Composite and inverse functions; functions in modelling – A-level Maths exam questions with mark scheme

1CoreCalculator5 marks

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f}(x)=3x-5,\ x\in\mathbb{R}\qquad \mathrm{g}(x)=\frac{2}{x+1},\ x\in\mathbb{R},\ x\neq -1$$

(a) Find $\mathrm{f}^{-1}(x)$. [2]

(b) Solve the equation $\mathrm{fg}(x)=1$. [3]

(a) Attempts to make $x$ the subject of $y=3x-5$ (or swaps $x$ and $y$ first), e.g. $x=\dfrac{y+5}{3}$M1
$\mathrm{f}^{-1}(x)=\dfrac{x+5}{3}$ (must be in terms of $x$; allow $\tfrac13x+\tfrac53$)A1
(b) Correct order for composite: $\mathrm{fg}(x)=3\left(\dfrac{2}{x+1}\right)-5$M1
Sets $\dfrac{6}{x+1}-5=1$ and proceeds to a value of $x$ (e.g. $\dfrac{6}{x+1}=6\Rightarrow x+1=1$)M1
$x=0$A1
Answer: (a) $\mathrm{f}^{-1}(x)=\dfrac{x+5}{3}$ (b) $x=0$
Examiner tip: $\mathrm{fg}(x)$ means apply g first – students who work out $\mathrm{g}(\mathrm{f}(x))$ lose both method marks in (b).

2StandardCalculator7 marks

The functions $\mathrm{f}$ and $\mathrm{g}$ are defined by $$\mathrm{f}(x)=\mathrm{e}^{2x}+3,\ x\in\mathbb{R}\qquad \mathrm{g}(x)=\ln(x-1),\ x\in\mathbb{R},\ x>1$$

(a) State the range of $\mathrm{f}$. [1]

(b) Find $\mathrm{f}^{-1}(x)$, stating its domain. [3]

(a) $\mathrm{f}(x)>3$ (accept $y>3$, $(3,\infty)$; not $x>3$)B1
(b) Attempts to change the subject: $\mathrm{e}^{2x}=y-3\Rightarrow 2x=\ln(y-3)$M1
$\mathrm{f}^{-1}(x)=\tfrac12\ln(x-3)$A1
Domain $x>3$ (follow through from their range in (a))B1
(c) Uses $\mathrm{e}^{2\ln(x-1)}=(x-1)^{2}$ to obtain $\mathrm{fg}(x)=(x-1)^{2}+3$M1
Solves $(x-1)^{2}+3=12$ to obtain at least one value of $x$M1
$x=4$ only, with $x=-2$ rejected (as $x>1$ in the domain of g)A1
Answer: (a) $\mathrm{f}(x)>3$ (b) $\mathrm{f}^{-1}(x)=\tfrac12\ln(x-3),\ x>3$ (c) $x=4$
Examiner tip: Always check solutions against the domain – leaving $x=-2$ in (c) costs the final A mark.

3HardCalculator7 marks

The function $\mathrm{f}$ is defined by $$\mathrm{f}(x)=\frac{2x+3}{x-2},\quad x\in\mathbb{R},\ x\neq 2$$

(a) Show that $\mathrm{ff}(x)=x$. [3]

(b) Hence write down $\mathrm{f}^{-1}(x)$. [1]

(d) Find the values of $x$ for which $\mathrm{f}(x)=x$, giving your answers in simplest surd form. [2]

(a) Substitutes f into itself: $\dfrac{2\left(\frac{2x+3}{x-2}\right)+3}{\frac{2x+3}{x-2}-2}$M1
Multiplies numerator and denominator by $(x-2)$ (or combines fractions): $\dfrac{4x+6+3(x-2)}{2x+3-2(x-2)}$M1
$=\dfrac{7x}{7}=x$ csoA1*
(b) $\mathrm{f}^{-1}(x)=\dfrac{2x+3}{x-2}$ (accept $\mathrm{f}^{-1}(x)=\mathrm{f}(x)$)B1
(c) $\mathrm{f}(x)\neq 2$ (accept $y\in\mathbb{R},\ y\neq2$)B1
(d) Sets $\dfrac{2x+3}{x-2}=x$ and obtains a 3-term quadratic $x^{2}-4x-3=0$, attempting to solveM1
$x=2\pm\sqrt7$A1
Answer: (b) $\mathrm{f}^{-1}(x)=\dfrac{2x+3}{x-2}$ (c) $\mathrm{f}(x)\neq2$ (d) $x=2\pm\sqrt7$
Examiner tip: In a 'show that' every line must appear – jumping from the compound fraction straight to $x$ loses the A1*.

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