| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Find inverse function after simplification |
| Difficulty | Standard +0.3 This question requires combining algebraic fractions using partial fractions in reverse, recognizing that 4x²-25 = (2x+5)(2x-5), then finding an inverse of a simple linear fractional function. While it involves multiple steps, each technique is standard C3 material with no novel insight required—slightly easier than average due to the straightforward algebraic manipulation and routine inverse function work. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.02y Partial fractions: decompose rational functions |
| Answer | Marks |
|---|---|
| \(4x^2 - 25 \rightarrow (2x+5)(2x-5)\) | B1 |
| \(\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{6(2x-5)+2(2x+5)+60}{(2x+5)(2x-5)}\) | M1 |
| \(= \frac{16x+40}{(2x+5)(2x-5)}\) | A1 |
| \(= \frac{8(2x+5)}{(2x+5)(2x-5)} = \frac{8}{2x-5}\) | A1 |
| Answer | Marks |
|---|---|
| \(f(x) = \frac{8}{2x-5} \Rightarrow y = \frac{8}{2x-5} \Rightarrow 2xy - 5y = 8 \Rightarrow x = \frac{8+5y}{2y}\) | M1 |
| \(\Rightarrow f^{-1}(x) = \frac{8+5x}{2x}\) oe | A1 |
| \(0 < x < \frac{8}{3}\) | B1ft |
| Answer | Marks |
|---|---|
| \(\frac{6}{2x+5} + \frac{2}{2x-5} = \frac{16x-20}{4x^2-25}\) | B1, M1 |
| \(\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{16x-20+60}{4x^2-25} = \frac{16x+40}{4x^2-25}\) | M1, A1 |
| \(= \frac{8(2x+5)}{(2x+5)(2x-5)} = \frac{8}{2x-5}\) | A1 |
| Answer | Marks |
|---|---|
| \(\frac{60}{4x^2-25} = \frac{-6}{2x+5} + \frac{6}{2x-5}\) | B1, M1 |
| \(\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{-6}{2x+5} + \frac{6}{2x-5}\) | A1 |
| \(= \frac{8}{(2x-5)}\) | A1 |
**(a)**
| $4x^2 - 25 \rightarrow (2x+5)(2x-5)$ | B1 |
| $\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{6(2x-5)+2(2x+5)+60}{(2x+5)(2x-5)}$ | M1 |
| $= \frac{16x+40}{(2x+5)(2x-5)}$ | A1 |
| $= \frac{8(2x+5)}{(2x+5)(2x-5)} = \frac{8}{2x-5}$ | A1 |
(4 marks)
**(b)**
| $f(x) = \frac{8}{2x-5} \Rightarrow y = \frac{8}{2x-5} \Rightarrow 2xy - 5y = 8 \Rightarrow x = \frac{8+5y}{2y}$ | M1 |
| $\Rightarrow f^{-1}(x) = \frac{8+5x}{2x}$ oe | A1 |
| $0 < x < \frac{8}{3}$ | B1ft |
(3 marks)
(7 marks total)
**Alternative solutions to part (a):**
**ALT I:**
| $\frac{6}{2x+5} + \frac{2}{2x-5} = \frac{16x-20}{4x^2-25}$ | B1, M1 |
| $\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{16x-20+60}{4x^2-25} = \frac{16x+40}{4x^2-25}$ | M1, A1 |
| $= \frac{8(2x+5)}{(2x+5)(2x-5)} = \frac{8}{2x-5}$ | A1 |
**ALT II:**
| $\frac{60}{4x^2-25} = \frac{-6}{2x+5} + \frac{6}{2x-5}$ | B1, M1 |
| $\frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{-6}{2x+5} + \frac{6}{2x-5}$ | A1 |
| $= \frac{8}{(2x-5)}$ | A1 |
---\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$\mathrm { f } ( x ) = \frac { 6 } { 2 x + 5 } + \frac { 2 } { 2 x - 5 } + \frac { 60 } { 4 x ^ { 2 } - 25 } , \quad x > 4$$
(a) Show that $\mathrm { f } ( x ) = \frac { A } { B x + C }$ where $A , B$ and $C$ are constants to be found.\\
(b) Find $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.\\
\hfill \mbox{\textit{Edexcel C3 2018 Q2 [7]}}