| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Find inverse function after simplification |
| Difficulty | Standard +0.3 This is a multi-part question requiring algebraic manipulation (partial fractions in reverse), finding an inverse function, and solving an equation. While it involves several steps, each technique is standard C3 material: simplifying rational expressions, the routine inverse function algorithm (swap x and y, rearrange), and straightforward equation solving. The algebraic manipulation in part (a) is slightly more involved than basic exercises but follows predictable patterns, making this slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02v Inverse and composite functions: graphs and conditions for existence1.02y Partial fractions: decompose rational functions |
| Answer | Marks |
|---|---|
| \(f(x) = 1 + \frac{4x}{2x - 5} - \frac{15}{(2x - 5)(x - 1)}\) | B1 |
| \(= \frac{2x^2 - 7x + 5 + 4x(x - 1) - 15}{(2x - 5)(x - 1)}\) | M1 A1 |
| \(= \frac{6x^2 - 11x - 10}{(2x - 5)(x - 1)} = \frac{(3x + 2)(2x - 5)}{(2x - 5)(x - 1)} = \frac{3x + 2}{x - 1}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{3x + 2}{x - 1}, \quad y(x - 1) = 3x + 2\) | M1 | |
| \(xy - 3x = y + 2\) | M1 | |
| \(x = \frac{y + 2}{y - 3}\) | ||
| \(\therefore f^{-1}(x) = \frac{x + 2}{x - 3}\) | A1 | |
| \(f(x) = 2 \Rightarrow x = f^{-1}(2) = -4\) | M1 A1 | |
| \(f(x) = 2 \Rightarrow x = f^{-1}(2) = -4\) | M1 A1 | (12) |
| Answer | Marks |
|---|---|
| \(f(x) = 2 \Rightarrow x = f^{-1}(2) = -4\) | M1 A1 |
**(a)**
$f(x) = 1 + \frac{4x}{2x - 5} - \frac{15}{(2x - 5)(x - 1)}$ | B1 |
$= \frac{2x^2 - 7x + 5 + 4x(x - 1) - 15}{(2x - 5)(x - 1)}$ | M1 A1 |
$= \frac{6x^2 - 11x - 10}{(2x - 5)(x - 1)} = \frac{(3x + 2)(2x - 5)}{(2x - 5)(x - 1)} = \frac{3x + 2}{x - 1}$ | M1 A1 |
**(b)**
$y = \frac{3x + 2}{x - 1}, \quad y(x - 1) = 3x + 2$ | M1 |
$xy - 3x = y + 2$ | M1 |
$x = \frac{y + 2}{y - 3}$ |
$\therefore f^{-1}(x) = \frac{x + 2}{x - 3}$ | A1 |
$f(x) = 2 \Rightarrow x = f^{-1}(2) = -4$ | M1 A1 |
$f(x) = 2 \Rightarrow x = f^{-1}(2) = -4$ | M1 A1 | (12)
**(c)**
$f(x) = 2 \Rightarrow x = f^{-1}(2) = -4$ | M1 A1 |7.
$$f ( x ) = 1 + \frac { 4 x } { 2 x - 5 } - \frac { 15 } { 2 x ^ { 2 } - 7 x + 5 } , \quad x \in \mathbb { R } , \quad x < 1$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$f ( x ) = \frac { 3 x + 2 } { x - 1 }$$
\item Find an expression for the inverse function $\mathrm { f } ^ { - 1 } ( x )$ and state its domain.
\item Solve the equation $\mathrm { f } ( x ) = 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q7 [12]}}