| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Standard +0.3 This is a slightly above-average C3 question. Part (a) requires algebraic manipulation with partial fractions (shown in reverse), part (b) is straightforward range finding, part (c) is standard inverse function technique, and part (d) involves function composition and solving a quadratic. All parts use routine methods with no novel insight required, though the initial algebraic simplification adds modest complexity. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x^2 - 2x - 3 = (x-3)(x+1)\) | B1 | |
| \(f(x) = \frac{2(x-1) - (x+1)}{(x-3)(x+1)}\) or \(\frac{2(x-1)}{(x-3)(x+1)} - \frac{x+1}{(x-3)(x+1)}\) | M1 A1 | |
| \(= \frac{x-3}{(x-3)(x+1)} = \frac{1}{x+1}\) | cso A1 | (4) |
| (b) Accept \(0 < y < \frac{1}{4}\), \(0 < f(x) < \frac{1}{4}\) etc. | B1 B1 | (2) |
| (c) Let \(y = f(x)\): \(y = \frac{1}{x+1}\) | ||
| \(x = \frac{1}{y+1}\) | ||
| \(yx + x = 1\) | ||
| \(y = \frac{1-x}{x}\) or \(\frac{1}{x} - 1\) | M1 A1 | |
| \(f^{-1}(x) = \frac{1-x}{x}\) | ||
| Domain of \(f^{-1}\) is \(\left(0, \frac{1}{4}\right)\) fit their part (b) | B1 ft | (3) |
| (d) \(fg(x) = \frac{1}{2x^2 - 3x + 1}\) | ||
| \(\frac{1}{2x^2 - 2} = \frac{1}{8}\) | M1 | |
| \(2x^2 - 2 = 8\) | ||
| \(x^2 = 5\) | A1 | |
| \(x = \pm\sqrt{5}\) both | A1 | (3) [12] |
**(a)** $x^2 - 2x - 3 = (x-3)(x+1)$ | B1 |
$f(x) = \frac{2(x-1) - (x+1)}{(x-3)(x+1)}$ or $\frac{2(x-1)}{(x-3)(x+1)} - \frac{x+1}{(x-3)(x+1)}$ | M1 A1 |
$= \frac{x-3}{(x-3)(x+1)} = \frac{1}{x+1}$ | cso A1 | (4)
**(b)** Accept $0 < y < \frac{1}{4}$, $0 < f(x) < \frac{1}{4}$ etc. | B1 B1 | (2)
**(c)** Let $y = f(x)$: $y = \frac{1}{x+1}$ | |
$x = \frac{1}{y+1}$ | |
$yx + x = 1$ | |
$y = \frac{1-x}{x}$ or $\frac{1}{x} - 1$ | M1 A1 |
$f^{-1}(x) = \frac{1-x}{x}$ | |
Domain of $f^{-1}$ is $\left(0, \frac{1}{4}\right)$ fit their part (b) | B1 ft | (3)
**(d)** $fg(x) = \frac{1}{2x^2 - 3x + 1}$ | |
$\frac{1}{2x^2 - 2} = \frac{1}{8}$ | M1 |
$2x^2 - 2 = 8$ | |
$x^2 = 5$ | A1 |
$x = \pm\sqrt{5}$ both | A1 | (3) [12]
---4. The function $f$ is defined by
$$f : x \mapsto \frac { 2 ( x - 1 ) } { x ^ { 2 } - 2 x - 3 } - \frac { 1 } { x - 3 } , \quad x > 3$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { 1 } { x + 1 } , \quad x > 3$.
\item Find the range of f.
\item Find $\mathrm { f } ^ { - 1 } ( x )$. State the domain of this inverse function.
The function $g$ is defined by
$$\mathrm { g } : x \mapsto 2 x ^ { 2 } - 3 , \quad x \in \mathbb { R }$$
\item Solve $\mathrm { fg } ( x ) = \frac { 1 } { 8 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q4 [12]}}