| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| 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 A-level question. Part (i) requires straightforward substitution and algebraic manipulation to show ff(x) = x (3 marks suggests routine working). Part (ii) is even more straightforward since recognizing that ff(x) = x immediately implies f is self-inverse, so f^{-1}(x) = f(x). While it tests understanding of inverse functions and composition, it requires no problem-solving insight and follows a standard template for this topic. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| \(ff(x) = \frac{\frac{x + 3}{2x - 1} + 3}{2\left(\frac{x + 3}{2x - 1}\right) - 1} = \frac{7x}{7} = x\) | B1 M1 A1 [3] | Replacing "\(x\)" twice - must be correct. Correct algebra – clearing \((2x - 1)\). AG – all correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to f^{-1}(x) = \frac{x + 3}{2x - 1}\) | M1 A1 [2] | Attempt to make \(x\) the subject and complete method. co |
**(i)** $f(x) = \frac{x + 3}{2x - 1}$
$ff(x) = \frac{\frac{x + 3}{2x - 1} + 3}{2\left(\frac{x + 3}{2x - 1}\right) - 1} = \frac{7x}{7} = x$ | B1 M1 A1 [3] | Replacing "$x$" twice - must be correct. Correct algebra – clearing $(2x - 1)$. AG – all correct.
**(ii)** $y = \frac{x + 3}{2x - 1}$
$\to 2xy - y = x + 3$
$\to x(2y - 1) = y + 3$
$\to f^{-1}(x) = \frac{x + 3}{2x - 1}$ | M1 A1 [2] | Attempt to make $x$ the subject and complete method. co
**or** since $ff(x) = x$,
$f^{-1}(x) = f(x) = \frac{x + 3}{2x - 1}$ (M1, A1)The function $f$ is defined by $f : x \mapsto \frac{x + 3}{2x - 1}$, $x \in \mathbb{R}$, $x \neq \frac{1}{2}$.
\begin{enumerate}[label=(\roman*)]
\item Show that $f f(x) = x$. [3]
\item Hence, or otherwise, obtain an expression for $f^{-1}(x)$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q6 [5]}}