| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine if inverse exists |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on piecewise functions and inverses. Parts (a), (b), and (d) involve routine substitution and algebraic manipulation. Part (c) requires understanding that a function needs to be one-to-one to have an inverse, which is standard A-level content. The piecewise nature adds minimal complexity as the cases are clearly defined and easy to work with. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(gg(0) = g((0-2)^2+1) = g(5) = 4(5)-7 = 13\) | M1 | Uses complete method to find \(gg(0)\); substituting \(x=0\) into \((0-2)^2+1\) and result into \(g(x)\), or attempts to substitute \(x=0\) into \(4((x-2)^2+1)-7\) or \(4(x-2)^2-3\) |
| \(gg(0) = 13\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solves either \((x-2)^2+1=28 \Rightarrow x=\ldots\) or \(4x-7=28 \Rightarrow x=\ldots\) | M1 | You can ignore inequality symbols for M marks |
| At least one critical value \(x=2-3\sqrt{3}\) or \(x=\frac{35}{4}\) is correct | A1 | Writing \(\approx -3.20\) or truncated \(-3.19\) or truncated \(-3.2\) in place of \(2-3\sqrt{3}\) accepted for A marks |
| Solves both \((x-2)^2+1=28 \Rightarrow x=\ldots\) and \(4x-7=28 \Rightarrow x=\ldots\) | M1 | If a 3TQ formed (e.g. \(x^2-4x-23=0\)) a correct method for solving is required |
| Correct final answer: \(x < 2-3\sqrt{3}\), \(x > \frac{35}{4}\) | A1 | Allow set notation e.g. \(\{x\in\mathbb{R}: x<2-3\sqrt{3} \cup x>8.75\}\); writing \(2-\sqrt{27}\) in place of \(2-3\sqrt{3}\) accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| h is a one-one function so has an inverse; g is a many-one function so does not have an inverse | B1 | Minimal acceptable reason is "h is a one-one and g is a many-one"; give B1 for "h is a one-one and g is not"; allow B1 for "g is a many-one and h is not" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\{h^{-1}(x)=-\frac{1}{2} \Rightarrow\right\}\) \(x=h\left(-\frac{1}{2}\right)\) | M1 (B1 on open) | Writes \(x=h\left(-\frac{1}{2}\right)\) |
| \(x=\left(-\frac{1}{2}-2\right)^2+1\); note: condone \(x=\left(\frac{1}{2}-2\right)^2+1\) | M1 | |
| \(\Rightarrow x=7.25\) only cso | A1 | Uses \(x=h\left(-\frac{1}{2}\right)\) to deduce \(x=7.25\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\)their \(h^{-1}(x)\} = \pm 2 \pm \sqrt{x\pm 1}\) | M1 | |
| Attempts to solve \(\pm 2 \pm \sqrt{x\pm 1} = -\frac{1}{2} \Rightarrow \pm\sqrt{x\pm 1} = \ldots\) | M1 | Give A0 cso for \(2+\sqrt{x-1}=-\frac{1}{2} \Rightarrow \sqrt{x-1}=-\frac{5}{2}\); give A1 cso for \(2\pm\sqrt{x-1}=-\frac{1}{2} \Rightarrow -\sqrt{x-1}=-\frac{5}{2} \Rightarrow x-1=\frac{25}{4} \Rightarrow x=7.25\) |
| \(\Rightarrow x=7.25\) only cso | A1 | Use correct \(h^{-1}(x)=2-\sqrt{x-1}\) to deduce \(x=7.25\) only |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $gg(0) = g((0-2)^2+1) = g(5) = 4(5)-7 = 13$ | M1 | Uses complete method to find $gg(0)$; substituting $x=0$ into $(0-2)^2+1$ and result into $g(x)$, or attempts to substitute $x=0$ into $4((x-2)^2+1)-7$ or $4(x-2)^2-3$ |
| $gg(0) = 13$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solves either $(x-2)^2+1=28 \Rightarrow x=\ldots$ or $4x-7=28 \Rightarrow x=\ldots$ | M1 | You can ignore inequality symbols for M marks |
| At least one critical value $x=2-3\sqrt{3}$ or $x=\frac{35}{4}$ is correct | A1 | Writing $\approx -3.20$ or truncated $-3.19$ or truncated $-3.2$ in place of $2-3\sqrt{3}$ accepted for A marks |
| Solves both $(x-2)^2+1=28 \Rightarrow x=\ldots$ and $4x-7=28 \Rightarrow x=\ldots$ | M1 | If a 3TQ formed (e.g. $x^2-4x-23=0$) a correct method for solving is required |
| Correct final answer: $x < 2-3\sqrt{3}$, $x > \frac{35}{4}$ | A1 | Allow set notation e.g. $\{x\in\mathbb{R}: x<2-3\sqrt{3} \cup x>8.75\}$; writing $2-\sqrt{27}$ in place of $2-3\sqrt{3}$ accepted |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| h is a one-one function so has an inverse; g is a many-one function so does not have an inverse | B1 | Minimal acceptable reason is "h is a one-one and g is a many-one"; give B1 for "h is a one-one and g is not"; allow B1 for "g is a many-one and h is not" |
### Part (d) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{h^{-1}(x)=-\frac{1}{2} \Rightarrow\right\}$ $x=h\left(-\frac{1}{2}\right)$ | M1 (B1 on open) | Writes $x=h\left(-\frac{1}{2}\right)$ |
| $x=\left(-\frac{1}{2}-2\right)^2+1$; note: condone $x=\left(\frac{1}{2}-2\right)^2+1$ | M1 | |
| $\Rightarrow x=7.25$ only **cso** | A1 | Uses $x=h\left(-\frac{1}{2}\right)$ to deduce $x=7.25$ only |
### Part (d) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{$their $h^{-1}(x)\} = \pm 2 \pm \sqrt{x\pm 1}$ | M1 | |
| Attempts to solve $\pm 2 \pm \sqrt{x\pm 1} = -\frac{1}{2} \Rightarrow \pm\sqrt{x\pm 1} = \ldots$ | M1 | Give A0 cso for $2+\sqrt{x-1}=-\frac{1}{2} \Rightarrow \sqrt{x-1}=-\frac{5}{2}$; give A1 cso for $2\pm\sqrt{x-1}=-\frac{1}{2} \Rightarrow -\sqrt{x-1}=-\frac{5}{2} \Rightarrow x-1=\frac{25}{4} \Rightarrow x=7.25$ |
| $\Rightarrow x=7.25$ only **cso** | A1 | Use correct $h^{-1}(x)=2-\sqrt{x-1}$ to deduce $x=7.25$ only |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fa4afaf4-fe5d-4f3a-b3de-9600d5502a49-12_728_1086_246_493}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the graph of $y = \mathrm { g } ( x )$, where
$$g ( x ) = \begin{cases} ( x - 2 ) ^ { 2 } + 1 & x \leqslant 2 \\ 4 x - 7 & x > 2 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\operatorname { gg } ( 0 )$.
\item Find all values of $x$ for which
$$\mathrm { g } ( x ) > 28$$
The function h is defined by
$$\mathrm { h } ( x ) = ( x - 2 ) ^ { 2 } + 1 \quad x \leqslant 2$$
\item Explain why h has an inverse but g does not.
\item Solve the equation
$$\mathrm { h } ^ { - 1 } ( x ) = - \frac { 1 } { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q6 [10]}}