| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.8 This is a straightforward composite and inverse functions question requiring only standard techniques: stating range, computing fg(x) by substitution, solving a simple equation, explaining why f has no inverse (not one-to-one), and finding g^{-1}(x) using the standard swap-and-rearrange method. All parts are routine textbook exercises 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 |
|---|---|
| (Range of \(f\)) \(\geqslant 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(fg(x) = \dfrac{1}{(x+2)^2}\) | B1 | OE; Maybe in part (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{1}{(x+2)^2} = 4 \Rightarrow (x+2)^2 = \dfrac{1}{4}\) | M1 | Or \(4(x+2)^2 = 1\) |
| \(x + 2 = (\pm)\dfrac{1}{2}\) | M1 | \((2x+5)(2x+3) = 0\) |
| \(x = -\dfrac{5}{2}, -\dfrac{3}{2}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Not one to one | E1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \dfrac{1}{y+2}\) | M1 | \(x \Leftrightarrow y\) |
| \(y + 2 = \dfrac{1}{x}\) | M1 | Attempt to isolate |
| \(y = \dfrac{1}{x} - 2 \quad \left(\dfrac{1-2x}{x}\right)\) | A1 |
## Question 8:
**Part (a):**
| (Range of $f$) $\geqslant 0$ | B1 | |
**Part (b)(i):**
| $fg(x) = \dfrac{1}{(x+2)^2}$ | B1 | OE; Maybe in part (ii) |
**Part (b)(ii):**
| $\dfrac{1}{(x+2)^2} = 4 \Rightarrow (x+2)^2 = \dfrac{1}{4}$ | M1 | Or $4(x+2)^2 = 1$ |
| $x + 2 = (\pm)\dfrac{1}{2}$ | M1 | $(2x+5)(2x+3) = 0$ |
| $x = -\dfrac{5}{2}, -\dfrac{3}{2}$ | A1 A1 | |
**Part (c)(i):**
| Not one to one | E1 | OE |
**Part (c)(ii):**
| $x = \dfrac{1}{y+2}$ | M1 | $x \Leftrightarrow y$ |
| $y + 2 = \dfrac{1}{x}$ | M1 | Attempt to isolate |
| $y = \dfrac{1}{x} - 2 \quad \left(\dfrac{1-2x}{x}\right)$ | A1 | |
---8 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = x ^ { 2 } & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 1 } { x + 2 } & \text { for real values of } x , x \neq - 2
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item \begin{enumerate}[label=(\roman*)]
\item Find fg(x).
\item Solve the equation $\operatorname { fg } ( x ) = 4$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Explain why the function f does not have an inverse.
\item The inverse of g is $\mathrm { g } ^ { - 1 }$. Find $\mathrm { g } ^ { - 1 } ( x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q8 [10]}}