| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| 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 manipulation (factorising a quadratic, finding common denominators, simplifying) followed by routine quotient rule differentiation. The 'show that' structure provides guidance, and while the algebra requires care, it involves standard techniques without novel insight. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\) at any stage | B1 | |
| \(f(x) = \frac{(2x+3)(2x-1)-(9+2x)}{(2x-1)(x+2)}\) | M1, A1√ | |
| f.t. on error in denominator factors | ||
| Simplifying numerator to quadratic form: \(\left[\frac{4x^2+4x-3-9-2x}{(2x-1)(x+2)}\right]\) | M1 | |
| Correct numerator: \(4x^2 + 2x - 12\) | A1 | |
| Factorising numerator, with a denominator: \(\frac{2(2x-3)(x+2)}{(2x-1)(x+2)}\) o.e. | M1 | |
| \(= \frac{4x-6}{2x-1}\) (✱) | A1 cso | (7 marks) |
| Alt.(a) \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\) at any stage | B1 | |
| \(f(x) = \frac{(2x+3)(2x^2+3x-2)-(9+2x)(x+2)}{(x+2)(2x^2+3x-2)}\) | M1A1 f.t. | |
| \(= \frac{4x^3+10x^2-8x-24}{(x+2)(2x^2+3x-2)}\) | ||
| \(= \frac{2(x+2)(2x^2+x-6)}{(x+2)(2x^2+3x-2)}\) or \(\frac{2(2x-3)(x^2+4x+4)}{(x+2)(2x^2+3x-2)}\) o.e. | M1, A1 | |
| Any one linear factor × quadratic factor in numerator | M1, A1 | |
| \(= \frac{2(x+2)(x+2)(2x-3)}{(x+2)(2x^2+3x-2)}\) o.e. | M1 | |
| \(= \frac{2(2x-3)}{2x-1}\) or \(\frac{4x-6}{2x-1}\) (✱) | A1 | |
| (b) Complete method for \(f'(x)\); e.g. \(f'(x) = \frac{(2x-1) \times 4 - (4x-6) \times 2}{(2x-1)^2}\) o.e | M1 A1 | |
| \(= \frac{8}{(2x-1)^2}\) or \(8(2x-1)^{-2}\) | A1 | (3 marks) |
| Not treating \(f^{-1}\) (for \(f'\)) as misread |
**(a)** $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ at any stage | B1 | |
$f(x) = \frac{(2x+3)(2x-1)-(9+2x)}{(2x-1)(x+2)}$ | M1, A1√ | |
f.t. on error in denominator factors | | |
Simplifying numerator to quadratic form: $\left[\frac{4x^2+4x-3-9-2x}{(2x-1)(x+2)}\right]$ | M1 | |
Correct numerator: $4x^2 + 2x - 12$ | A1 | |
Factorising numerator, with a denominator: $\frac{2(2x-3)(x+2)}{(2x-1)(x+2)}$ o.e. | M1 | |
$= \frac{4x-6}{2x-1}$ (**✱**) | A1 cso | (7 marks) |
**Alt.(a)** $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ at any stage | B1 | |
$f(x) = \frac{(2x+3)(2x^2+3x-2)-(9+2x)(x+2)}{(x+2)(2x^2+3x-2)}$ | M1A1 f.t. | |
$= \frac{4x^3+10x^2-8x-24}{(x+2)(2x^2+3x-2)}$ | | |
$= \frac{2(x+2)(2x^2+x-6)}{(x+2)(2x^2+3x-2)}$ or $\frac{2(2x-3)(x^2+4x+4)}{(x+2)(2x^2+3x-2)}$ o.e. | M1, A1 | |
Any one linear factor × quadratic factor in numerator | M1, A1 | |
$= \frac{2(x+2)(x+2)(2x-3)}{(x+2)(2x^2+3x-2)}$ o.e. | M1 | |
$= \frac{2(2x-3)}{2x-1}$ or $\frac{4x-6}{2x-1}$ (**✱**) | A1 | |
**(b)** Complete method for $f'(x)$; e.g. $f'(x) = \frac{(2x-1) \times 4 - (4x-6) \times 2}{(2x-1)^2}$ o.e | M1 A1 | |
$= \frac{8}{(2x-1)^2}$ or $8(2x-1)^{-2}$ | A1 | (3 marks) |
Not treating $f^{-1}$ (for $f'$) as misread | | |
**Notes:**
- 1st M1 in either version is for correct method
- 1st A1: Allow $\frac{2x+3(2x-1)-(9+2y)}{(2x-1)(x+2)}$ or $\frac{(2x+3)(2x-1)-(9+2x)}{(2x-1)(x+2)}$ or $\frac{2x+3(2x-1)-9+2x}{(2x-1)(x+2)}$ (fractions)
- 2nd M1 in (main a) is for forming 3 term quadratic in numerator
- 3rd M1 is for factorising resulting quadratic (usual rules): factor of 2 need not be extracted
- (**✱**) A1 is given answer so is cso
- Alt (a) 3rd M1 is for factorising resulting quadratic
- Notice that B1 likely to be scored very late but on ePen scored first
- SC: For M allow $\pm$ given expression or one error in product rule
- Alt: Attempt at $f(x) = 2 - 4(2x-1)^{-1}$ and diff. M1; $k(2x-1)^{-2}$ A1; A1 as above
- Accept $8(4x^2-4x+1)^{-1}$. Differentiating original function – mark as scheme.
---$$f ( x ) = \frac { 2 x + 3 } { x + 2 } - \frac { 9 + 2 x } { 2 x ^ { 2 } + 3 x - 2 } , \quad x > \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = \frac { 4 x - 6 } { 2 x - 1 }$.
\item Hence, or otherwise, find $\mathrm { f } ^ { \prime } ( x )$ in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q2 [10]}}