OCR A Level Mathematics A: Modulus, functions, inverses and transformations (H240) – 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 the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, sections 1.02l, 1.02s, 1.02u, 1.02v, 1.02w and 1.02x. Single transformations (1.02w) are Stage 1, common with AS (H230). Everything else here is Stage 2, A Level only. Teaching is in the study guide; a condensed version is in the revision notes. Course pages: OCR A Level Mathematics hub and the printable checklist. Try the free 10-minute diagnostics to find weak spots first.

Questions

1. Solve |5x + 2| < 13. [3]

2. Solve |2x + 7| = 3x − 1. [4]

3. Solve |2x − 3| ≤ |x + 9|. [5]

4. (a) Sketch y = |3x + 12|, labelling the points where it meets the axes. [3] (b) Use your sketch to solve |3x + 12| < 6. [3]

5. A function is given by f(x) = x² + 4x + 11 for −6 ≤ x ≤ 1. (a) Find the range of f. [3] (b) Explain why f does not have an inverse. [1]

6. For each rule, state whether it is a function. If it is, say whether it is one-one or many-one. (a) x ↦ x³ for x ∈ ℝ [1] (b) x ↦ |x − 2| for x ∈ ℝ [1] (c) the rule giving every y with y² = x, for x > 0 [1] (d) State the range of the function in part (b). [1]

7. f(x) = 5/(x − 2) for x ≠ 2, and g(x) = x² + 1 for x ∈ ℝ. (a) Find fg(x) and say which real values of x give no output. [3] (b) Find gf(x). [1] (c) Solve fg(x) = 1. [2]

8. f(x) = (x + 7)/(2x − 3) for x ≠ 3/2. (a) Find f⁻¹(x). [3] (b) State the domain and the range of f⁻¹. [2] (c) Find f⁻¹(5). [1]

9. The function f is given by f(x) = 3x² + 18x + 4 for x ≥ k. (a) Express f(x) as a(x + p)² + q. [2] (b) What is the least k that allows f to have an inverse? [1] (c) Taking that value of k, obtain f⁻¹(x), with its domain. [3]

10. f(x) = √(2x − 1) for x ≥ 1/2. (a) Find f⁻¹(x) and state its domain. [3] (b) Find the coordinates of any point where y = f(x) meets y = f⁻¹(x). [4]

11. The highest point of the curve y = f(x) is P(−1, 4), and the curve also passes through the point (3, 0). (a) Describe the single transformation that maps y = f(x) to each of (i) y = f(x) − 2, (ii) y = f(x + 5), (iii) y = f(x/4). [3] (b) Give the images of P and of (3, 0) on the curve y = f(2x). [2] (c) Give the image of P on the curve y = −3f(x). [1]

12. f(x) = x² − 8x + 3. The curve C has equation y = 3f(2x − 6) + 1. (a) Describe a sequence of four transformations that maps y = f(x) to C. [4] (b) Determine where the lowest point of C is. [2] (c) Show that C has equation y = 12(x − 5)² − 38. [2]

Answers

1. Write −13 < 5x + 2 < 13 [1]. Subtract 2: −15 < 5x < 11 [1]. Divide by 5: −3 < x < 11/5 [1]. [3] Examiner insight: give the answer as one chain or as two inequalities joined by “and”; a “less than” modulus gives a single interval, so “or” would be the wrong link.

2. First case: 2x + 7 = 3x − 1, so x = 8 [1]. Second case: −(2x + 7) = 3x − 1, so −2x − 7 = 3x − 1 and x = −6/5 [1]. Test x = 8: |23| = 23 and 3(8) − 1 = 23, so it works [1]. For x = −6/5 the right side is −23/5, which is negative and cannot equal a modulus, so reject it [1]. x = 8 [4] Examiner insight: writing both roots without a check leaves a wrong answer standing, so show the substitution for each candidate.

3. Both sides are moduli, so square: (2x − 3)² ≤ (x + 9)² [1]. Expand: 4x² − 12x + 9 ≤ x² + 18x + 81 [1]. Collect: 3x² − 30x − 72 ≤ 0, so x² − 10x − 24 ≤ 0 [1]. Factorise: (x − 12)(x + 2) ≤ 0 [1]. The quadratic is below zero between its roots: −2 ≤ x ≤ 12 [1]. [5] Examiner insight: squaring is the efficient route here, but the inequality must stay “≤”, and the closed end points should be included because the question uses “≤”.

4. (a) The line y = 3x + 12 meets the axes at (−4, 0) and (0, 12) [1]. For x < −4 reflect the part below the axis: the left arm is y = −3x − 12 [1]. The sketch is a V with vertex (−4, 0), passing through (0, 12), arms of gradient 3 and −3 [1]. (b) Draw y = 6. The right arm meets it where 3x + 12 = 6, so x = −2 [1]. The left arm meets it where −3x − 12 = 6, so x = −6 [1]. The V lies below y = 6 between these: −6 < x < −2 [1]. [6] Examiner insight: label the vertex as a point on the x-axis, not just a corner, and mark the y-intercept, since a sketch should show its key features rather than accurate plotting.

5. (a) Complete the square: f(x) = (x + 2)² + 7 [1]. The minimum value 7 occurs at x = −2, which is inside the domain [1]. f(−6) = 23 and f(1) = 16, so the greatest value is 23: 7 ≤ f(x) ≤ 23 [1]. (b) f is many-one: for example f(−3) = f(−1) = 8, so it is not one-one and has no inverse [1]. [4] Examiner insight: the range must be stated in terms of f(x), and stopping at f(−6) and f(1) would miss the turning point inside the domain.

6. (a) A function; one-one [1]. (b) A function; many-one, for example x = 1 and x = 3 both give 1 [1]. (c) Not a function: an input such as x = 4 has two images, 2 and −2, so it is one-many [1]. (d) The range is f(x) ≥ 0 [1]. [4] Examiner insight: give a reason with a specific example input in parts (b) and (c); the label alone does not show why.

7. (a) fg(x) = f(x² + 1) = 5/(x² + 1 − 2) [1] = 5/(x² − 1) [1]. It is not defined at x = 1 and x = −1 [1]. (b) gf(x) = (5/(x − 2))² + 1 = 25/(x − 2)² + 1 [1]. (c) 5/(x² − 1) = 1 gives x² − 1 = 5 [1], so x = ±√6 [1]. Both are allowed since neither is 1 or −1. [6] Examiner insight: fg means g first, so the 5 in the numerator comes from f; reversing the order gives a different function, gf, and a wrong answer to part (a).

8. (a) Write y = (x + 7)/(2x − 3), so y(2x − 3) = x + 7 [1]. Then 2xy − 3y = x + 7 and x(2y − 1) = 3y + 7 [1]. So f⁻¹(x) = (3x + 7)/(2x − 1) [1]. (b) f(x) = 1/2 + (17/2)/(2x − 3), so the range of f omits 1/2. The domain of f⁻¹ is x ≠ 1/2 [1]. The range of f⁻¹ is the domain of f: f⁻¹(x) ≠ 3/2 [1]. (c) f⁻¹(5) = (15 + 7)/(10 − 1) = 22/9 [1]. [6] Examiner insight: the domain of f⁻¹ comes from the range of f, not from the denominator you happen to see in the formula, though here the two agree.

9. (a) 3(x² + 6x) + 4 = 3((x + 3)² − 9) + 4 [1] = 3(x + 3)² − 23 [1]. (b) The vertex is at x = −3, so f is one-one only if the domain starts there or later: k = −3 [1]. (c) Write y + 23 = 3(x + 3)², so (x + 3)² = (y + 23)/3 [1]. Since x ≥ −3, take the positive root: x = −3 + √((y + 23)/3) [1]. So f⁻¹(x) = −3 + √((x + 23)/3), domain x ≥ −23 [1]. [6] Examiner insight: say why you choose the positive root, linking it to the domain of f, because a bare formula with the wrong sign is a common slip on restricted quadratics.

10. (a) y² = 2x − 1, so x = (y² + 1)/2 [1]. f⁻¹(x) = (x² + 1)/2 [1], with domain x ≥ 0, the range of f [1]. (b) The graphs are reflections in y = x and f is increasing, so any meeting point lies on y = x [1]. Solve √(2x − 1) = x, so 2x − 1 = x² [1]. Then (x − 1)² = 0, so x = 1 [1]. f(1) = 1, so the point is (1, 1) [1]. [7] Examiner insight: explain why solving f(x) = x is enough; solving f(x) = f⁻¹(x) directly leads to a quartic that is much harder.

11. (a) (i) Translation by the vector (0, −2) [1]. (ii) Translation by (−5, 0) [1]. (iii) Stretch, scale factor 4, parallel to the x-axis [1]. (b) Halve the x-coordinates: P goes to (−1/2, 4) [1] and (3, 0) goes to (3/2, 0) [1]. (c) Multiply the y-coordinate by −3: (−1, −12) [1]. [6] Examiner insight: a description needs the type, the size and the direction or axis; “shift left” or “stretch by 4” alone leaves out part of what the question asks for.

12. (a) Translation by (6, 0), giving f(x − 6) [1]. Stretch, scale factor 1/2, parallel to the x-axis, giving f(2x − 6) [1]. Stretch, scale factor 3, parallel to the y-axis [1]. Translation by (0, 1) [1]. (b) f(x) = (x − 4)² − 13 has its minimum at (4, −13) [1]. The image has X = (4 + 6)/2 = 5 and Y = 3(−13) + 1 = −38: (5, −38) [1]. (c) f(2x − 6) = (2x − 10)² − 13 = 4(x − 5)² − 13 [1]. So y = 3(4(x − 5)² − 13) + 1 = 12(x − 5)² − 38 [1]. [8] Examiner insight: when the result is given, show the line where the bracket is rewritten and the line that reaches the stated form, so the route from f to C is visible.

Where marks are usually lost

• Accepting a root of |linear| = linear without checking the sign of the right side.
• Writing “or” where a single interval is needed, or the reverse.
• Forgetting to reverse an inequality when multiplying or dividing by a negative.
• Giving a range from the end points of the domain only.
• Applying composites in the wrong order: fg(x) uses g first.
• Missing excluded values in the domain of a composite or an inverse.
• Taking the wrong root when inverting a restricted quadratic.
• Moving f(x + a) to the right, or giving scale factor a for f(ax).
• Describing a transformation without a vector, scale factor or axis.

Next steps

• Read the study guide for the full method.
• Condense with the revision notes.
• Tick off progress on the printable checklist and see the OCR A Level Mathematics hub.
• Take the free 10-minute diagnostics.
• Want help in person or online? Book a free trial class.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), first assessment 2018, published by OCR.