AQA A-level Mathematics: Graphs of Functions, Composite Functions and Inverse Functions (7357) – 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 follow content B7 (graphs of functions) and B8 (composite and inverse functions) of the AQA A-level Mathematics 7357 specification, version 1.3. Use gf(x) to mean g(f(x)).

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Questions

1. Sketch y = (x + 3)(x − 1)², marking every intercept with the axes. [4]

2. Solve |3x + 6| = 4 − x. [5]

3. On one set of axes sketch y = −4/x and y = 6/x², showing asymptotes. Find the coordinates of the point where the curves meet. [5]

4. The table shows pairs of values that are thought to satisfy y = kx³.

x	1	2	3	4
y	2.5	20	67.5	160

(a) State what to plot to test this, and what the graph should show. [1] (b) Show that the data fit and find k. [2] (c) Find y when x = 5. [1]

5. By rearranging x³ + 2x − 6 = 0 and sketching two curves, show that it has exactly one real root. Then show that the root lies between 1 and 2. [5]

6. f(x) = 2x + 1 and g(x) = x² − 3, for x ∈ ℝ.

(a) Find gf(x) and fg(x) in expanded form. [2] (b) Show that gf(x) = fg(x) has no real solutions. [3]

7. The function f is x ↦ x² − 4x on ℝ, and g is x ↦ √x on x ≥ 0.

(a) Find the domain of gf. [2] (b) Find fg(x) and its range. [4]

8. f(x) = (5 − x)/(2x + 1) for x ≠ −1/2.

(a) Find f⁻¹(x). [3] (b) Describe what your answer shows about f, and state the range of f. [2]

9. f(x) = 3x² − 12x + 4 for x ≥ k.

(a) Write f(x) in the form a(x − p)² + q. [2] (b) Explain why f has no inverse when x ∈ ℝ. [1] (c) Take k = 2. Find f⁻¹(x) and state its domain. [4]

10. f(x) = √(3x + 4) for x ≥ −4/3.

(a) Find f⁻¹(x), with its domain. [3] (b) Find the coordinates of the common point of the graphs of f and f⁻¹. [3]

11. f(x) = (x − 2)² + 1 for x ≥ 2, and g(x) = 6/x for x ≠ 0.

(a) Find the range of f. [1] (b) Find f⁻¹(x) and its domain. [3] (c) Find gf(x) and its range. [3] (d) Solve gf(x) = 2. [3]

12. y is proportional to x². (a) x increases by 20 per cent. Find the percentage change in y. [2] (b) z is inversely proportional to √w. Find the effect on z of multiplying w by 4. [1]

Answers

1. Roots: x = −3 and x = 1 [1]. y-intercept: (3)(1) = 3, so (0, 3) [1]. The factor (x − 1)² means the curve touches the x-axis at (1, 0) [1]. A positive cubic comes from bottom left, crosses at (−3, 0), rises through (0, 3), falls to touch at (1, 0), then rises [1]. Examiner insight: Give coordinates for each intercept and show the touch clearly, because a crossing drawn at a double root contradicts your own factorisation.

2. Right arm: 3x + 6 = 4 − x [1], so x = −1/2 [1]. Left arm: −(3x + 6) = 4 − x [1], so −2x = 10 and x = −5 [1]. Check: |4.5| = 4.5 and |−9| = 9 = 4 − (−5), so both work [1]. x = −1/2 or x = −5 Examiner insight: Checking in the original equation shows you have not kept a root from the wrong arm, and it protects the answer.

3. y = −4/x lies in quadrants 2 and 4, y = 6/x² lies above the x-axis, and both have asymptotes x = 0 and y = 0 [2]. Equating: −4/x = 6/x², so −4x = 6 [1], x = −3/2 [1], y = −4/(−3/2) = 8/3. The point is (−3/2, 8/3) [1]. Examiner insight: Put both asymptotes on the sketch and keep the curves away from the axes. The single intersection should be visible where the left branches meet.

4. (a) Plot y against x³; a straight line through the origin [1]. (b) y/x³ = 2.5, 2.5, 2.5, 2.5 for all four rows [1], so k = 2.5 [1]. (c) y = 2.5 × 125 = 312.5 [1]. Examiner insight: A statement about the graph needs both “straight line” and “through the origin”. Check every row, not just one.

5. x = 0 is not a root, so divide by x: x² + 2 = 6/x [1]. For x < 0, 6/x is negative and x² + 2 is positive, so no intersection [1]. For x > 0 the hyperbola decreases from large values while the parabola increases from 2, so they meet exactly once [1]. Let h(x) = x³ + 2x − 6. h(1) = −3 and h(2) = 6 [1]. The sign changes and h is continuous, so the root lies between 1 and 2 [1]. Examiner insight: Say why x = 0 may be divided out, and explain the sign change in words. A sketch alone will not justify “exactly one”.

6. (a) gf(x) = (2x + 1)² − 3 = 4x² + 4x − 2 [1]. fg(x) = 2(x² − 3) + 1 = 2x² − 5 [1]. (b) 4x² + 4x − 2 = 2x² − 5 gives 2x² + 4x + 3 = 0 [1]. Discriminant = 16 − 24 = −8 [1]. This is negative, so there are no real solutions [1]. Examiner insight: “Show that” needs the quadratic written out, the discriminant evaluated, and a concluding sentence.

7. (a) gf(x) = √(x² − 4x), which needs x² − 4x ≥ 0 [1]. So x(x − 4) ≥ 0, giving x ≤ 0 or x ≥ 4 [1]. (b) fg(x) = (√x)² − 4√x = x − 4√x [1]. Complete the square in √x: (√x − 2)² − 4 [1]. The minimum is −4, at x = 4 [1]. Range: fg(x) ≥ −4 [1]. Examiner insight: Test the domain of gf for the inner output, not the original rule. For a range, write down the minimum and where it occurs.

8. (a) y(2x + 1) = 5 − x [1]. 2xy + y = 5 − x, so x(2y + 1) = 5 − y [1]. f⁻¹(x) = (5 − x)/(2x + 1) [1]. (b) f⁻¹ = f, so f is self-inverse [1]. The range of f is the domain of f⁻¹, so f(x) ≠ −1/2 [1]. Examiner insight: If you notice f⁻¹ = f, say so explicitly. Reading the range from the domain of the inverse is a quick check.

9. (a) 3(x² − 4x) + 4 = 3(x − 2)² − 12 + 4 = 3(x − 2)² − 8 [2]. (b) It is many-one: for example f(1) = f(3) = −5 [1]. (c) (x − 2)² = (y + 8)/3 [1]. Since x ≥ 2 take the positive root: x = 2 + √((y + 8)/3) [1]. So f⁻¹(x) = 2 + √((x + 8)/3) [1], with domain x ≥ −8 [1]. Examiner insight: Justify the sign of the root from the domain of f. Quote the inverse’s domain, which is the range of f.

10. (a) y² = 3x + 4 [1], so f⁻¹(x) = (x² − 4)/3 [1], with domain x ≥ 0 [1]. (b) f is increasing, so the graphs meet on y = x: √(3x + 4) = x [1]. Then 3x + 4 = x² gives (x − 4)(x + 1) = 0 [1]. Since √ cannot be negative, x = −1 is rejected, leaving (4, 4) [1]. Examiner insight: After squaring, test each root in the original equation. Rejecting x = −1 with a reason keeps the final answer secure.

11. (a) f(x) ≥ 1 [1]. (b) x − 2 = √(y − 1), taking the positive root as x ≥ 2 [1]. f⁻¹(x) = 2 + √(x − 1) [1], with domain x ≥ 1 [1]. (c) gf(x) = 6/((x − 2)² + 1) [1]. As f takes every value in [1, ∞) [1], gf takes every value from just above 0 up to 6: 0 < gf(x) ≤ 6 [1]. (d) (x − 2)² + 1 = 3 [1], so (x − 2)² = 2 [1]. As x ≥ 2, x = 2 + √2 [1]. Examiner insight: Link each range to the previous one. Mention that 2 − √2 is rejected by the domain.

12. (a) y is multiplied by 1.2² = 1.44 [1], a 44 per cent increase [1]. (b) z ∝ w^(−1/2), so z is multiplied by 4^(−1/2) = 1/2 [1], that is, halved. Examiner insight: Convert the multiplier to a percentage change at the end. Do not simply double the 20 per cent.

Where marks are usually lost

• Not checking roots from a modulus equation in the original equation.
• Drawing a double root as a crossing, or omitting the y-intercept.
• Leaving asymptotes off reciprocal curves.
• Writing the composite in the wrong order.
• Stating f⁻¹ without its domain, or taking the wrong sign of the root after a restriction.
• Dividing by x and not noting that x = 0 is lost.
• Giving percentage change for proportion as simple multiples rather than squared or inverse factors.
• Leaving a “show that” without the final concluding sentence.

Next steps

• Revise with the revision notes.
• Reread the study guide.
• Go to the course hub and the checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class with a Marlbridge teacher.

Official syllabus

AQA, A-level Mathematics (7357) specification, version 1.3, 31 January 2018, content B7 and B8.