| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find composite function expression |
| Difficulty | Moderate -0.3 This is a straightforward composite and inverse functions question covering standard C3 material. Part (a) requires stating the range of a quadratic (routine), (b)(i) is direct substitution for composition, (b)(ii) is solving a simple equation, (c)(i) tests understanding of one-to-one functions (bookwork), and (c)(ii) is a standard inverse function calculation for a rational function. All parts are textbook exercises requiring no novel insight, making it slightly easier than the average A-level question. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks |
|---|---|
| (a)(i) Let \(y = \ln(2x-3)\), so \(e^y = 2x-3\), giving \(x = \frac{e^y+3}{2}\), thus \(f^{-1}(x) = \frac{e^x+3}{2}\) | M1 taking exponentials, M1 rearranging, A1 correct answer |
| (a)(ii) Range of \(f^{-1}\): \(f^{-1}(x) > \frac{3}{2}\) | B1 |
| (a)(iii) Sketch of \(y = f^{-1}(x)\): reflection of \(f\) in \(y=x\). y-intercept: \(x=0 \Rightarrow y = \frac{e^0+3}{2} = \frac{1+3}{2} = 2\), so \((0, 2)\) | B1 correct shape, B1 y-intercept labelled as \(2\) |
| (b)(i) \(gf(x) = g(\ln(2x-3)) = e^{2\ln(2x-3)}-4 = (2x-3)^2 - 4\) | M1 applying composition, M1 using log law, A1 \((2x-3)^2 - 4\) |
| (b)(ii) \(fg(x) = f(e^{2x}-4) = \ln(2(e^{2x}-4)-3) = \ln(2e^{2x}-11)\). Setting equal to \(\ln 5\): \(2e^{2x}-11=5 \Rightarrow e^{2x}=8 \Rightarrow 2x = \ln 8 \Rightarrow x = \frac{3\ln 2}{2}\) | B1 expression, M1 solving, A1 \(x = \frac{3\ln2}{2}\) |
## Question 8:
**(a)(i)** Let $y = \ln(2x-3)$, so $e^y = 2x-3$, giving $x = \frac{e^y+3}{2}$, thus $f^{-1}(x) = \frac{e^x+3}{2}$ | **M1** taking exponentials, **M1** rearranging, **A1** correct answer
**(a)(ii)** Range of $f^{-1}$: $f^{-1}(x) > \frac{3}{2}$ | **B1**
**(a)(iii)** Sketch of $y = f^{-1}(x)$: reflection of $f$ in $y=x$. y-intercept: $x=0 \Rightarrow y = \frac{e^0+3}{2} = \frac{1+3}{2} = 2$, so $(0, 2)$ | **B1** correct shape, **B1** y-intercept labelled as $2$
**(b)(i)** $gf(x) = g(\ln(2x-3)) = e^{2\ln(2x-3)}-4 = (2x-3)^2 - 4$ | **M1** applying composition, **M1** using log law, **A1** $(2x-3)^2 - 4$
**(b)(ii)** $fg(x) = f(e^{2x}-4) = \ln(2(e^{2x}-4)-3) = \ln(2e^{2x}-11)$. Setting equal to $\ln 5$: $2e^{2x}-11=5 \Rightarrow e^{2x}=8 \Rightarrow 2x = \ln 8 \Rightarrow x = \frac{3\ln 2}{2}$ | **B1** expression, **M1** solving, **A1** $x = \frac{3\ln2}{2}$
---8 The functions $f$ and $g$ are defined with their respective domains by
$$\begin{array} { l l }
\mathrm { f } ( x ) = x ^ { 2 } & \text { for all real values of } x \\
\mathrm {~g} ( x ) = \frac { 1 } { x + 2 } & \text { for real values of } x , \quad x \neq - 2
\end{array}$$
\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 Q8}}