| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.3 This is a standard C3 composite and inverse functions question requiring routine techniques: finding range of a quadratic, finding inverse of a rational function, composing functions, and determining domains. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| \(f(x) \leq 3\) | M1A1 | M1 for \(f < 3\), \(x \leq 3\); condone \(y\), \(f\), range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \dfrac{2}{x+1}\) | ||
| \(x + 1 = \dfrac{2}{y}\) | M1 | Attempt to obtain \(x\) as a function of \(y\) or \(y\) as a function of \(x\) |
| \(x = \dfrac{2}{y} - 1\) | M1 | \(x \leftrightarrow y\) at any stage |
| \(g^{-1}(x) = \dfrac{2}{x} - 1 = \dfrac{2-x}{x}\) | A1 | Any correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(g^{-1}(x)\right) \neq -1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(h(x) = \dfrac{2}{3 - x^2 + 1} = \dfrac{2}{4 - x^2} = \dfrac{2}{(2-x)(2+x)}\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x \in \mathbb{R})\), \(x \neq +2\), \(x \neq -2\) | B1 | Condone omit '\(x\) is real'; allow \(x^2 \neq 4\) |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) \leq 3$ | M1A1 | M1 for $f < 3$, $x \leq 3$; condone $y$, $f$, range |
### Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \dfrac{2}{x+1}$ | | |
| $x + 1 = \dfrac{2}{y}$ | M1 | Attempt to obtain $x$ as a function of $y$ or $y$ as a function of $x$ |
| $x = \dfrac{2}{y} - 1$ | M1 | $x \leftrightarrow y$ at any stage |
| $g^{-1}(x) = \dfrac{2}{x} - 1 = \dfrac{2-x}{x}$ | A1 | Any correct form |
### Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(g^{-1}(x)\right) \neq -1$ | B1 | |
### Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $h(x) = \dfrac{2}{3 - x^2 + 1} = \dfrac{2}{4 - x^2} = \dfrac{2}{(2-x)(2+x)}$ | M1, A1 | |
### Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x \in \mathbb{R})$, $x \neq +2$, $x \neq -2$ | B1 | Condone omit '$x$ is real'; allow $x^2 \neq 4$ |
**Total: 9 marks**
---3 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = 3 - x ^ { 2 } , & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 2 } { x + 1 } , & \text { for real values of } x , x \neq - 1
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the range of f.
\item The inverse of g is $\mathrm { g } ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { g } ^ { - 1 } ( x )$.
\item State the range of $\mathrm { g } ^ { - 1 }$.
\end{enumerate}\item The composite function gf is denoted by h .
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { h } ( x )$, simplifying your answer.
\item State the greatest possible domain of h .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2007 Q3 [9]}}