Functions: inverse and composite - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 29 exam-style questions on Functions: inverse and composite for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

29 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

\(h(x) = \dfrac{\sqrt{x + 6}}{2x}\)

Work out \(h(4.5)\)
Give your answer correct to 3 significant figures.

Model answer

\(h(4.5) = \dfrac{\sqrt{4.5 + 6}}{2 \times 4.5} = \dfrac{\sqrt{10.5}}{9}\)
\(= \dfrac{3.24037...}{9} = 0.360041...\)
\(h(4.5) = 0.360\) (3 s.f.)

Explanation

Replace \(x\) with 4.5 in both the top and the bottom of the fraction. Work out the numerator and denominator separately (or use the fraction key) so the calculator divides the whole square root by the whole of \(2x\). The zero in 0.360 is significant and must be written to give 3 significant figures.

Mark scheme

1 mark
M1 for \(\dfrac{\sqrt{10.5}}{9}\) or \(\dfrac{3.24...}{9}\)
oe, e.g. \(\dfrac{\sqrt{4.5 + 6}}{2 \times 4.5}\)
1 mark
A1 for 0.360
awrt 0.360 (0.360041...); correct answer with no working scores full marks

Question 2

3 marks

The function \(f\) converts a temperature of \(x\) degrees Celsius into degrees Fahrenheit.
\[f(x) = 1.8x + 32\]
Find \(f^{-1}(x)\), and use it to convert a temperature of 451 degrees Fahrenheit into degrees Celsius.
Give your answer correct to 1 decimal place.

Model answer

Let \(y = 1.8x + 32\)
\(y - 32 = 1.8x\)
\(x = \dfrac{y - 32}{1.8}\)
So \(f^{-1}(x) = \dfrac{x - 32}{1.8}\)
\(f^{-1}(451) = \dfrac{451 - 32}{1.8} = \dfrac{419}{1.8} = 232.777...\)
= 232.8 °C (1 d.p.)

Explanation

\(f\) turns Celsius into Fahrenheit, so the inverse function \(f^{-1}\) turns Fahrenheit back into Celsius: subtract 32 first, then divide by 1.8. Dividing first, \(\frac{451}{1.8} - 32\), reverses the steps in the wrong order.

Mark scheme

1 mark
M1 for a correct first step to rearrange \(y = 1.8x + 32\), e.g. \(y - 32 = 1.8x\)
allow with \(x\) and \(y\) swapped; correct answer with no working scores full marks
1 mark
A1 for \(f^{-1}(x) = \dfrac{x - 32}{1.8}\)
oe (exact), e.g. \(\dfrac{5(x - 32)}{9}\)
1 mark
B1 for 232.8 (°C)
awrt 232.8 (232.777...); ft their \(f^{-1}(451)\) if their \(f^{-1}\) is linear

Question 3

3 marks

\(f(x) = kx^2 - 3\), where \(k\) is a constant.
\(g(x) = x + 1.5\)

Given that \(fg(0.5) = gf(0.5)\), work out the value of \(k\).

Model answer

\(g(0.5) = 2\), so \(fg(0.5) = f(2) = 4k - 3\)
\(f(0.5) = 0.25k - 3\), so \(gf(0.5) = 0.25k - 3 + 1.5 = 0.25k - 1.5\)
\(4k - 3 = 0.25k - 1.5\)
\(3.75k = 1.5\)
\(k = 0.4\)

Explanation

Work out each composite at \(x = 0.5\), keeping \(k\) as a letter, then form and solve an equation. For \(fg(0.5)\) apply \(g\) first; for \(gf(0.5)\) apply \(f\) first. Check: with \(k = 0.4\), both sides equal \(-1.4\).

Mark scheme

1 mark
M1 for \(fg(0.5) = f(2) = 4k - 3\)
oe, e.g. \(k \times 2^2 - 3\); correct answer with no working scores full marks
1 mark
M1 for \(gf(0.5) = 0.25k - 3 + 1.5\) and setting the two expressions equal
oe, e.g. \(4k - 3 = 0.25k - 1.5\)
1 mark
A1 for \(k = 0.4\)
oe (exact), e.g. \(\frac{2}{5}\)

Question 4

4 marks

The function \(f(x) = 0.85x\) gives the value, in pounds, of a car one year after it was worth £\(x\).

Raj bought a car exactly 3 years ago. It is now worth £7676.56

Find \(fff(x)\) in its simplest form, and use it to work out how much Raj paid for the car.
Give your answer to the nearest pound.

Model answer

\(fff(x) = 0.85 \times 0.85 \times 0.85x = 0.614125x\)
After 3 years the value is \(0.614125x\)
\(0.614125x = 7676.56\)
\(x = 7676.56 \div 0.614125 = 12499.99...\)
Raj paid £12 500.

Explanation

Each application of \(f\) is one year of 15% depreciation, so three years is \(fff(x) = 0.85^3x\). To find the original price, reverse the function by dividing by 0.614125. A common mistake is to add 15% back on three times (multiplying by 1.15), which does not undo a 15% decrease.

Mark scheme

1 mark
M1 for \(fff(x) = 0.85^3x\)
oe, e.g. \(0.85 \times 0.85 \times 0.85x\); correct answer with no working scores full marks
1 mark
A1 for \(fff(x) = 0.614125x\)
cao
1 mark
M1 for \(7676.56 \div 0.614125\)
oe, e.g. \(7676.56 \div 0.85\) three times, or solving \(0.614125x = 7676.56\)
1 mark
A1 for £12 500
accept 12 499.99... or £12 500.00; do not accept answers from 7676.56 × 1.15³

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