OCR A Level Mathematics A: Graphs, sketching, intersections and proportionality (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), for first assessment in 2018, section 1.02 items 1.02m to 1.02r (Stage 1) and graphical solution of inequalities as an application of intersections. Modulus graphs (the A Level only parts of 1.02s and 1.02t) are in the separate modulus unit. Pure Mathematics content is assessed on all three papers.

Learn the method first in the study guide or the revision notes. The course hub and printable checklist show where this unit fits, and the free 10-minute diagnostics will find your weak spots.

Questions

1. Sketch the curve y = (x + 2)(x − 1)(x − 5), labelling every point where it meets the axes. [3]

2. Sketch y = (x − 1)²(x + 2)², labelling the points where it meets the axes and the coordinates of its maximum point. [4]

3. A cubic curve touches the x-axis at x = −1, crosses it at x = 3 and passes through (0, −9). Find its equation in the form y = ax³ + bx² + cx + d. [4]

4. Sketch y = 6x² − x³, finding the coordinates of its stationary points and stating their nature. [5]

5. On one set of axes, sketch y = −8/x and y = 8/x², showing the asymptotes. [3] Find the coordinates of the point where the two curves meet. [2]

6. (a) Solve x³ − 9x = 3x. [3] (b) Explain what the solutions to part (a) represent on the graphs of y = x³ − 9x and y = 3x. [1] (c) Hence solve x³ − 9x < 3x. [2]

7. (a) Sketch y = eˣ and y = 4 − x on one set of axes and explain why eˣ = 4 − x has exactly one real solution. [2] (b) Show that the solution lies between x = 1 and x = 1.5. [3]

8. By sketching y = sin x and y = x/3 for x in radians, find how many solutions sin x = x/3 has. [3]

9. y is proportional to √x. When x = 16, y = 12. Find y when x = 25, and find x when y = 21. [4]

10. The table shows values of x and y.

x	1	2	4	5
y	24	12	6	4.8

(a) Show that y is inversely proportional to x. [2] (b) State what to plot against y to obtain a straight line through the origin, and give the gradient. [2] (c) Find y when x = 8. [1]

11. The curve C has equation y = x²(x + 3) and the line L has equation y = 4x. (a) Sketch C and L on one set of axes, labelling the intercepts of C and the coordinates of its maximum point. [3] (b) Find the coordinates of all the points where C meets L. [3] (c) Hence solve x²(x + 3) ≤ 4x. [2]

Answers

1. Roots at x = −2, 1 and 5, so the curve crosses the x-axis at (−2, 0), (1, 0) and (5, 0). [1] Put x = 0: (2)(−1)(−5) = 10, so the y-intercept is (0, 10). [1] Positive cubic: the curve comes from bottom left, crosses at −2, peaks above the axis, crosses at 1, dips below, and crosses at 5 heading up to the top right. [1] [3] Examiner insight: A sketch is easier to credit when each axis crossing carries its coordinates and the curve clearly continues beyond the last root instead of stopping at it.

2. Both roots are repeated, so the curve touches the x-axis at (1, 0) and (−2, 0) without crossing. [1] At x = 0, y = (1)(4) = 4, so the y-intercept is (0, 4). [1] dy/dx = 2(x − 1)(x + 2)² + 2(x − 1)²(x + 2) = 2(x − 1)(x + 2)(2x + 1). [1] The middle stationary point is x = −1/2, where y = (−3/2)²(3/2)² = 81/16, so the maximum is (−1/2, 81/16). The curve is W-shaped, with both ends rising. [1] [4] Examiner insight: A product-rule derivative is easiest to check when you factorise it fully. Naming the point as a maximum, not just giving its coordinates, shows you have read the shape.

3. The curve touches at −1 and crosses at 3, so y = k(x + 1)²(x − 3). [1] At (0, −9): k(1)(−3) = −9. [1] k = 3. [1] Expanding: 3(x² + 2x + 1)(x − 3) = y = 3x³ − 3x² − 15x − 9. [1] [4] Examiner insight: Begin with the factorised form, since the roots are given. Leaving out k, or squaring the wrong factor, loses the whole structure of the answer.

4. y = x²(6 − x). The curve touches the x-axis at (0, 0) and crosses at (6, 0). [1] The leading term is −x³, so it comes from top left and finishes at bottom right. [1] dy/dx = 12x − 3x² = 3x(4 − x), zero at x = 0 and x = 4. [1] x = 0: y = 0, and d²y/dx² = 12 − 6x = 12 > 0, so minimum at (0, 0). [1] x = 4: y = 96 − 64 = 32, and d²y/dx² = −12 < 0, so maximum at (4, 32). [1] [5] Examiner insight: A stationary point question usually asks for its nature as well as its position. A short reason, such as the sign of the second derivative, supports the label you give.

5. y = −8/x: two branches in quadrants 2 and 4. [1] y = 8/x²: two branches above the x-axis, symmetric about the y-axis. [1] Both have the asymptotes x = 0 and y = 0 drawn and the curves approach them without meeting. [1] Equate: −8/x = 8/x², so −8x = 8 and x = −1. [1] y = 8/1 = 8, so (−1, 8). [1] [5] Examiner insight: On the same axes the two curves overlap in quadrant 2. Check that your sketch shows the intersection there, matching the algebra.

6. (a) x³ − 12x = 0. [1] x(x² − 12) = 0. [1] x = 0, x = 2√3, x = −2√3. [1] (b) They are the x-coordinates of the points where the curve y = x³ − 9x meets the line y = 3x. [1] (c) The curve is below the line when x³ − 12x < 0. For x > 2√3 the cubic is positive, so it is negative between 0 and 2√3, positive between −2√3 and 0, and negative below −2√3. [1] x < −2√3 or 0 < x < 2√3. [1] [6] Examiner insight: Dividing the first equation by x would lose the solution x = 0. Factorise instead and use the sketch to read off the intervals.

7. (a) eˣ is increasing and always positive, passing through (0, 1). The line 4 − x is decreasing. [1] An increasing curve and a decreasing line can meet only once. [1] (b) Let f(x) = eˣ + x − 4. f(1) = e − 3 ≈ −0.28, which is negative. [1] f(1.5) = e^1.5 + 1.5 − 4 ≈ 1.98, which is positive. [1] The sign changes, so the intersection (the solution) lies between 1 and 1.5. [1] [5] Examiner insight: Quote the two values and the sign of each. Stating only that “it changes sign” without the numbers leaves your claim without evidence.

8. Sketch y = sin x, which oscillates between −1 and 1, and the line y = x/3 through the origin with gradient 1/3. [1] Since |sin x| ≤ 1, any intersection has |x/3| ≤ 1, so |x| ≤ 3. At x = π/2 the sine is 1 and the line is about 0.52; at x = π the sine is 0 and the line is about 1.05, so they cross once in this range. [1] By symmetry the same happens for negative x, and they also meet at the origin: 3 solutions. [1] [3] Examiner insight: A convincing count uses a stated bound, such as |sin x| ≤ 1, to rule out any intersection beyond x = 3.

9. y = k√x. [1] 12 = k × 4, so k = 3. [1] x = 25: y = 3 × 5 = 15. [1] y = 21: √x = 7, so x = 49. [1] [4] Examiner insight: Find k from the given pair, write it into the equation, and only then use it. Squaring the wrong side when finding x is a common slip.

10. (a) xy = 24, 24, 24 and 24 for the four pairs. [1] The product is constant, so y = 24/x and y is inversely proportional to x. [1] (b) Plot y against 1/x. [1] The graph is a straight line through the origin with gradient 24. [1] (c) y = 24/8 = 3. [1] [5] Examiner insight: Show the product for every pair. A single check does not demonstrate that a relationship holds for all the data.

11. (a) C touches the x-axis at (0, 0) and crosses it at (−3, 0). [1] dy/dx = 3x² + 6x = 3x(x + 2), so the maximum is at (−2, 4). [1] L is a straight line through the origin with gradient 4, shown on the same axes. [1] (b) x²(x + 3) = 4x gives x(x² + 3x − 4) = 0. [1] x(x + 4)(x − 1) = 0, so x = −4, 0, 1. [1] The points are (−4, −16), (0, 0) and (1, 4). [1] (c) C is below or on L when x(x + 4)(x − 1) ≤ 0. [1] x ≤ −4 or 0 ≤ x ≤ 1. [1] [8] Examiner insight: Part (c) follows from the sketch. Use the strict or non-strict inequality that matches the question, and include the intersections when the sign is “less than or equal”.

Where marks are usually lost

• Drawing a repeated root as a crossing, or a single root as a touch.
• Leaving out the y-intercept or giving an x-value without the corresponding y-value.
• Getting the ends of a cubic or quartic the wrong way round for a negative leading coefficient.
• Dividing by x, or by another factor containing x, and losing a solution.
• Giving stationary points without saying whether each is a maximum or a minimum.
• Drawing reciprocal curves that cross an axis or bend back from an asymptote.
• Reading an inequality from a graph without testing a value in each interval.
• Writing only x-values when the question asks for coordinates.
• Treating a linear graph with a non-zero intercept as a proportional relationship.

Next steps

• Review the method in the revision notes.
• Read the full explanations in the study guide.
• Go to the course hub or tick topics on the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

Source: OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. Section 1.02, items 1.02m to 1.02r. The graphical inequalities in 1.02t involve the modulus function and are in the separate modulus unit.