AQA A-level Mathematics: Linear, Quadratic and Rational Inequalities and Regions (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 B5 of the AQA A-level Mathematics 7357 specification (version 1.3, 31 January 2018): linear and quadratic inequalities, inequalities with brackets and fractions, answers in “and”/“or” or set notation, and regions. Work through the study guide first if any topic is new. Do every question by hand and show all working.

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Questions

1. Solve 7x − 2(x + 4) ≥ 3(2x − 5). [3]

2. Solve −7 ≤ 5 − 4x < 13. [3]

3. Solve 3x² − 7x − 6 ≤ 0. [4]

4. Solve x² − 8x + 3 > 0. Give the answer in exact form and in set notation. [4]

5. Find the set of values of k for which x² + kx + 9 > 0 for every real value of x. [4]

6. Solve (2x − 1)(x + 2) > 4 − x². [4]

7. Solve 7/(2x − 1) ≥ 1. [5]

8. Solve x + 4 > 12/x. [6]

9. The curve C has equation y = x² − 4x − 5 and the line L has equation y = x − 5.

(a) Find the coordinates of the points where C meets L. [3] (b) Find the values of x for which C is below L, using set notation. [2] (c) Hence write down the solution of x² − 5x ≥ 0. [1]

10. The region R is defined by y > x² − 6x + 5 and y ≤ 5.

(a) Show that the line y = 5 meets the curve y = x² − 6x + 5 where x = 0 and x = 6. [2] (b) State the x-intercepts, the y-intercept and the turning point of the curve. [3] (c) State which boundary is dashed and which is solid, with a reason. [1] (d) Determine whether (3, 0) and (6, 5) lie in R. [2]

11. The triangle T has vertices (0, 0), (0, 6) and (4, 4). Its sides on the y-axis and on the line x + 2y = 12 are included in T, and its side on the line y = x is not.

(a) Show that (0, 6) and (4, 4) both lie on x + 2y = 12. [1] (b) Write three inequalities that define T. [3] (c) Find the area of the triangle. [2] (d) State, with a reason, whether (4, 4) is in T. [1]

12. A rectangular garden has width x metres and length (x + 6) metres. Its area must be at least 40 m² and its perimeter at most 40 m.

(a) Show that the area condition gives x² + 6x − 40 ≥ 0 and solve it for x > 0. [4] (b) Find the inequality given by the perimeter condition. [2] (c) Write the set of possible values of x in set notation. [1]

Answers

1. Expand: 7x − 2x − 8 ≥ 6x − 15, so 5x − 8 ≥ 6x − 15 [1]. Collect: −x ≥ −7 [1]. Divide by −1 and reverse the sign: x ≤ 7 [1]. [3] Examiner insight: Expanding the negative sign in −2(x + 4) correctly is where most slips happen. Substituting x = 0 (−8 ≥ −15, true) is a quick check that the direction is right.

2. Subtract 5 from each part: −12 ≤ −4x < 8 [1]. Divide each part by −4, reversing both signs: 3 ≥ x > −2 [1]. Written in order: −2 < x ≤ 3 [1]. [3] Examiner insight: The reversal has to be applied to both signs of the double inequality. Rewriting the answer with the smaller number on the left makes the strict and non-strict ends easy to check.

3. 3x² − 7x − 6 = (3x + 2)(x − 3) [1]. The roots are x = −2/3 and x = 3 [1]. The curve is a ∪-shape, so it is below or on the axis between the roots [1]. −2/3 ≤ x ≤ 3 [1]. [4] Examiner insight: A quick sketch shows which part of the axis is required. Answers that list the roots without a conclusion about the interval leave the key decision unsupported.

4. x = (8 ± √(64 − 12))/2 [1] = (8 ± √52)/2 = 4 ± √13 [1]. The curve is a ∪-shape, so it is above the axis outside the roots [1]. {x: x < 4 − √13} ∪ {x: x > 4 + √13} [1]. [4] Examiner insight: “Exact form” asks you to keep surds, so decimals such as 0.39 and 7.61 would not answer the question as set. Simplify √52 = 2√13 before halving.

5. For the curve to stay above the x-axis it must have no real roots, so the discriminant is negative [1]. k² − 4(1)(9) < 0, so k² − 36 < 0 [1]. (k − 6)(k + 6) < 0 [1]. −6 < k < 6 [1]. [4] Examiner insight: Explaining why the discriminant condition is used earns more than quoting it. Keep k as the variable in the final inequality, and remember that a repeated root (k = ±6) gives a curve touching the axis, so the ends are excluded.

6. Expand the left side: 2x² + 3x − 2 > 4 − x² [1]. Collect: 3x² + 3x − 6 > 0, so x² + x − 2 > 0 [1]. (x + 2)(x − 1) > 0 [1]. x < −2 or x > 1 [1]. [4] Examiner insight: Move everything to one side before factorising. Reading factors from (2x − 1)(x + 2) > 4 − x² directly has no valid basis for an inequality.

7. Note x ≠ 1/2. Multiply both sides by (2x − 1)², which is positive [1]: 7(2x − 1) ≥ (2x − 1)² [1]. Collect: (2x − 1)(7 − (2x − 1)) ≥ 0, so (2x − 1)(8 − 2x) ≥ 0 [1]. This is a ∩-shape with roots 1/2 and 4, so it is non-negative between them [1]. Exclude 1/2: 1/2 < x ≤ 4 [1]. [5] Examiner insight: Multiplying by the square of the denominator keeps the sign safe. Check one value from each region: x = 1 gives 7 ≥ 1, true, while x = 0 gives −7 ≥ 1, false.

8. Note x ≠ 0. Multiply by x², which is positive: x²(x + 4) > 12x [1]. Collect: x³ + 4x² − 12x > 0, so x(x² + 4x − 12) > 0 [1]. x(x + 6)(x − 2) > 0 [1]. The cubic has a positive x³ term and roots −6, 0 and 2, so it starts below the axis and is positive between −6 and 0 and to the right of 2 [1]. Hence −6 < x < 0 or x > 2 [1]. A check at x = 3 (7 > 4) and x = 1 (5 > 12, false) supports this [1]. [6] Examiner insight: Cancelling x from both sides would lose the solutions where x is negative. Write the factorised cubic and the sign pattern explicitly, because the sketch or sign table is the evidence for the intervals.

9. (a) x² − 4x − 5 = x − 5 [1], so x² − 5x = 0 and x(x − 5) = 0, giving x = 0 or x = 5 [1]. Points (0, −5) and (5, 0) [1]. (b) C is a ∪-shape that crosses L at x = 0 and x = 5, so it is below L between these points [1]: {x: 0 < x < 5} [1]. (c) x² − 5x ≥ 0 means C is on or above L: x ≤ 0 or x ≥ 5 [1]. [6] Examiner insight: Part (c) is the same working as part (b) with the inequality reversed, so use the earlier result instead of starting again. The ends are now included because of the “≥”.

10. (a) Set x² − 6x + 5 = 5 [1], so x² − 6x = 0, giving x = 0 and x = 6 [1]. (b) x² − 6x + 5 = (x − 1)(x − 5), so x-intercepts at 1 and 5 [1]. y-intercept 5 [1]. Completing the square gives (x − 3)² − 4, so the turning point is (3, −4) [1]. (c) The curve is dashed because y > is strict; the line y = 5 is solid because y ≤ includes equality [1]. (d) (3, 0): 0 > −4 is true and 0 ≤ 5 is true, so (3, 0) is in R [1]. (6, 5): the curve gives 5 at x = 6, and 5 > 5 is false, so (6, 5) is not in R [1]. [8] Examiner insight: A point on a dashed curve is not in the region, even when it satisfies the other inequality. Mark the intercepts and turning point when sketching, because they let a reader check the boundary at a glance.

11. (a) 0 + 2(6) = 12 and 4 + 2(4) = 12 [1]. (b) The y-axis side gives x ≥ 0 [1]. The line y = x with the point (1, 3) inside: 3 > 1, so y > x [1]. The line x + 2y = 12 with (1, 3): 1 + 6 = 7 ≤ 12, so x + 2y ≤ 12 [1]. (c) Take the side on the y-axis as the base: length 6 [1]. The height is the distance from (4, 4) to the y-axis, 4, so area = ½ × 6 × 4 = 12 square units [1]. (d) No: at (4, 4), y > x gives 4 > 4, which is false, because the point is on the dashed side [1]. [7] Examiner insight: Test a point strictly inside the shape to decide each sign, not a vertex. Writing the three inequalities with “and” between them, not as one chain, avoids contradictory-looking statements.

12. (a) Area = x(x + 6) ≥ 40, so x² + 6x − 40 ≥ 0 [1]. Factorise: (x + 10)(x − 4) ≥ 0 [1]. The ∪-shape gives x ≤ −10 or x ≥ 4 [1]. A width must be positive, so x ≥ 4 [1]. (b) Perimeter = 2(x + x + 6) = 4x + 12 [1]. 4x + 12 ≤ 40, so x ≤ 7 [1]. (c) Both conditions must hold, so {x: 4 ≤ x ≤ 7} [1]. [7] Examiner insight: Say why the branch x ≤ −10 is rejected, in the context of a width. Joining the two conditions with “and” gives one interval; using “or” here would wrongly allow every positive width.

Where marks are usually lost

• Failing to reverse the inequality when dividing by a negative number, as in questions 1 and 2.
• Writing the “between the roots” answer for a ∪-shaped curve that must be above the axis, as in questions 4 and 6.
• Leaving an “or” answer as one chain of inequalities.
• Dividing by x or multiplying by (2x − 1) in a fraction inequality instead of using the square or a single fraction.
• Forgetting to exclude the value that makes the denominator zero, as in question 7.
• Giving decimal roots when the question asks for exact form.
• Using a solid boundary for a strict inequality, or the reverse, and counting points on a dashed boundary as inside the region.
• Choosing the region from a point on the boundary, or not testing at all.
• Dropping the context in a modelling question, so a negative width stays in the answer.

Next steps

• Re-read the revision notes for the method boxes.
• Revisit the study guide for any question you could not start.
• See the course hub and the checklist for the rest of Section B.
• Take the free 10-minute diagnostics to find weak spots.
• Want one-to-one help? Book a free trial class.

Official syllabus

AQA, A-level Mathematics 7357 specification, for first teaching in 2017 (version 1.3, 31 January 2018), content B5. Published by AQA.