Edexcel International GCSE Mathematics A 4MA1: Sequences, functions and graphs – 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 Topic 3, Sequences, functions and graphs (sections 3.1–3.4), of the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Questions 1–3 are on content for both tiers, except 2(b) and 2(c), which are Higher tier only. Questions 4–12 are Higher tier only. A calculator may be used on every 4MA1 paper, but show your working: most marks are for method.

Learn the content first in the study guide and the revision notes. The course hub is Edexcel IGCSE Mathematics and the printable checklist lists every statement.

Questions

1. (Both tiers) Here are the first four terms of a sequence: 3, 10, 17, 24, …

(a) Write down the next two terms. [1] (b) Find an expression for the nth term. [2] (c) Show whether 150 is a term of the sequence. [2]

2. (Both tiers; (b) and (c) Higher tier only) A is the point (−4, 3) and B is the point (6, −2).

(a) Find the coordinates of the midpoint of AB. [2] (b) Calculate the gradient of AB. [2] (c) Find the equation of the line through A and B in the form y = mx + c. [2]

3. (Both tiers) Amira leaves home at 09:00 and cycles 12 km to a lake at a constant speed, arriving at 09:40. She stays at the lake until 10:00, then cycles straight home at a constant speed, arriving at 11:00.

(a) Calculate her speed, in km/h, from home to the lake. [2] (b) Calculate her speed, in km/h, on the journey home. [2] (c) Describe the part of her distance–time graph from 09:40 to 10:00. [1]

4. (Higher tier only) The 4th term of an arithmetic sequence is 19 and the 9th term is 44.

(a) Find the first term a and the common difference d. [3] (b) Find an expression for the nth term. [1] (c) Calculate the sum of the first 30 terms. [2]

5. (Higher tier only) f(x) = 2x − 3, g(x) = x² + 4 and h(x) = 5/(x + 1).

(a) Find fg(3). [2] (b) Find gf(x), simplifying your answer. [2] (c) Find f⁻¹(x). [2] (d) State the value of x that must be excluded from the domain of h. [1]

6. (Higher tier only) f(x) = (2x + 1)/(x − 3), x ≠ 3. Find f⁻¹(x). [4]

7. (Higher tier only)

(a) The line y = 3x − 1 meets the curve y = x² + x − 4 at the points A and B. Find the coordinates of A and B. [5] (b) The graphs of y = x³ − 2x² and y = 4 − x are drawn on the same axes. Write down the equation, in the form x³ + px² + qx + r = 0, whose solutions are the x-coordinates of their points of intersection. [1]

8. (Higher tier only) f(x) = x² − 4x.

(a) Find the coordinates of the turning point of y = f(x). [2] (b) Write down the coordinates of the turning point of y = f(x + 3). [1] (c) Write down the coordinates of the turning point of y = f(x) − 5. [1] (d) Use the symmetry of the graph of y = cos x to solve cos x = −0.5 for 0° ≤ x ≤ 360°. [2]

9. (Higher tier only) The line L₁ has equation 2x + 5y = 10.

(a) Find the gradient of L₁. [1] (b) The line L₂ is perpendicular to L₁ and passes through (4, 3). Find the equation of L₂. [3] (c) L₂ crosses the x-axis at P. Find the coordinates of P. [1]

10. (Higher tier only) A curve has equation y = x³ − 6x² + 9x + 2.

(a) Find dy/dx. [2] (b) Find the coordinates of the two turning points and state which is the maximum. [5] (c) Find the gradient of the curve at the point where x = 4. [1]

11. (Higher tier only) A particle moves in a straight line. Its displacement, s metres, from a fixed point O after t seconds is s = 2t³ − 21t² + 60t, for 0 ≤ t ≤ 6.

(a) Find an expression for the velocity v in terms of t. [2] (b) Find the times when the particle is at rest. [3] (c) Find the acceleration when t = 5. [2] (d) Find the displacement from O when the particle is first at rest. [1]

12. (Higher tier only) An open box (no lid) has a square base of side x cm and height h cm. Its volume is 4000 cm³.

(a) Show that the total outside surface area, S cm², is S = x² + 16000/x. [3] (b) Find the value of x for which S is a minimum, and the minimum value of S. [5] (c) Explain, by considering the shape of the graph of S against x for x > 0, why this value is a minimum. [1]

Answers

1. (a) 31, 38 [1] (b) Difference is 7, so the expression starts 7n [1]; 7n − 4 [1] → 7n − 4 (c) 7n − 4 = 150 [1]; 7n = 154, n = 22, a whole number, so 150 is the 22nd term [1] Examiner insight: In (c) a bare “yes” scores nothing; the mark needs the equation and the whole-number value of n.

2. (a) ((−4 + 6)/2, (3 + (−2))/2) [1] = (1, 0.5) [1] (b) (−2 − 3)/(6 − (−4)) [1] = −5/10 = −1/2 [1] (c) 3 = (−1/2)(−4) + c, so c = 1 [1]; y = −(1/2)x + 1 [1] Examiner insight: The method mark in (b) needs the change in y over the change in x; the inverted fraction −2 earns no marks, and a wrong gradient can still gain follow-through in (c).

3. (a) 40 minutes = 40/60 = 2/3 hour [1]; 12 ÷ (2/3) = 18 km/h [1] (b) 12 km in 1 hour (10:00 to 11:00) [1] = 12 km/h [1] (c) A horizontal line at a distance of 12 km from home [1] Examiner insight: Dividing 12 by 40 (minutes) and giving 0.3 km/h scores 0 for accuracy; the time must be converted to hours before dividing.

4. (a) a + 3d = 19 and a + 8d = 44 [1]; subtracting, 5d = 25, d = 5 [1]; a = 4 [1] (b) 4 + 5(n − 1) = 5n − 1 [1] (c) S₃₀ = (30/2)[2(4) + 29(5)] [1] = 15 × 153 = 2295 [1] Examiner insight: Writing the 4th term as a + 4d loses the method mark; each wrong starting equation loses the method mark and everything built on it.

5. (a) g(3) = 13 [1]; f(13) = 23 [1] (b) (2x − 3)² + 4 [1] = 4x² − 12x + 13 [1] (c) y = 2x − 3, so x = (y + 3)/2 [1]; f⁻¹(x) = (x + 3)/2 [1] (d) x = −1 [1] Examiner insight: In (b) the unsimplified (2x − 3)² + 4 earns the method mark only; the question says “simplifying”, so the expanded form is needed for the accuracy mark.

6. y(x − 3) = 2x + 1 [1]; xy − 2x = 3y + 1 [1]; x(y − 2) = 3y + 1 [1]; f⁻¹(x) = (3x + 1)/(x − 2) [1] Examiner insight: The key method mark is for collecting the x terms on one side and factorising; an answer left as x = (3y + 1)/(y − 2) loses the final mark.

7. (a) x² + x − 4 = 3x − 1 [1]; x² − 2x − 3 = 0 [1]; (x − 3)(x + 1) = 0, x = 3 or x = −1 [1]; y = 3(3) − 1 = 8 and y = 3(−1) − 1 = −4 [1]; (3, 8) and (−1, −4) [1] (b) x³ − 2x² − (4 − x) = 0, so x³ − 2x² + x − 4 = 0 [1] Examiner insight: Stopping at x = 3 and x = −1 loses the last two marks, because the question asks for coordinates of points.

8. (a) x² − 4x = (x − 2)² − 4 [1], so the turning point is (2, −4) [1] (b) (−1, −4) [1] (c) (2, −9) [1] (d) x = cos⁻¹(−0.5) = 120° [1]; the graph is symmetrical about x = 180°, so the other solution is 360° − 120° = 240°; x = 120°, 240° [1] Examiner insight: In (b), (5, −4) scores 0; f(x + 3) moves the graph 3 units in the negative x-direction.

9. (a) y = −(2/5)x + 2, so gradient = −2/5 [1] (b) Perpendicular gradient = 5/2 [1]; 3 = (5/2)(4) + c, so c = −7 [1]; y = (5/2)x − 7 [1] (c) 0 = (5/2)x − 7, x = 2.8; P = (2.8, 0) [1] Examiner insight: Reading the gradient of 2x + 5y = 10 as 2 scores 0; rearrange to y = mx + c before quoting m, or every later mark depends on follow-through.

10. (a) 3x² − 12x [1] + 9 [1] → dy/dx = 3x² − 12x + 9 (b) 3x² − 12x + 9 = 0 [1]; 3(x − 1)(x − 3) = 0, x = 1 or x = 3 [1]; x = 1 gives y = 1 − 6 + 9 + 2 = 6 [1]; x = 3 gives y = 27 − 54 + 27 + 2 = 2 [1]; positive x³ term, so (1, 6) is the maximum and (3, 2) is the minimum [1] (c) 3(16) − 12(4) + 9 = 9 [1] Examiner insight: The specification only asks you to use the shape of the graph to classify; a clear reason (positive cubic, maximum comes first) is enough for the final mark.

11. (a) v = ds/dt [1] = 6t² − 42t + 60 [1] (b) 6t² − 42t + 60 = 0 [1]; t² − 7t + 10 = 0, (t − 2)(t − 5) = 0 [1]; t = 2 s and t = 5 s [1] (c) a = dv/dt = 12t − 42 [1]; a = 60 − 42 = 18 m/s² [1] (d) s = 2(8) − 21(4) + 60(2) = 52 m [1] Examiner insight: Units are part of the answer in kinematics; substituting t = 5 into v instead of a gives 0 and earns no marks in (c).

12. (a) x²h = 4000, so h = 4000/x² [1]; S = x² + 4xh (base plus four sides) [1]; S = x² + 4x(4000/x²) = x² + 16000/x [1] (b) dS/dx = 2x [1] − 16000/x² [1]; 2x − 16000/x² = 0 gives x³ = 8000 [1]; x = 20 [1]; S = 400 + 16000/20 = 1200 cm² [1] (c) S is very large when x is close to 0 and when x is large, and there is only one turning point for x > 0, so it must be a minimum [1] Examiner insight: On a “show that” every line must appear; jumping from S = x² + 4xh straight to the printed answer without substituting h loses the final mark.

Where marks are usually lost

• Leaving an nth term as “+7” (a term-to-term rule) when the question asks for an expression in n.
• Setting up a + nd instead of a + (n − 1)d for a given term.
• Working time in minutes when the speed is asked for in km/h.
• Composing functions in the wrong order: fg means apply g first.
• Stopping at x-values when the question asks for points or coordinates.
• Moving the graph the wrong way for f(x + a), or changing the y-coordinate as well.
• Taking the coefficient of x in ax + by = c as the gradient.
• Substituting a turning-point x into dy/dx instead of into y.
• Missing units (m/s, m/s², cm²) in practical calculus questions.
• Skipping the substitution line in a “show that” question.

Next steps

• Refresh the formulas: revision notes
• Re-learn anything you missed: study guide
• Course hub: Edexcel IGCSE Mathematics
• Track coverage: printable checklist
• Check other topics: all free 10-minute diagnostics
• Want help from a teacher? Book a free trial class

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 3, Sequences, functions and graphs, sections 3.1–3.4 (Foundation and Higher tier content).