About Functions in A Level Mathematics
Functions is one of the 8 main topics in our A Level Mathematics (Cambridge (CAIE) 9709) question bank, and about 3% of its real questions come from it. Tayyari Karo has 15 real questions on this topic from 2021 to 2026, from Paper 1. By difficulty: 9 medium, 5 hard, 1 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 6 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2026 · Paper 1 · Functions
Functions f and g are defined by f(x) = x − 3 for x ∈ ℝ and g(x) = x³ − 8a for x ∈ ℝ, where a is a positive constant.
(a) Find an expression for g⁻¹(x). [2]
(b) It is given that gf(−1) + 20a² = 0.
(i) Find the value of a. [4]
(ii) Find the set of values of the constant k for which the equation gf(x) = fg(x) + k has no real solutions. [5]
Show model answer
(a) y = x³ − 8a gives x = ∛(y + 8a), so g⁻¹(x) = ∛(x + 8a).
(b)(i) f(−1) = −4, g(−4) = −64 − 8a. So −64 − 8a + 20a² = 0, i.e. 5a² − 2a − 16 = 0, (5a + 8)(a − 2) = 0, so a = 2 (a > 0).
(ii) With a = 2: gf(x) = (x − 3)³ − 16 and fg(x) = x³ − 16 − 3 = x³ − 19.
(x − 3)³ − 16 = x³ − 19 + k gives 9x² − 27x + 24 + k = 0. No real roots requires 27² − 4(9)(24 + k) < 0, so k > −3.75 (−15/4).
Rearrange for the inverse; substitute f(−1) into g to form a quadratic in a; for part (ii) equate gf and fg + k, expand and use the discriminant < 0.
October/November 2025 · Paper 1 · Functions
Functions f and g are defined by
f(x) = (x + 3)² − 12 for x ≥ 0,
g(x) = 2x − 5 for x ∈ ℝ.
(a) State the range of f. [1]
(b) Find an expression for f⁻¹(x). [2]
(c) Solve the equation gf(x) = 69. [4]
Show model answer
(a) At x = 0, f(0) = 9 − 12 = −3, and f is increasing for x ≥ 0 (right of vertex x = −3), so range is f(x) ≥ −3.
(b) y = (x + 3)² − 12 → x + 3 = √(y + 12) (taking the positive root since x ≥ 0) → x = √(y + 12) − 3. So f⁻¹(x) = √(x + 12) − 3.
(c) gf(x) = 2[(x + 3)² − 12] − 5 = 2(x + 3)² − 29. Set equal to 69: 2(x + 3)² − 29 = 69 → (x + 3)² = 49 → x + 3 = ±7 → x = 4 or x = −10. Since the domain requires x ≥ 0, x = 4 only.
Substitute f(x) into g and solve the resulting quadratic, discarding any solution outside the domain x ≥ 0.
May/June 2023 · Paper 1 · Functions
The functions f and g are defined as follows, where a and b are constants.
f(x) = 1 + 2a/(x−a) for x > a
g(x) = bx − 2 for x ∈ ℝ
(a) Given that f(7) = 5/2 and gf(5) = 4, find the values of a and b. [4]
For the rest of this question, use the value of a found in (a).
(b) Find the domain of f⁻¹. [1]
(c) Find an expression for f⁻¹(x). [3]
Show model answer
(a) f(7) = 1 + 2a/(7−a) = 5/2 → 2a/(7−a) = 3/2 → 4a = 3(7−a) = 21−3a → 7a = 21 → a = 3.
f(5) = 1 + 2(3)/(5−3) = 1 + 3 = 4.
gf(5) = g(4) = 4b − 2 = 4 → 4b = 6 → b = 3/2.
a = 3, b = 3/2
(b) The range of f is the domain of f⁻¹. As x→a⁺, f(x)→∞, and as x→∞, f(x)→1⁺, so range of f is f(x) > 1.
Domain of f⁻¹: x > 1
(c) y = 1 + 6/(x−3) → y−1 = 6/(x−3) → x−3 = 6/(y−1) → x = 3 + 6/(y−1).
f⁻¹(x) = 3 + 6/(x−1)
Substitute to form simultaneous equations for a and b; then swap x and y and rearrange to find the inverse.
May/June 2022 · Paper 1 · Functions
The function f is defined as follows: f(x) = (x² − 4)/(x² + 4) for x > 2.
(a) Find an expression for f⁻¹(x). [3]
(b) Show that 1 − 8/(x² + 4) can be expressed as (x² − 4)/(x² + 4), and hence state the range of f. [4]
(c) Explain why the composite function ff cannot be formed. [1]
Show model answer
(a) Let y = (x² − 4)/(x² + 4).
y(x² + 4) = x² − 4
yx² + 4y = x² − 4
x²(y − 1) = −4 − 4y
x² = 4(1 + y)/(1 − y)
x = 2√[(1 + y)/(1 − y)]
Swapping variables: f⁻¹(x) = 2√[(1 + x)/(1 − x)], equivalently √[(4x + 4)/(1 − x)]
(b) 1 − 8/(x² + 4) = [(x² + 4) − 8]/(x² + 4) = (x² − 4)/(x² + 4) ✓, which equals f(x).
For x > 2, x² + 4 > 8, so 0 < 8/(x² + 4) < 1.
As x → 2⁺, x² + 4 → 8⁺, so 8/(x² + 4) → 1⁻, giving f(x) → 0⁺ (but never equal to 0).
As x → ∞, 8/(x² + 4) → 0, so f(x) → 1⁻ (but never reaches 1).
Range of f: 0 < f(x) < 1
(c) The range of f, 0 < f(x) < 1, does not lie within the domain of f, x > 2 (there is no overlap), so the output of f can never be fed back into f — ff cannot be formed.
Rearrange y = f(x) to make x the subject to find the inverse; for the range, rewrite f(x) in the given alternative form and consider what happens to 8/(x²+4) as x ranges over x > 2; for part (c) compare the range of f against its own domain.
October/November 2022 · Paper 1 · Functions
The function f is defined by f(x) = 2 - 3/(4x - p) for x > p/4, where p is a constant.
(a) Find f'(x) and hence determine whether f is an increasing function, a decreasing function or neither. [3]
(b) Express f⁻¹(x) in the form p/a - b/(cx - d), where a, b, c and d are integers. [4]
(c) Hence state the value of p for which f⁻¹(x) = f(x). [1]
Show model answer
(a) f'(x) = -3×(-1)×4×(4x - p)⁻² = 12/(4x - p)². Since (4x - p)² > 0 for all x in the domain, f'(x) > 0, so f is an increasing function.
(b) Let y = 2 - 3/(4x - p). Then (y - 2)(4x - p) = -3 → 4xy - py = 8x - 2p - 3 → 4x(y - 2) = p(y - 2) - 3 → x = p/4 - 3/[4(y - 2)]. So f⁻¹(x) = p/4 - 3/(4x - 8).
(c) Comparing f⁻¹(x) = p/4 - 3/(4x - 8) with f(x) = 2 - 3/(4x - p), f⁻¹(x) = f(x) when p = 8.
Differentiate to test monotonicity; rearrange y = f(x) to make x the subject for the inverse; then compare the two forms for part (c).
May/June 2021 · Paper 1 · Functions
Functions f and g are defined as follows:
f(x) = (x - 2)² - 4 for x ≥ 2,
g(x) = ax + 2 for x ∈ ℝ,
where a is a constant.
(a) State the range of f. [1]
(b) Find f⁻¹(x). [2]
(c) Given that a = -5/3, solve the equation f(x) = g(x). [3]
(d) Given instead that gg(f⁻¹(12)) = 62, find the possible values of a. [5]
Show model answer
(a) Since x ≥ 2, (x-2)² ≥ 0, so the range of f is f(x) ≥ -4.
(b) y = (x - 2)² - 4
y + 4 = (x - 2)²
x - 2 = √(y + 4) (taking the positive root since x ≥ 2)
x = 2 + √(y + 4)
f⁻¹(x) = 2 + √(x + 4)
(c) With a = -5/3:
(x - 2)² - 4 = -5/3 x + 2
x² - 4x = -5/3 x + 2
x² - 4x + 5/3 x - 2 = 0
x² - 7/3 x - 2 = 0
Multiply by 3: 3x² - 7x - 6 = 0
(3x + 2)(x - 3) = 0
x = -2/3 or x = 3
Since f requires x ≥ 2, only x = 3 is valid.
(d) f⁻¹(12) = 2 + √(12 + 4) = 2 + 4 = 6
g(f⁻¹(12)) = g(6) = 6a + 2
gg(f⁻¹(12)) = g(6a + 2) = a(6a + 2) + 2 = 6a² + 2a + 2
Set equal to 62:
6a² + 2a + 2 = 62
6a² + 2a - 60 = 0
3a² + a - 30 = 0
a = [-1 ± √(1 + 360)]/6 = [-1 ± 19]/6
a = 3 or a = -10/3
Rearrange y=(x-2)²-4 to isolate x for part (b)'s inverse; for later parts substitute into f(x)=g(x), or chain g(g(f⁻¹(x))) step by step and solve the resulting equation in a.
