OCR A Level Mathematics A: Indices, surds, simultaneous equations, quadratics and inequalities (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), sections 1.02a to 1.02i. All content is Stage 1. The mark splits in square brackets are Marlbridge’s own indicative allocations for self-marking.

Links: study guide, revision notes, course hub, printable checklist, free diagnostics, Pure Mathematics practice.

Questions

1. Evaluate, without a calculator:

(a) 343^(2/3) [1] (b) (25/4)^(−3/2) [2]

2.

(a) Simplify (81a¹² b⁻⁸)^(−1/4), giving your answer with positive indices. [3] (b) Solve 4x^(3/2) = 108. [2]

3.

(a) Show that √12 + 15/√3 − √27 can be written as 4√3. [3] (b) Rationalise the denominator of (6 − √5)/(3 + √5), giving your answer in the form (p − q√5)/r. [3]

4. A rectangle has length (3 + √2) cm and width (3 − √2) cm.

(a) Show that its area is exactly 7 cm². [2] (b) Find the exact length of a diagonal. [3]

5. Solve the simultaneous equations x² + 2xy = 8 and 2x + y = 7. [6]

6. A line has equation y = 3x + k and a curve has equation y = x² − x + 6.

(a) Show that any intersection satisfies x² − 4x + (6 − k) = 0. [2] (b) Find the value of k for which the line just touches the curve. [3] (c) Find where they touch. [2]

7. f(x) = −2x² + 8x − 3.

(a) Complete the square, writing f(x) as a(x + p)² + q. [3] (b) State the coordinates of the maximum point and the equation of the line of symmetry. [2] (c) Solve f(x) = 0, giving exact answers. [2]

8. Find the set of values of k for which 3x² + kx + 12 = 0 has real roots. Give your answer in set notation and in interval notation. [4]

9. Solve:

(a) 3x⁴ − 13x² + 4 = 0 [4] (b) 3/(x + 2)² + 2/(x + 2) = 1 [3]

10.

(a) Solve (2x − 1)/3 − (x + 2)/6 < 1. [3] (b) Solve 2x² + x − 15 ≥ 0. [2] (c) Find the values of x that satisfy both inequalities, in set notation. [2]

11. The region R is defined by y ≥ x² + 2x − 3 and y ≤ x + 3.

(a) Find the coordinates of the points where the boundaries meet. [3] (b) State, with a reason, whether the boundary lines should be solid or dashed. [1] (c) Find the turning point of the parabola. [2] (d) Determine whether the point (1, 2) lies in R. [2]

12. The curve C has equation y = x² + 6x + 14 and the line L has equation y = mx + 5.

(a) Express x² + 6x + 14 in the form (x + p)² + q and state the minimum point of C. [3] (b) Show that the x-coordinates of intersection satisfy x² + (6 − m)x + 9 = 0. [2] (c) Find the values of m for which L is a tangent to C. [3] (d) For the larger value of m, find the coordinates of the point of contact. [2]

Answers

1. (a) ³√343 = 7, so 343^(2/3) = 49 [1] (b) Flip: (4/25)^(3/2) [1] = (2/5)³ = 8/125 [1] Examiner insight: Show the root and the power as separate steps. A negative index is easy to mishandle in a single line, and written steps let a slip be traced.

2. (a) 81^(−1/4) = 1/3 [1]; (a¹²)^(−1/4) = a⁻³ [1]; (b⁻⁸)^(−1/4) = b² [1]. Answer: b²/(3a³) (b) x^(3/2) = 27 [1]; x = 27^(2/3) = 9 [1] Examiner insight: “Positive indices” is an instruction, so a result containing a⁻³ would not fully answer the question. Deal with the numerical coefficient and each letter separately.

3. (a) √12 = 2√3 [1]; 15/√3 = 5√3 [1]; √27 = 3√3, so 2√3 + 5√3 − 3√3 = 4√3 [1] (b) Multiply by (3 − √5): numerator (6 − √5)(3 − √5) = 18 − 6√5 − 3√5 + 5 = 23 − 9√5 [2]; denominator 9 − 5 = 4, so (23 − 9√5)/4 [1] Examiner insight: On a “show that” question, state each simplification. A bare end result gives the reader nothing to follow, because the working is the answer.

4. (a) Area = (3 + √2)(3 − √2) [1] = 9 − 2 = 7 cm² [1] (b) d² = (3 + √2)² + (3 − √2)² [1] = (11 + 6√2) + (11 − 6√2) = 22 [1]; d = √22 cm [1] Examiner insight: Expand each bracket fully before adding. The √2 terms cancel, and this is easiest to see when both expansions are written out.

5. y = 7 − 2x [1]; x² + 2x(7 − 2x) = 8 [1]; 3x² − 14x + 8 = 0 [1]; (3x − 2)(x − 4) = 0 [1]; x = 2/3, y = 17/3 [1]; x = 4, y = −1 [1]. Answer: (2/3, 17/3) and (4, −1) Examiner insight: Rearranging the linear equation first usually keeps the algebra shorter. Finish by substituting back into the linear equation so each x is paired with the correct y.

6. (a) x² − x + 6 = 3x + k [1]; rearrange to x² − 4x + (6 − k) = 0 [1] (b) Discriminant 16 − 4(6 − k) = 4k − 8 [1]; set equal to 0 [1]; k = 2 [1] (c) x² − 4x + 4 = 0, so x = 2 [1]; y = 6 + 2 = 8, point (2, 8) [1] Examiner insight: A tangent means one repeated root, so the discriminant is set to zero. Writing that reason explicitly makes your method clear.

7. (a) −2(x² − 4x) − 3 [1]; = −2[(x − 2)² − 4] − 3 [1]; = −2(x − 2)² + 5 [1] (b) Maximum point (2, 5) [1]; line of symmetry x = 2 [1] (c) (x − 2)² = 5/2 [1]; x = 2 ± √10/2 [1] Examiner insight: When a is negative the turning point is a maximum. Take care with the sign when you divide by −2 as you rearrange.

8. Discriminant k² − 144 [1]; real roots need k² − 144 ≥ 0 [1]; k ≤ −12 or k ≥ 12 [1]; set notation {k : k ≤ −12} ∪ {k : k ≥ 12}, interval notation (−∞, −12] ∪ [12, ∞) [1] Examiner insight: “Real roots” includes the repeated case, so the inequality is not strict. Check the end values and brackets in both notations.

9. (a) Let u = x²: 3u² − 13u + 4 = 0 [1]; (3u − 1)(u − 4) = 0 [1]; u = 1/3 or u = 4 [1]; x = ±√3/3 or x = ±2 [1] (b) Let u = 1/(x + 2): 3u² + 2u − 1 = 0 [1]; (3u − 1)(u + 1) = 0, so u = 1/3 or −1 [1]; x + 2 = 3 or −1, so x = 1 or x = −3 [1] Examiner insight: Each value of u gives its own pair of values of x. Forgetting the negative root in part (a) is an easy way to lose solutions.

10. (a) Multiply by 6: 2(2x − 1) − (x + 2) < 6 [1]; 3x − 4 < 6 [1]; x < 10/3 [1] (b) (2x − 5)(x + 3) ≥ 0 [1]; x ≤ −3 or x ≥ 5/2 [1] (c) Both: x ≤ −3 or 5/2 ≤ x < 10/3 [1]; {x : x ≤ −3} ∪ {x : 5/2 ≤ x < 10/3} [1] Examiner insight: Multiplying every term, including the right-hand side, by the common denominator avoids fraction errors. For part (c), a number line makes the overlap easy to see.

11. (a) x² + 2x − 3 = x + 3 [1]; x² + x − 6 = 0 so x = −3 or 2 [1]; (−3, 0) and (2, 5) [1] (b) Solid, because both inequalities are non-strict (≥ and ≤) [1] (c) y = (x + 1)² − 4 [1]; turning point (−1, −4) [1] (d) Parabola at x = 1: 1 + 2 − 3 = 0, and 2 ≥ 0 [1]; line at x = 1: 4, and 2 ≤ 4, so (1, 2) is in R [1] Examiner insight: To justify membership, test the point in every inequality and say so. A statement that “it looks inside” is weaker than showing both conditions hold.

12. (a) (x + 3)² [1] + 5 [1]; minimum point (−3, 5) [1] (b) x² + 6x + 14 = mx + 5 [1]; rearrange: x² + (6 − m)x + 9 = 0 [1] (c) (6 − m)² − 36 = 0 [1]; 6 − m = ±6 [1]; m = 0 or m = 12 [1] (d) m = 12: x² − 6x + 9 = 0, so x = 3 [1]; y = 12(3) + 5 = 41, point (3, 41) [1] Examiner insight: Substitute into the line to find y, then confirm it against the curve. Agreement on both is a quick check that the contact point is right.

Where marks are usually lost

• Treating a negative index as a negative number, for example 8^(−2/3) = −4.
• Leaving out the cross term when squaring a bracket that contains a surd.
• Multiplying by the original denominator instead of its conjugate.
• Finding x but not y in simultaneous equations.
• Substituting into the quadratic instead of the linear equation and creating false pairs.
• Using b² − 4ac > 0 when the question says “real roots”, which includes equal roots.
• Losing the negative root when solving u = x².
• Forgetting that dividing by a negative reverses an inequality.
• Writing a solution to an “or” inequality as one chained inequality.
• Using solid and dashed boundaries the wrong way round.

Next steps

• Review the revision notes for any method you missed.
• Reread the study guide for the worked examples.
• Return to the course hub and tick off the checklist.
• Take one of our free 10-minute diagnostics to find the next weak area.
• Book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Content references 1.02a to 1.02i.