| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Differentiation of Simplified Fractions |
| Difficulty | Standard +0.3 This is a slightly above-average C3 question requiring algebraic fraction manipulation (factorising the quadratic denominator, finding common denominators, simplifying) followed by straightforward quotient rule differentiation. The steps are methodical and standard for C3, though the algebra requires care. The 'hence show that' in part (b) confirms students are on the right track. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2x+2}{x^2-2x-3} - \frac{x+1}{x-3} = \frac{2x+2}{(x-3)(x+1)} - \frac{x+1}{x-3}\) | M1 A1 | |
| \(= \frac{2x+2-(x+1)(x+1)}{(x-3)(x+1)}\) | M1 | |
| \(= \frac{(x+1)(1-x)}{(x-3)(x+1)}\) | M1 | |
| \(= \frac{1-x}{x-3}\) | Accept \(-\frac{x-1}{x-3}\), \(\frac{x-1}{3-x}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}\left(\frac{1-x}{x-3}\right) = \frac{(x-3)(-1)-{(1-x)}1}{(x-3)^2}\) | M1 A1 | |
| \(= \frac{-x+3-1+x}{(x-3)^2} = \frac{2}{(x-3)^2}\) | cso | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2x+2}{x^2-2x-3} - \frac{2(x+1)}{(x-3)(x+1)} - \frac{2}{x-3}\) | M1 A1 | |
| \(\frac{2}{x-3} - \frac{x+1}{x-3} = \frac{2-(x+1)}{x-3}\) | M1 | |
| \(= \frac{1-x}{x-3}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = -1 - \frac{2}{x-3} = -1 - 2(x-3)^{-1}\) | M1 A1 | |
| \(f'(x) = (-1)(-2)(x-3)^{-2} = \frac{2}{(x-3)^2}\) | cso | A1 |
| \(f(x) = (1-x)(x-3)^{-1}\) | M1 | |
| \(f'(x) = (-1)(x-3) + (1-x)(-1)(x-3)^{-2}\) | M1 | |
| \(= \frac{-1}{x-3} - \frac{1-x}{(x-3)^2} = \frac{-(x-3)-(1-x)}{(x-3)^2}\) | A1 | |
| \(= \frac{2}{(x-3)^2}\) | A1 | (3) |
**(a)**
$\frac{2x+2}{x^2-2x-3} - \frac{x+1}{x-3} = \frac{2x+2}{(x-3)(x+1)} - \frac{x+1}{x-3}$ | M1 A1
$= \frac{2x+2-(x+1)(x+1)}{(x-3)(x+1)}$ | M1
$= \frac{(x+1)(1-x)}{(x-3)(x+1)}$ | M1
$= \frac{1-x}{x-3}$ | Accept $-\frac{x-1}{x-3}$, $\frac{x-1}{3-x}$ | A1 | (4)
**(b)**
$\frac{d}{dx}\left(\frac{1-x}{x-3}\right) = \frac{(x-3)(-1)-{(1-x)}1}{(x-3)^2}$ | M1 A1
$= \frac{-x+3-1+x}{(x-3)^2} = \frac{2}{(x-3)^2}$ | cso | A1 | (3) [7]
**Alternative to (a):**
$\frac{2x+2}{x^2-2x-3} - \frac{2(x+1)}{(x-3)(x+1)} - \frac{2}{x-3}$ | M1 A1
$\frac{2}{x-3} - \frac{x+1}{x-3} = \frac{2-(x+1)}{x-3}$ | M1
$= \frac{1-x}{x-3}$ | A1 | (4)
**Alternatives to (b):**
$f(x) = -1 - \frac{2}{x-3} = -1 - 2(x-3)^{-1}$ | M1 A1
$f'(x) = (-1)(-2)(x-3)^{-2} = \frac{2}{(x-3)^2}$ | cso | A1 | (3)
$f(x) = (1-x)(x-3)^{-1}$ | M1
$f'(x) = (-1)(x-3) + (1-x)(-1)(x-3)^{-2}$ | M1
$= \frac{-1}{x-3} - \frac{1-x}{(x-3)^2} = \frac{-(x-3)-(1-x)}{(x-3)^2}$ | A1
$= \frac{2}{(x-3)^2}$ | A1 | (3)
---2.
$$f ( x ) = \frac { 2 x + 2 } { x ^ { 2 } - 2 x - 3 } - \frac { x + 1 } { x - 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Express $\mathrm { f } ( x )$ as a single fraction in its simplest form.
\item Hence show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { ( x - 3 ) ^ { 2 } }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2009 Q2 [7]}}