| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Moderate -0.3 This is a straightforward C3 composite/inverse functions question requiring standard techniques: stating range of x², finding inverse of a rational function (routine algebraic manipulation), and solving fg(x) = 9 by substitution. All parts are textbook exercises with no novel insight 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 |
|---|---|---|
| \(f(x) \geq 0\) | B1 | Allow \(f \geq 0\), \(y \geq 0\), \(\geq 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2x-3}\); \(x = \frac{1}{2y-3}\); \(x(2y-3) = 1\); \(2xy - 3x = 1\); \(2xy = 1 + 3x\); \(y = \frac{1+3x}{2x} = g^{-1}(x)\) o.e. | M1, M1, M1, A1 | Swap \(x\) and \(y\); attempt to isolate; w.n.f.e |
| Answer | Marks |
|---|---|
| \((g^{-1}(x)) = \frac{3}{2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{1}{2x-3}\right)^2 = 9\); \(2x - 3 = \pm\frac{1}{3}\) | M1 | Square root and invert (condone missing ±); alternative: attempt to solve a quadratic that comes from \(4x^2 - 12x + 9 = \frac{1}{9}\) o.e. |
| \(x = \frac{5}{3}, \frac{4}{3}\) o.e. | A1 |
**4(a)**
| $f(x) \geq 0$ | B1 | Allow $f \geq 0$, $y \geq 0$, $\geq 0$ |
**4(b)(i)**
| $y = \frac{1}{2x-3}$; $x = \frac{1}{2y-3}$; $x(2y-3) = 1$; $2xy - 3x = 1$; $2xy = 1 + 3x$; $y = \frac{1+3x}{2x} = g^{-1}(x)$ o.e. | M1, M1, M1, A1 | Swap $x$ and $y$; attempt to isolate; w.n.f.e |
**4(b)(ii)**
| $(g^{-1}(x)) = \frac{3}{2}$ | B1 | |
**4(c)**
| $\left(\frac{1}{2x-3}\right)^2 = 9$; $2x - 3 = \pm\frac{1}{3}$ | M1 | Square root and invert (condone missing ±); alternative: attempt to solve a quadratic that comes from $4x^2 - 12x + 9 = \frac{1}{9}$ o.e. |
| $x = \frac{5}{3}, \frac{4}{3}$ o.e. | A1 | |
**Total: 8 marks**
---4 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = x ^ { 2 } , & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 1 } { 2 x - 3 } , & \text { for real values of } x , x \neq \frac { 3 } { 2 }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item State the range of f.
\item \begin{enumerate}[label=(\roman*)]
\item The inverse of g is $\mathrm { g } ^ { - 1 }$. Find $\mathrm { g } ^ { - 1 } ( x )$.
\item State the range of $\mathrm { g } ^ { - 1 }$.
\end{enumerate}\item Solve the equation $\operatorname { fg } ( x ) = 9$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q4 [8]}}