AS Level · Mathematics 9709 · Pure Mathematics 1

Functions: AS Level Mathematics questions with answers

Free Functions questions for AS Level Mathematics (9709), part of Pure Mathematics 1. Each question has the answer and the marking points an examiner would look for.

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Functions practice questions

Question 1Structured · 6 marksA* challenge

The function f is defined by f(x) = 5x + 3x − 5 for x > 5.

(a)

Find f−1(x).

[3]
(b)

Hence explain what your answer to (a) shows about the graph of y = f(x), and find ff(x).

[2]
(c)

State the range of f.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)5x + 3x − 5
y(x − 5) = 5x + 3 rearranged M1
x terms collected and factorised M1
correct inverse A1
3
(b)f is self-inverse: the graph is symmetrical about y = x; ff(x) = x
f−1 = f, symmetrical in y = x B1
ff(x) = x B1
2
(c)f(x) > 5
f(x) > 5 B1
1
Question 2Structured · 8 marksA* challenge

The function f is defined by f(x) = 6x + 2 − 3 for x > −2.

(a)

State the range of f.

[1]
(b)

Find an expression for f−1(x) and state its domain.

[4]
(c)

Solve the equation f−1(x) = 1.

[2]
(d)

Explain why the graphs of y = f(x) and y = f−1(x) are reflections of each other in y = x.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)f(x) > −3
f(x) > −3 B1
1
(b)f−1(x) = 6x + 3 − 2, x > −3
y = … rearranged to make x the subject M1
x + 2 = 6y + 3 A1
f−1(x) = 6x + 3 − 2 A1
domain x > −3 B1
4
(c)x = −1
x = f(1) M1
x = −1 A1
2
(d)inverse swaps x and y
(x, y) on f ⇔ (y, x) on f−1: swapping x and y is reflection in y = x B1
1
Question 3Structured · 4 marks

Functions f and g are defined by f(x) = 3x + 4 and g(x) = 5x for x ≠ 0.

(a)

Find fg(x) and gf(x).

[2]
(b)

Solve fg(x) = 232.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)fg(x) = 15x + 4; gf(x) = 53x + 4
fg correct B1
gf correct B1
2
(b)x = 2
15x = 152 M1
x = 2 A1
2
Question 4Structured · 2 marks

The graph of y = f(x) is transformed by a stretch parallel to the x-axis with factor ½ followed by a translation of 0−1. Write down the equation of the new graph in terms of f.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
y = f(2x) − 1
one transformation correct B1
y = f(2x) − 1 B1
2
Question 5Structured · 9 marksA* challenge

The functions f and g are defined for x ∈ ℝ by f(x) = 3x − 5 and g(x) = x2 − 3.

(a)

Find expressions for fg(x) and gf(x).

[2]
(b)

Solve the equation fg(x) = gf(x).

[3]
(c)

Find the set of values of x for which gf(x) > fg(x).

[2]
(d)

Find f−1(x) and solve f−1(x) = f(x).

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)fg(x) = 3x2 − 14; gf(x) = 9x2 − 30x + 22
fg(x) = 3x2 − 14 B1
gf(x) = (3x − 5)2 − 3 = 9x2 − 30x + 22 B1
2
(b)x = 2, x = 3
6x2 − 30x + 36 = 0 M1
x2 − 5x + 6 = 0 A1
x = 2, 3 A1
3
(c)x < 2 or x > 3
gf(x) − fg(x) = 6(x − α)(x − β) > 0 M1
x < 2, x > 3 A1
2
(d)f−1(x) = x + 53; x = 52
f−1(x) = x + 53 B1
x = 52 (where f(x) = x) B1
2
Question 6Structured · 8 marksA* challenge

The function f is defined by f(x) = x2 − 6x + 7. The graph of y = f(x) is transformed to the graph of y = g(x) by a translation of −20, followed by a stretch parallel to the y-axis with factor 3, followed by a translation of 02.

(a)

Find g(x), giving your answer in the form ax2 + bx + c.

[4]
(b)

Find the coordinates of the vertex of the graph of y = g(x).

[2]
(c)

The order of the stretch and the final translation is reversed. Find the new vertex and explain why it differs.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)g(x) = 3x2 − 6x − 1
f(x + 2) = (x − 1)2 − 2 B1
× 3 M1
+ 2 added M1
3x2 − 6x − 1 A1
4
(b)(1, −4)
vertex of f is (3, −2) transformed M1
(1, −4) A1
2
(c)(1, 0)
(1, 0) B1
the translation is now also stretched by the factor 3 B1
2

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in