| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Find inverse function after simplification |
| Difficulty | Standard +0.3 This question involves routine partial fractions manipulation and standard inverse function techniques. Part (a) requires algebraic simplification over a common denominator, part (b) is straightforward function inversion (swap x and y, rearrange), and part (c) is simple substitution and solving a quadratic. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{5x+1}{(x+2)(x-1)} - \frac{3}{x+2}\) | B1 | |
| \(= \frac{5x+1-3(x-1)}{(x+2)(x-1)}\) | M1 | M1 for combining fractions even if denominator is not lowest common |
| \(= \frac{2x+4}{(x+2)(x-1)} = \frac{2(x+2)}{(x+2)(x-1)} = \frac{2}{x-1}\) * | M1 A1 cso | M1 must have linear numerator |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(y = \frac{2}{x-1} \Rightarrow xy - y = 2 \Rightarrow xy = 2 + y\) | M1A1 | |
| \(f^{-1}(x) = \frac{2+x}{x}\) | A1 | o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(fg(x) = \frac{2}{x^2+4}\) (attempt) \(\left[\frac{2}{"g"-1}\right]\) | M1 | |
| Setting \(\frac{2}{x^2+4} = \frac{1}{4}\) and finding \(x^2 = \ldots\); \(x = \pm 2\) | M1; A1 |
## Question 3:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{5x+1}{(x+2)(x-1)} - \frac{3}{x+2}$ | B1 | |
| $= \frac{5x+1-3(x-1)}{(x+2)(x-1)}$ | M1 | M1 for combining fractions even if denominator is not lowest common |
| $= \frac{2x+4}{(x+2)(x-1)} = \frac{2(x+2)}{(x+2)(x-1)} = \frac{2}{x-1}$ * | M1 A1 cso | M1 must have linear numerator |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = \frac{2}{x-1} \Rightarrow xy - y = 2 \Rightarrow xy = 2 + y$ | M1A1 | |
| $f^{-1}(x) = \frac{2+x}{x}$ | A1 | o.e. |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $fg(x) = \frac{2}{x^2+4}$ (attempt) $\left[\frac{2}{"g"-1}\right]$ | M1 | |
| Setting $\frac{2}{x^2+4} = \frac{1}{4}$ and finding $x^2 = \ldots$; $x = \pm 2$ | M1; A1 | |
---3. The function $f$ is defined by
$$f : x \rightarrow \frac { 5 x + 1 } { x ^ { 2 } + x - 2 } - \frac { 3 } { x + 2 } , x > 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { 2 } { x - 1 } , x > 1$.
\item Find $\mathrm { f } ^ { - 1 } ( x )$.
The function $g$ is defined by
$$\mathrm { g } : x \rightarrow x ^ { 2 } + 5 , \quad x \in \mathbb { R }$$
\item Solve $\operatorname { fg } ( x ) = \frac { 1 } { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2005 Q3 [10]}}