| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Verify composite identity |
| Difficulty | Moderate -0.3 This is a slightly below-average C3 question. Part (a) requires algebraic manipulation to find the range (rearranging to make x the subject), part (b) is straightforward substitution and simplification showing f is self-inverse, and part (c) is immediate from part (b). All techniques are standard with no novel insight required, though the self-inverse property makes it easier than typical inverse function questions. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x) = \frac{2(x-2)+1}{x-2} = 2 + \frac{1}{x-2}\) | M1 | |
| \(x > 2 \therefore f(x) > 2\) | A1 | |
| (b) \(f(f(x)) = f\left(\frac{2x-3}{x-2}\right) = \frac{2(\frac{2x-3}{x-2})-3}{\frac{2x-3}{x-2}-2} = \frac{2(2x-3)-3(x-2)}{2x-3-2(x-2)} = \frac{4x-6-3x+6}{2x-3-2x+4} = x\) | M1 A1 | |
| (c) \(f^{-1}(x) = \frac{2x-3}{x-2}\) | B1 | (6 marks) |
**(a)** $f(x) = \frac{2(x-2)+1}{x-2} = 2 + \frac{1}{x-2}$ | M1 |
$x > 2 \therefore f(x) > 2$ | A1 |
**(b)** $f(f(x)) = f\left(\frac{2x-3}{x-2}\right) = \frac{2(\frac{2x-3}{x-2})-3}{\frac{2x-3}{x-2}-2} = \frac{2(2x-3)-3(x-2)}{2x-3-2(x-2)} = \frac{4x-6-3x+6}{2x-3-2x+4} = x$ | M1 A1 |
**(c)** $f^{-1}(x) = \frac{2x-3}{x-2}$ | B1 | (6 marks)$f(x) \equiv \frac{2x-3}{x-2}$, $x \in \mathbb{R}$, $x > 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the range of $f$. [2]
\item Show that $f(f(x) = x$ for all $x > 2$. [3]
\item Hence, write down an expression for $f^{-1}(x)$. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q1 [6]}}