Edexcel International GCSE Maths Functions Past Paper Questions and Solutions (4MA1 Higher)

By /
Edexcel International GCSE Maths A (4MA1) · Higher · Sequences, Functions and Graphs

Functions: Composite and Inverse: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every functions: composite and inverse question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 11 questions worth 53 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

14 marksHigherJune 2022 · Paper 1H (4MA1/1H) · Q14

The function f is defined as f : x ↦ 2x⁄x − 6, x ≠ 6

(a) Find f(10) (1)

(b) Express the inverse function f−1 in the form f−1 : x ↦ … (3)

Show solution

Part (a)

  1. 2 × 10⁄10 − 6 = 20⁄4

Answer: 5

Part (b)

  1. Let y = 2x⁄x − 6 and make x the subject: y(x − 6) = 2x, so xy − 6y = 2x
  2. xy − 2x = 6y, so x(y − 2) = 6y and x = 6y⁄y − 2
  3. Swap y for x.

Answer: f−1 : x ↦ 6x⁄x − 2

24 marksHigherJune 2024 · Paper 1H (4MA1/1H) · Q15

The function f is defined as f : x ↦ 3x + 1⁄x − 2

(a) State the value of x that cannot be included in any domain of the function f. (1)

(b) Express the inverse function f−1 in the form f−1(x) = … (3)

Show solution

Part (a)

  1. You cannot divide by zero, so the denominator x − 2 cannot be 0.

Answer: x = 2

Part (b)

  1. Write y = 3x + 1⁄x − 2 and rearrange to make x the subject.
  2. Multiply both sides by (x − 2): y(x − 2) = 3x + 1, so xy − 2y = 3x + 1
  3. Get all the x terms on one side: xy − 3x = 2y + 1
  4. Factorise out x: x(y − 3) = 2y + 1, so x = 2y + 1⁄y − 3
  5. Finally, swap y for x to write the inverse function.

Answer: f−1(x) = 2x + 1⁄x − 3

39 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q16

The functions f and g are defined as f : x ↦ 5x − 7 and g : x ↦ 5x⁄x + 4

(a) Write down the value of x that must be excluded from any domain of g. (1)

(b) Find gf(2.6) (2)

(c) Solve fg(x) = 2 (3)

(d) Express the inverse function g−1 in the form g−1 : x ↦ … (3)

Show solution

Part (a)

  1. The denominator cannot be zero: x + 4 = 0 when x = −4.

Answer: x = −4

Part (b)

  1. f first: f(2.6) = 5 × 2.6 − 7 = 6
  2. Then g: g(6) = 30⁄10

Answer: 3

Part (c)

  1. fg(x) = f(5x⁄x + 4) = 5 × 5x⁄x + 4 − 7 = 25x⁄x + 4 − 7
  2. Set equal to 2: 25x⁄x + 4 = 9, so 25x = 9x + 36
  3. 16x = 36

Answer: x = 2.25

Part (d)

  1. Let y = 5x⁄x + 4. Then xy + 4y = 5x, so xy − 5x = −4y
  2. x(y − 5) = −4y, so x = −4y⁄y − 5 = 4y⁄5 − y
  3. Swap y for x.

Answer: g−1 : x ↦ 4x⁄5 − x

46 marksHigherJanuary 2023 · Paper 1HR (4MA1/1HR) · Q16

The function f is such that f(x) = 2⁄3x − 5 where x ≠ 5⁄3

(a) Find f(1⁄3) (1)

(b) Find f−1(x) (2)

The function g is such that g(x) = 5x2 − 20x + 23

(c) Express g(x) in the form a(x − b)2 + c (3)

Show solution

Part (a)

  1. f(1⁄3) = 2⁄3 × 1⁄3 − 5 = 2⁄1 − 5 = 2⁄−4

Answer: −½

Part (b)

  1. Write y = 2⁄3x − 5 and rearrange for x: y(3x − 5) = 2
  2. 3xy − 5y = 2, so 3xy = 2 + 5y and x = 2 + 5y⁄3y

Answer: f−1(x) = 5x + 2⁄3x

Part (c)

  1. Take out the factor 5 from the x terms: 5(x2 − 4x) + 23
  2. Complete the square: x2 − 4x = (x − 2)2 − 4
  3. 5[(x − 2)2 − 4] + 23 = 5(x − 2)2 − 20 + 23

Answer: 5(x − 2)2 + 3

54 marksHigherJune 2025 · Paper 1H (4MA1/1H) · Q16

f(x) = 9 − √x where x ≥ 0, and g(x) = 4x2

(a) Find f(9) (1)

(b) Solve fg(x) < 0. Show clear algebraic working. (3)

Show solution

Part (a)

  1. f(9) = 9 − √9 = 9 − 3

Answer: 6

Part (b)

  1. fg(x) means put g(x) into f: fg(x) = f(4x2) = 9 − √4x2
  2. Working with x ≥ 0 (as in the definition of f), √4x2 = 2x, so fg(x) = 9 − 2x
  3. Solve 9 − 2x < 0: add 2x to both sides to get 9 < 2x
  4. Divide by 2: 4.5 < x

Answer: x > 4.5

Tip: This is the mark scheme answer. If negative values of x were allowed, x < −4.5 would also make fg(x) negative, because 4x2 is the same for x and −x.
64 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q17

The function f is such that f(x) = (x − 4)2 for all values of x.

(a) Find f(1) (1)

(b) State the range of the function f. (1)

The function g is such that g(x) = 4⁄x + 3, x ≠ −3

(c) Work out fg(2) (2)

Show solution

Part (a)

  1. (1 − 4)2 = (−3)2

Answer: 9

Part (b)

  1. A square can never be negative, and it equals 0 when x = 4.

Answer: f(x) ≥ 0

Part (c)

  1. g(2) = 4⁄5 = 0.8
  2. f(0.8) = (0.8 − 4)2 = (−3.2)2

Answer: 10.24

74 marksHigherJune 2023 · Paper 1H (4MA1/1H) · Q17

The functions g and h are such that g(x) = 11⁄2x − 5 and h(x) = x2 + 4, x ≥ 0

(a) What value of x must be excluded from any domain of g? (1)

(b) Solve gh(x) = 1 (3)

Show solution

Part (a)

  1. The denominator cannot be zero: 2x − 5 = 0 when x = 2.5

Answer: x = 2.5

Part (b)

  1. gh(x) means apply h first, then g: gh(x) = g(x2 + 4) = 11⁄2(x2 + 4) − 5 = 11⁄2x2 + 3
  2. Set equal to 1: 11⁄2x2 + 3 = 1, so 2x2 + 3 = 11
  3. 2x2 = 8, so x2 = 4 and x = ±2
  4. h is only defined for x ≥ 0, so reject x = −2.

Answer: x = 2

84 marksHigherJanuary 2023 · Paper 1H (4MA1/1H) · Q18

The function f is such that f(x) = k⁄x where x ≠ 0 and k is an integer.

(a) Express the inverse function f−1 in the form f−1(x) = … (1)

The function g is such that g(x) = 2 − 3x4 where x ≠ 0. The function h is such that h(x) = 3x⁄2 − x where x ≠ 2.

(b)(i) Find g(−2) (1)

(ii) Express the composite function hg in the form hg(x) = … Give your answer in its simplest form. (2)

Show solution

Part (a)

  1. y = k⁄x rearranges to x = k⁄y, so f is its own inverse.

Answer: f−1(x) = k⁄x

Part (b)(i)

  1. (−2)4 = 16, so g(−2) = 2 − 48

Answer: −46

Part (b)(ii)

  1. hg(x) = h(2 − 3x4) = 3(2 − 3x4)⁄2 − (2 − 3x4) = 6 − 9x4⁄3x4
  2. Divide top and bottom by 3.

Answer: hg(x) = 2 − 3x4⁄x4

95 marksHigherJanuary 2020 · Paper 1H (4MA1/1H) · Q21

The functions f and g are such that f(x) = x2 − 2x and g(x) = x + 3. The function h is such that h(x) = fg(x) for x ≥ −2. Express the inverse function h−1(x) in the form h−1(x) = …

Show solution
  1. h(x) = f(x + 3) = (x + 3)2 − 2(x + 3) = x2 + 6x + 9 − 2x − 6 = x2 + 4x + 3
  2. Complete the square: x2 + 4x + 3 = (x + 2)2 − 1
  3. Let y = (x + 2)2 − 1. Then (x + 2)2 = y + 1, so x + 2 = √y + 1 (positive root because x ≥ −2)
  4. x = √y + 1 − 2. Swap y for x.

Answer: h−1(x) = √x + 1 − 2

103 marksHigherJanuary 2021 · Paper 1H (4MA1/1H) · Q22

The function f is such that f(x) = x2 − 8x + 5 where x ≤ 4. Express the inverse function f−1 in the form f−1(x) = …

Show solution
  1. Complete the square: x2 − 8x + 5 = (x − 4)2 − 16 + 5 = (x − 4)2 − 11
  2. Let y = (x − 4)2 − 11. Then (x − 4)2 = y + 11
  3. Since x ≤ 4, x − 4 is zero or negative, so take the negative square root: x − 4 = −√y + 11
  4. x = 4 − √y + 11. Swap y for x.

Answer: f−1(x) = 4 − √x + 11

116 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q24

The functions f and g are defined as f(x) = 5x2 − 10x + 7 where x ≥ 1, and g(x) = 7x − 6

(a) Find fg(2) (2)

(b) Express the inverse function f−1 in the form f−1(x) = … (4)

Show solution

Part (a)

  1. Work from the inside out: g(2) = 7 × 2 − 6 = 8
  2. f(8) = 5 × 64 − 10 × 8 + 7 = 320 − 80 + 7

Answer: 247

Part (b)

  1. To rearrange, first complete the square: 5x2 − 10x + 7 = 5(x2 − 2x) + 7 = 5[(x − 1)2 − 1] + 7 = 5(x − 1)2 + 2
  2. Let y = 5(x − 1)2 + 2. Then (x − 1)2 = y − 2⁄5
  3. x − 1 = √y − 2⁄5 (positive root, because x ≥ 1)
  4. x = 1 + √y − 2⁄5. Swap y for x.

Answer: f−1(x) = 1 + √x − 2⁄5

More Sequences, Functions and Graphs questions

Need help with Edexcel International GCSE Maths? The Vector Academy runs small-batch online classes for 4MA1 Higher, taught by experienced tutors. Book a free demo class.

Book a free trial class

Tell us a little about your requirement.