OxfordAQA International AS and A-level Mathematics: Functions, composites, inverses and modulus (9660) – 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 the first part of content area P2.1 (Algebra and functions) in the OxfordAQA International AS and A-level Mathematics specification, Version 5.2: functions, domain and range, composite and inverse functions, and the modulus. This content is in International A-level (A2) Unit P2 (Pure Maths) and is not examined at International AS.

Explanations are in the study guide and the revision notes. See the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. State the range of each function. (a) f(x) = 3 − 2x for −1 ≤ x < 4 [1] (b) g(x) = |x − 2| + 3 for all real x [1] (c) h(x) = x² − 4x + 7 for 0 ≤ x ≤ 5 [1]

2. f(x) = 4 − 3x and g(x) = x² + 5, both for all real x.

(a) Find fg(x) and gf(x), each in expanded form. [3] (b) Solve gf(x) = 30. [3]

3. f(x) = 1/(x − 3) for x ≠ 3, and g(x) = x² − 1 for all real x.

(a) Find fg(x) and state the values of x for which it is undefined. [3] (b) State the range of fg. [2]

4. f(x) = (5x − 3)/(x + 1) for x ≠ −1.

(a) Find f⁻¹(x). [3] (b) State the domain of f⁻¹. [1] (c) Solve f(x) = f⁻¹(x). [3]

5. f(x) = (x − 2)² + 5 for x ≤ 2.

(a) State the range of f. [1] (b) Find f⁻¹(x), stating its domain. [3] (c) Find f⁻¹(9). [1]

6. f(x) = x/2 + 4 and g(x) = 3x − 1. Find (fg)⁻¹(x) and show that it equals g⁻¹f⁻¹(x). [4]

7. Solve |2x + 1| = 7 − 3x. [4]

8. Solve |2x − 1| < 3|x + 4|. [6]

9. f(x) = x² − 2x − 8.

(a) Sketch y = |f(x)|, giving the axis intercepts and the highest point of the reflected part. [3] (b) Solve |f(x)| = 5. [2] (c) Hence solve |f(x)| < 5. [2]

10. f(x) = x² − 6x + 17 for x ≥ 3, and g(x) = 2x − 1 for all real x.

(a) State the range of f. [1] (b) Find f⁻¹(x), stating its domain. [3] (c) State the domain of fg. [1] (d) Solve fg(x) = 24. [3]

11. g(x) = |2x − 5| − 3 for all real x.

(a) State the range of g. [1] (b) Sketch y = g(x), giving the vertex and the axis intercepts. [3] (c) Explain why g has no inverse as defined. The domain is now restricted to x ≥ 5/2. Find g⁻¹(x) and state its domain. [3] (d) Solve g(x) < x. [3]

Answers

1. (a) f is decreasing, f(−1) = 5 is reached, f(4) = −5 is not: −5 < f(x) ≤ 5 [1] (b) |x − 2| ≥ 0, so g(x) ≥ 3 [1] (c) h(x) = (x − 2)² + 3, minimum 3 at x = 2 inside the domain, h(0) = 7, h(5) = 12: 3 ≤ h(x) ≤ 12 [1] [3] Examiner insight: Write the inequality signs carefully. Strict versus non-strict ends are where answers to range questions most often lose credit.

2. (a) fg(x) = 4 − 3(x² + 5) [1] = −3x² − 11 [1]. gf(x) = (4 − 3x)² + 5 = 9x² − 24x + 21 [1] [3] (b) (4 − 3x)² + 5 = 30, so (4 − 3x)² = 25 [1]. 4 − 3x = 5 or −5 [1]. x = −1/3 or x = 3 [1] [3] Examiner insight: Solving gf(x) = 30 before expanding is shorter and cuts the risk of an arithmetic slip. Showing the substitution step makes the order of composition clear.

3. (a) fg(x) = 1/(x² − 1 − 3) [1] = 1/(x² − 4) [1]. Undefined when x² = 4, so x = 2 or x = −2 [1] [3] (b) x² − 4 takes every value from −4 upwards except 0 [1]. So fg(x) ≤ −1/4 or fg(x) > 0 [1] [2] Examiner insight: A range answer that lists only one of the two pieces is likely to be incomplete. Check the output at the smallest value of the inner expression.

4. (a) y(x + 1) = 5x − 3 [1]. x(y − 5) = −3 − y, so x = (y + 3)/(5 − y) [1]. f⁻¹(x) = (x + 3)/(5 − x) [1] [3] (b) x ≠ 5 [1] [1] (c) (5x − 3)(5 − x) = (x + 3)(x + 1) [1]. 6x² − 24x + 18 = 0, so x² − 4x + 3 = 0 [1]. x = 1 or x = 3 [1] [3] Examiner insight: The domain of the inverse is the range of f, not a value you need to calculate again. Checking that your solutions are not the excluded values −1 and 5 is a quick final step.

5. (a) f(x) ≥ 5 [1] [1] (b) (x − 2)² = y − 5 [1]. Since x ≤ 2, x − 2 = −√(y − 5) [1]. f⁻¹(x) = 2 − √(x − 5), x ≥ 5 [1] [3] (c) 2 − √4 = 0 [1] [1] Examiner insight: Choosing the negative root and linking it to the domain x ≤ 2 is the reasoning to show. A bare formula with the wrong sign is easy to spot with f⁻¹(5) = 2.

6. fg(x) = (3x − 1)/2 + 4 = (3x + 7)/2 [1]. Let y = (3x + 7)/2, so x = (2y − 7)/3 and (fg)⁻¹(x) = (2x − 7)/3 [1]. f⁻¹(x) = 2x − 8 and g⁻¹(x) = (x + 1)/3 [1]. g⁻¹f⁻¹(x) = (2x − 8 + 1)/3 = (2x − 7)/3, the same [1] [4] Examiner insight: On a “show that” the final comparison must be stated, not implied. Writing both expressions side by side helps.

7. Case 1: 2x + 1 = 7 − 3x gives x = 6/5 [1]. Case 2: −(2x + 1) = 7 − 3x gives x = 8 [1]. For x = 8 the right side is −17, which a modulus cannot equal, so reject [1]. For x = 6/5 both sides are 17/5, so accept: x = 6/5 [1] [4] Examiner insight: Both candidates look plausible until tested. Writing the check beside each case shows why one is rejected.

8. Equation |2x − 1| = 3|x + 4|: 2x − 1 = 3x + 12 gives x = −13 [1]. 2x − 1 = −3x − 12 gives x = −11/5 [1]. Test x = 0: 1 < 12 is true [1]. Test x = −5: 11 < 3 is false [1]. Test x = −20: 41 < 48 is true [1]. x < −13 or x > −11/5 [1] [6] Examiner insight: A sketch of the two V shapes (vertices at 1/2 and −4) supports the test values and shows the steeper line is above the other on both outer regions.

9. (a) f(x) = (x − 4)(x + 2), so the axis crossings are (−2, 0) and (4, 0) [1]. f(0) = −8, so |f| meets the y-axis at (0, 8) [1]. The reflected part peaks at (1, 9) [1] [3] (b) f(x) = 5: x² − 2x − 13 = 0, so x = 1 ± √14 [1]. f(x) = −5: x² − 2x − 3 = 0, so x = −1 or 3 [1] [2] (c) 1 − √14 < x < −1 or 3 < x < 1 + √14 [1][1] [2] Examiner insight: Solving both f(x) = 5 and f(x) = −5 is the step that carries the question. Reading the answer from your sketch is a good way to confirm the order of the four values.

10. (a) f(x) = (x − 3)² + 8, so f(x) ≥ 8 [1] [1] (b) (x − 3)² = y − 8 [1]. Since x ≥ 3, x = 3 + √(y − 8) [1]. f⁻¹(x) = 3 + √(x − 8), x ≥ 8 [1] [3] (c) g(x) must be at least 3, so 2x − 1 ≥ 3 and x ≥ 2 [1] [1] (d) fg(x) = (2x − 4)² + 8 = 4x² − 16x + 24 [1]. 4x² − 16x = 0, so x = 0 or 4 [1]. x = 0 is outside the domain of fg, so x = 4 [1] [3] Examiner insight: Linking part (c) to part (d) is where the question is decided. A solution found by algebra needs checking against the domain you have stated.

11. (a) g(x) ≥ −3 [1] [1] (b) Vertex (5/2, −3) [1]. y-intercept g(0) = 2, so (0, 2) [1]. |2x − 5| = 3 gives x = 1 or 4, so (1, 0) and (4, 0) [1] [3] (c) g is many-one, for example g(1) = g(4) = 0 [1]. For x ≥ 5/2, g(x) = 2x − 8 [1]. g⁻¹(x) = (x + 8)/2, x ≥ −3 [1] [3] (d) |2x − 5| < x + 3 [1]. Equation: 2x − 5 = x + 3 gives x = 8, and 5 − 2x = x + 3 gives x = 2/3 [1]. 2/3 < x < 8 [1] [3] Examiner insight: An example showing two inputs with one output is a clear way to explain why there is no inverse. For part (d), say which region your test value shows is correct.

Where marks are usually lost

• Reading fg(x) as “g after f” and composing in the wrong order.
• Giving the domain of an inverse from the formula instead of the range of f.
• Using the positive square root when inverting a function with domain x ≤ k.
• Not testing the answers to a modulus equation where one side is not a modulus.
• Stating a range with an end point that the domain does not reach, or missing a gap in it.
• Solving only one of the two equations f(x) = k and f(x) = −k when finding where |f(x)| = k.
• Giving crossing points when the question asks for an inequality, or joining a “between” answer with “or”.
• Leaving out the check that an algebraic solution lies in the domain of a composite.

Next steps

• Revisit the revision notes for the methods and the quick self-test.
• Re-read the study guide for any worked example you found hard.
• Go to the course hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA, International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), content area P2.1.