Claiming a composite exists without comparing the inner range with the outer domain.
• Keeping a root of an equation that lies outside the domain.
• Using scale factor 2 instead of ½ for the stretch in y = f(2x).
• A sketch of f and f⁻¹ with no line y = x, or with end points that do not swap coordinates.
• Giving decimals when the question asks for exact answers.

Next steps

• Functions revision notes
• Functions study guide
• Pure 1 mixed practice
• A Level Mathematics hub
• Printable 9709 checklist
• Free AS diagnostic and the 9709 self-check bank
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027, Version 4, Cambridge Assessment International Education (part of Cambridge University Press & Assessment). Subject content, topic 1 Pure Mathematics 1 (for Paper 1), section 1.2 Functions.