Pearson Edexcel International Advanced Level Mathematics: Inequalities, polynomials and graphs (YMA01) – 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 cover Unit P1: Pure Mathematics 1, topics 1.6 to 1.12 of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. The specification makes P1 and P2 compulsory for IAS Mathematics, IAL Mathematics and IAL Pure Mathematics. Read the study guide first, or use the revision notes for a quick refresher.

Course pages: the Edexcel A Level Mathematics hub, the printable checklist and the free 10-minute diagnostics. For sketches, label intercepts, turning points given in the question and asymptote equations.

Questions

1. Solve the simultaneous equations y = 3 − x and x² + 2y² = 33. [5]

2. The curve C has equation y = x² + 2x − 8 and the line L has equation y = 3x − 2.

(a) Find the coordinates of the points where C and L meet. [4] (b) Hence find the set of values of x for which C is below L. [2]

3. Solve each inequality.

(a) 7(x − 2) < 3x + 6 [2] (b) 6x² − 7x − 20 ≥ 0 [3] (c) 9/x ≥ 4 [4]

4. Find the set of values of x for which both 3x − 2 > 7 and x² − 2x − 8 < 0. [4]

5. The region R satisfies y ≤ 8 − x² and y > 2x.

(a) Work out where the two boundaries intersect. [3] (b) Describe how you would draw R, stating which boundary is solid, which is dotted, and which side of each boundary is shaded. [3]

6. (a) Factorise fully 3x³ − 12x. [2] (b) Factorise fully x³ + 3x² − 16x − 48. [3] (c) Hence solve x³ + 3x² − 16x − 48 = 0. [1]

7. The function f is given by f(x) = 2x² − 3x + 1.

(a) Find f(−2). [1] (b) Expand and simplify f(x + 1). [2] (c) Solve f(x + 1) = 0. [1]

8. (a) Sketch the curve y = (2 − x)(x + 1)(x + 4), showing where it crosses the axes. [4] (b) Use your sketch to write down the values of x for which y > 0. [1]

9. (a) Sketch y = 12/x², stating the equations of its asymptotes. [2] (b) On the same axes sketch y = x + 1, and explain why the equation x³ + x² − 12 = 0 has exactly one real solution. [2] (c) Show that x = 2 is a solution of x³ + x² − 12 = 0, and show algebraically that there are no others. [3]

10. The curve y = f(x) has a maximum at (−2, 5), a minimum at (4, −3) and passes through the origin. Write down the coordinates of the maximum, the minimum and the point corresponding to the origin on each curve.

(a) y = f(x) + 4 [2] (b) y = f(x + 3) [2] (c) y = f(x/2) [2] (d) y = −f(x) (maximum and minimum only) [1]

11. (a) Sketch y = 1/(x − 2) + 3, giving the equations of the asymptotes and the coordinates where the curve meets the axes. [4] (b) Sketch y = 3 sin 2x for 0° ≤ x ≤ 360°, marking the maximum and minimum points. [3] (c) Sketch y = tan(x + 30°) for 0° ≤ x ≤ 360°, giving the asymptotes, the x-intercepts and the y-intercept. [3]

12. A rectangle has perimeter 46 cm and diagonal 17 cm. Find its length and width. [5]

Answers

1. The linear equation already gives y = 3 − x [1]. Substitute: x² + 2(3 − x)² = 33, so 3x² − 12x + 18 = 33 [1]. Then x² − 4x − 5 = 0 [1], so (x − 5)(x + 1) = 0 and x = 5 or −1 [1]. Then y = −2 or y = 4: (5, −2) and (−1, 4) [1]. [5] Examiner insight: Giving the y-values with the correct x-value is part of the answer; a list of x-values alone is usually incomplete.

2. (a) Equate: x² + 2x − 8 = 3x − 2, so x² − x − 6 = 0 [1]. Factorise: (x − 3)(x + 2) = 0 [1]. So x = 3 or −2 [1]. Points: (3, 7) and (−2, −8) [1]. [4] (b) C is below L between the intersection points [1]: −2 < x < 3 [1]. [2] Examiner insight: Part (b) is easiest to defend with a sketch; state in words that the upward curve lies below the line between the two meeting points.

3. (a) 7x − 14 < 3x + 6 [1]. Then 4x < 20, so x < 5 [1]. [2] (b) (2x − 5)(3x + 4) ≥ 0 [1], with roots x = 5/2 and x = −4/3 [1]. The curve is above or on the axis outside the roots: x ≤ −4/3 or x ≥ 5/2 [1]. [3] (c) Multiply by x² (positive, x ≠ 0): 9x ≥ 4x² [1]. Then 4x² − 9x ≤ 0, so x(4x − 9) ≤ 0 [1]. This gives 0 ≤ x ≤ 9/4 [1]. As x ≠ 0, 0 < x ≤ 9/4 [1]. [4] Examiner insight: In a fraction inequality, multiplying by x alone without considering its sign is a common route to a wrong answer; excluding x = 0 explicitly is easy to forget.

4. 3x > 9, so x > 3 [1]. For the quadratic: (x − 4)(x + 2) < 0 [1], so −2 < x < 4 [1]. Both hold for 3 < x < 4 [1]. [4] Examiner insight: A number line showing both ranges makes the overlap visible and reduces the chance of quoting only one inequality.

5. (a) 8 − x² = 2x, so x² + 2x − 8 = 0 [1]. (x + 4)(x − 2) = 0, so x = −4 or 2 [1]. Points: (−4, −8) and (2, 4) [1]. [3] (b) y ≤ 8 − x²: the parabola is solid and the shaded side is below it [1]. y > 2x: the line is dotted and the shaded side is above it [1]. R lies between the line and the parabola, between the two meeting points, and the meeting points are not in R because they lie on the dotted line [1]. [3] Examiner insight: Naming the test point you used, such as (0, 1), shows why you chose a side and is a good habit when the question allows working to be shown.

6. (a) Take out 3x: 3x(x² − 4) [1]. Then 3x(x − 2)(x + 2) [1]. [2] (b) Group: x²(x + 3) − 16(x + 3) [1]. Take out the bracket: (x + 3)(x² − 16) [1]. So (x + 3)(x − 4)(x + 4) [1]. [3] (c) x = −4, −3 or 4 [1]. [1] Examiner insight: The question asks for a fully factorised answer, so stopping at (x + 3)(x² − 16) leaves the final step undone.

7. (a) f(−2) = 2(4) + 6 + 1 = 15 [1]. [1] (b) f(x + 1) = 2(x + 1)² − 3(x + 1) + 1 [1] = 2x² + 4x + 2 − 3x − 3 + 1 = 2x² + x [1]. [2] (c) x(2x + 1) = 0, so x = 0 or x = −1/2 [1]. [1] Examiner insight: Replace the whole of x by (x + 1) in every term, including the brackets; a slip such as squaring only the x is a frequent source of lost credit.

8. (a) Roots at x = −4, −1 and 2 [1]. The y-intercept is (2)(1)(4) = 8, so (0, 8) [1]. The coefficient of x³ is −1, so the curve starts high on the left and ends low on the right [1]. Crossing at all three roots, with intercepts labelled, gives a correct sketch [1]. [4] (b) From the sketch, y > 0 where the curve is above the axis: x < −4 or −1 < x < 2 [1]. [1] Examiner insight: A cubic with a negative leading coefficient needs the end behaviour shown clearly; a sketch that rises to the right will not match this function.

9. (a) The graph has two branches above the x-axis, symmetric about the y-axis [1]. Asymptotes x = 0 and y = 0 [1]. [2] (b) The line y = x + 1 crosses the right-hand branch once [1]. On the left, for x < 0 the curve is at least 12 when −1 ≤ x < 0, while the line is between 0 and 1, and for x < −1 the line is negative but the curve is positive, so there is no crossing there. Hence one solution [1]. [2] (c) Substitute x = 2: 8 + 4 − 12 = 0 [1]. Factorise: (x − 2)(x² + 3x + 6) = 0 [1]. For the quadratic, b² − 4ac = 9 − 24 = −15, which is negative, so there are no other real solutions [1]. [3] Examiner insight: Questions asking you to show something need each step written out; stating only that the quadratic has no real roots, without the working, may not be enough.

10. (a) Maximum (−2, 9), minimum (4, 1), origin to (0, 4) [2]. [2] (b) Maximum (−5, 5), minimum (1, −3), origin to (−3, 0) [2]. [2] (c) Maximum (−4, 5), minimum (8, −3), origin stays at (0, 0) [2]. [2] (d) Maximum becomes (−2, −5) and minimum (4, 3); the former maximum becomes a minimum [1]. [1] Examiner insight: Give coordinates for every labelled point; naming which transformation each equation represents can help you check direction before you write numbers.

11. (a) The graph is y = 1/x translated 2 right and 3 up [1]. Asymptotes x = 2 and y = 3 [1]. At x = 0, y = −1/2 + 3 = 5/2, so the y-intercept is (0, 5/2) [1]. For y = 0, 1/(x − 2) = −3, so x = 5/3: (5/3, 0) [1]. [4] (b) Period 180° [1]. Maximum points (45°, 3) and (225°, 3), minimum points (135°, −3) and (315°, −3) [1]. The curve passes through the x-axis at 0°, 90°, 180°, 270° and 360° [1]. [3] (c) Shift of tan x by 30° left [1]. Asymptotes x = 60° and x = 240°, x-intercepts at 150° and 330° [1], y-intercept (0, tan 30°) = (0, 1/√3) [1]. [3] Examiner insight: For a trigonometric sketch, state the interval and label the key values; a curve with the right shape but no coordinates is unlikely to earn full credit.

12. Let the sides be x and y. Perimeter: x + y = 23 [1]. Diagonal: x² + y² = 289 [1]. Substitute y = 23 − x: x² + (23 − x)² = 289, so 2x² − 46x + 240 = 0 [1]. Then x² − 23x + 120 = 0, so (x − 8)(x − 15) = 0 [1]. The sides are 8 cm and 15 cm [1]. [5] Examiner insight: Define your letters at the start and give the answer with units; the second solution is the same rectangle labelled the other way.

Where marks are usually lost

• Stating x-values without the matching y-values in simultaneous equations.
• Forgetting to reverse the sign when dividing an inequality by a negative.
• Multiplying a fraction inequality by x, not x², and missing x ≠ 0.
• Writing two outer ranges as a single inequality.
• Using a solid line for a strict inequality or a dotted line for ≤ and ≥.
• Stopping a factorisation before it is complete.
• Moving the graph of f(x + a) in the wrong direction, or doubling x-coordinates for f(x/2).
• Omitting intercepts, asymptote equations or turning points from sketches.
• Treating a double root as a crossing point.

Next steps

• Review the revision notes and the study guide.
• Go to the Edexcel A Level Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics to find weak topics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3, April 2019, published by Pearson Education Limited. Unit P1, topics 1.6 to 1.12.