| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.8 This is a straightforward composite and inverse function question requiring routine algebraic manipulation. Part (a) involves simple function composition, part (b)(i) uses the standard method of swapping x and y then rearranging, and part (b)(ii) requires recognizing that the range of h^{-1} equals the domain of h. All techniques are standard C3 material with no problem-solving insight required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(fg = h = \frac{6}{x+3}-2\) | M1 | Correct order |
| \(\left(= \frac{6-2x-6}{x+3} = \frac{-2x}{x+3}\right)\) | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x = \frac{-2y}{y+3}\); \(xy+3x = -2y\); \(y(x+2) = -3x\) | M1 | Attempt to isolate \(x\) or \(y\). Alt: \(y+2 = \frac{6}{x+3}\), \(x+3 = \frac{6}{y+2}\), \(x = \frac{6}{y+2}-3\) |
| \(h^{-1}(x) = y = \frac{-3x}{(x+2)}\) | M1, A1 | Total: 3 marks. Alt: \(h^{-1}(x) = \frac{6}{x+2}-3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| (Range) \(\neq -3\) | B1 | Total: 1 mark |
## Question 2:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $fg = h = \frac{6}{x+3}-2$ | M1 | Correct order |
| $\left(= \frac{6-2x-6}{x+3} = \frac{-2x}{x+3}\right)$ | A1 | Total: 2 marks |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|---------|
| $x = \frac{-2y}{y+3}$; $xy+3x = -2y$; $y(x+2) = -3x$ | M1 | Attempt to isolate $x$ or $y$. Alt: $y+2 = \frac{6}{x+3}$, $x+3 = \frac{6}{y+2}$, $x = \frac{6}{y+2}-3$ |
| $h^{-1}(x) = y = \frac{-3x}{(x+2)}$ | M1, A1 | Total: 3 marks. Alt: $h^{-1}(x) = \frac{6}{x+2}-3$ |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|---------|
| (Range) $\neq -3$ | B1 | Total: 1 mark |
---2 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = x - 2 & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 6 } { x + 3 } & \text { for real values of } x , \quad x \neq - 3
\end{array}$$
The composite function fg is denoted by h .
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { h } ( x )$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { h } ^ { - 1 } ( x )$, where $\mathrm { h } ^ { - 1 }$ is the inverse of h .
\item Find the range of $\mathrm { h } ^ { - 1 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q2 [6]}}