| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Evaluate composite at point |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on composite and inverse functions. Part (a) requires simple substitution into a composition; part (b) is routine inverse function finding; part (c) requires solving an equation using the inverse, which is a standard technique. All parts use familiar P3 content with no novel insights required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(fg(e^2) = f\left(\frac{5}{2}\ln e^2\right) = \frac{12}{\frac{5}{2}\ln e^2 + 1} = 2\) | M1, A1 | (2) |
| (b) \(f(x) = \frac{12}{x+1}\) | ||
| \(f^{-1}(x) = \frac{12}{x} - 1\) | M1 A1 | |
| \(0 < x < 12\) | B1 | (3) |
| (c) \(\frac{12}{x+1} = \frac{12}{x} - 1 \Rightarrow 12x = 12(x+1) - x(x+1)\) \(\Rightarrow x^2 + x - 12 = 0 \Rightarrow x = ...\) | M1 dM1 | Must be 3TQ |
| \(x = 3\) only | A1 | (3) |
| Alt (c): Solves \(f^{-1}(x) = x \Rightarrow \frac{12}{x} - 1 = x\) leading to quadratic equation, or solves \(f(x) = x \Rightarrow \frac{12}{x+1} = x\) leading to quadratic equation \(\Rightarrow x^2 + x - 12 = 0 \Rightarrow x = ...\) | M1 dM1 | Must be 3TQ |
| \(x = 3\) only | A1 | (3) |
**(a)** $fg(e^2) = f\left(\frac{5}{2}\ln e^2\right) = \frac{12}{\frac{5}{2}\ln e^2 + 1} = 2$ | M1, A1 | (2)
**(b)** $f(x) = \frac{12}{x+1}$ |
$f^{-1}(x) = \frac{12}{x} - 1$ | M1 A1 |
$0 < x < 12$ | B1 | (3)
**(c)** $\frac{12}{x+1} = \frac{12}{x} - 1 \Rightarrow 12x = 12(x+1) - x(x+1)$ $\Rightarrow x^2 + x - 12 = 0 \Rightarrow x = ...$ | M1 dM1 | Must be 3TQ |
$x = 3$ only | A1 | (3)
**Alt (c):** Solves $f^{-1}(x) = x \Rightarrow \frac{12}{x} - 1 = x$ leading to quadratic equation, or solves $f(x) = x \Rightarrow \frac{12}{x+1} = x$ leading to quadratic equation $\Rightarrow x^2 + x - 12 = 0 \Rightarrow x = ...$ | M1 dM1 | Must be 3TQ |
$x = 3$ only | A1 | (3)
**Total: 8 marks**
---2. The function $f$ and the function $g$ are defined by
$$\begin{array} { l l }
\mathrm { f } ( x ) = \frac { 12 } { x + 1 } & x > 0 , x \in \mathbb { R } \\
\mathrm {~g} ( x ) = \frac { 5 } { 2 } \ln x & x > 0 , x \in \mathbb { R }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find, in simplest form, the value of $\mathrm { fg } \left( \mathrm { e } ^ { 2 } \right)$
\item Find f-1
\item Hence, or otherwise, find all real solutions of the equation
$$\mathrm { f } ^ { - 1 } ( x ) = \mathrm { f } ( x )$$
\begin{center}
\end{center}
\begin{center}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2020 Q2 [8]}}