| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | State domain or range |
| Difficulty | Standard +0.3 This is a slightly-above-routine question on rational functions requiring composition evaluation, range determination from a restricted domain, and finding an inverse with domain. While it involves multiple parts, each step uses standard techniques (substitution, algebraic manipulation, swap-and-solve method) with no novel insight required. The restricted domain adds minor complexity beyond typical textbook exercises. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(g(5) = \frac{2(5)+5}{5-3} = 7.5 \Rightarrow gg(5) = \frac{2("7.5")+5}{"7.5"-3}\) | M1 | 1.1b - Full method of attempting \(g(5)\) and substituting the result into \(g\) |
| \(gg(5) = \frac{40}{9}\) (or \(4\frac{4}{9}\) or \(4.\dot{4}\)) | A1 | 1.1b - Obtains \(\frac{40}{9}\) or \(4\frac{4}{9}\) or \(4.\dot{4}\) or exact equivalent. Give A0 for 4.4 or 4.444... without reference to \(\frac{40}{9}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(gg(x) = \frac{2\left(\frac{2x+5}{x-3}\right)+5}{\left(\frac{2x+5}{x-3}\right)-3} \Rightarrow gg(5) = \frac{2\left(\frac{2(5)+5}{(5)-3}\right)+5}{\left(\frac{2(5)+5}{(5)-3}\right)-3}\) | M1 | 1.1b - Attempts to substitute \(x=5\) into expression; note \(gg(x) = \frac{9x-5}{14-x}\) |
| \(gg(5) = \frac{40}{9}\) (or \(4\frac{4}{9}\) or \(4.\dot{4}\)) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\{\)Range:\(\}\ 2 < y \leq \frac{15}{2}\) | B1 | 1.1b - Accept any of \(2 < g \leq \frac{15}{2}\), \(2 < g(x) \leq \frac{15}{2}\), \(\left(2, \frac{15}{2}\right]\). Accept \(g(x)>2\) and \(g(x) \leq \frac{15}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{2x+5}{x-3} \Rightarrow yx - 3y = 2x+5 \Rightarrow yx - 2x = 3y+5\) | M1 | 1.1b - Correct method of cross multiplication followed by attempt to collect terms in \(x\) |
| \(x(y-2) = 3y+5 \Rightarrow x = \frac{3y+5}{y-2}\) \(\left\{\text{or } y = \frac{3x+5}{x-2}\right\}\) | M1 | 2.1 - Complete method to find the inverse |
| \(g^{-1}(x) = \frac{3x+5}{x-2}\), \(\ 2 < x \leq \frac{15}{2}\) | A1ft | 2.5 - Uses correct notation; domain stated correctly. Allow \(2 < x \leq \frac{15}{2}\) or \(\left(2, \frac{15}{2}\right]\). Do not allow \(2 < g \leq \frac{15}{2}\) or \(2 < g^{-1}(x) \leq \frac{15}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{2x-6+11}{x-3} \Rightarrow y = 2 + \frac{11}{x-3} \Rightarrow y-2 = \frac{11}{x-3}\) | M1 | 1.1b - Writes \(y = \frac{2x+5}{x-3}\) in form \(y = 2 \pm \frac{k}{x-3}\), \(k \neq 0\) and rearranges to isolate \(y\) and 2 on one side |
| \(x - 3 = \frac{11}{y-2} \Rightarrow x = \frac{11}{y-2} + 3\) \(\left\{\text{or } y = \frac{11}{x-2}+3\right\}\) | M1 | 2.1 - Complete method to find the inverse |
| \(g^{-1}(x) = \frac{11}{x-2} + 3\), \(\ 2 < x \leq \frac{15}{2}\) | A1ft | 2.5 - As in Way 1 |
# Question 1:
## Part (a) - Way 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $g(5) = \frac{2(5)+5}{5-3} = 7.5 \Rightarrow gg(5) = \frac{2("7.5")+5}{"7.5"-3}$ | M1 | 1.1b - Full method of attempting $g(5)$ and substituting the result into $g$ |
| $gg(5) = \frac{40}{9}$ (or $4\frac{4}{9}$ or $4.\dot{4}$) | A1 | 1.1b - Obtains $\frac{40}{9}$ or $4\frac{4}{9}$ or $4.\dot{4}$ or exact equivalent. Give A0 for 4.4 or 4.444... without reference to $\frac{40}{9}$ |
## Part (a) - Way 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $gg(x) = \frac{2\left(\frac{2x+5}{x-3}\right)+5}{\left(\frac{2x+5}{x-3}\right)-3} \Rightarrow gg(5) = \frac{2\left(\frac{2(5)+5}{(5)-3}\right)+5}{\left(\frac{2(5)+5}{(5)-3}\right)-3}$ | M1 | 1.1b - Attempts to substitute $x=5$ into expression; note $gg(x) = \frac{9x-5}{14-x}$ |
| $gg(5) = \frac{40}{9}$ (or $4\frac{4}{9}$ or $4.\dot{4}$) | A1 | 1.1b |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{$Range:$\}\ 2 < y \leq \frac{15}{2}$ | B1 | 1.1b - Accept any of $2 < g \leq \frac{15}{2}$, $2 < g(x) \leq \frac{15}{2}$, $\left(2, \frac{15}{2}\right]$. Accept $g(x)>2$ **and** $g(x) \leq \frac{15}{2}$ |
## Part (c) - Way 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{2x+5}{x-3} \Rightarrow yx - 3y = 2x+5 \Rightarrow yx - 2x = 3y+5$ | M1 | 1.1b - Correct method of cross multiplication followed by attempt to collect terms in $x$ |
| $x(y-2) = 3y+5 \Rightarrow x = \frac{3y+5}{y-2}$ $\left\{\text{or } y = \frac{3x+5}{x-2}\right\}$ | M1 | 2.1 - Complete method to find the inverse |
| $g^{-1}(x) = \frac{3x+5}{x-2}$, $\ 2 < x \leq \frac{15}{2}$ | A1ft | 2.5 - Uses correct notation; domain stated correctly. Allow $2 < x \leq \frac{15}{2}$ or $\left(2, \frac{15}{2}\right]$. Do not allow $2 < g \leq \frac{15}{2}$ or $2 < g^{-1}(x) \leq \frac{15}{2}$ |
## Part (c) - Way 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{2x-6+11}{x-3} \Rightarrow y = 2 + \frac{11}{x-3} \Rightarrow y-2 = \frac{11}{x-3}$ | M1 | 1.1b - Writes $y = \frac{2x+5}{x-3}$ in form $y = 2 \pm \frac{k}{x-3}$, $k \neq 0$ and rearranges to isolate $y$ and 2 on one side |
| $x - 3 = \frac{11}{y-2} \Rightarrow x = \frac{11}{y-2} + 3$ $\left\{\text{or } y = \frac{11}{x-2}+3\right\}$ | M1 | 2.1 - Complete method to find the inverse |
| $g^{-1}(x) = \frac{11}{x-2} + 3$, $\ 2 < x \leq \frac{15}{2}$ | A1ft | 2.5 - As in Way 1 |
---1.
$$\operatorname { g } ( x ) = \frac { 2 x + 5 } { x - 3 } \quad x \geqslant 5$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { gg } ( 5 )$.
\item State the range of g.
\item Find $\mathrm { g } ^ { - 1 } ( x )$, stating its domain.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q1 [6]}}