| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polynomial Division & Manipulation |
| Type | Differentiation of Simplified Fractions |
| Difficulty | Standard +0.3 This is a straightforward algebraic manipulation question requiring polynomial division/simplification followed by routine quotient rule differentiation. Part (a) involves factoring a quadratic and combining fractions (standard C3 skill), part (b) is direct application of quotient rule, and part (c) requires simple sign analysis of the derivative. All steps are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 - 3kx + 2k^2 = (x-2k)(x-k)\) | B1 | Seeing this factorisation anywhere in solution |
| \(2 - \frac{(x-5k)(x-k)}{(x-2k)(x-k)} = 2 - \frac{(x-5k)}{(x-2k)} = \frac{2(x-2k)-(x-5k)}{(x-2k)}\) | M1 | Writing as single term or two terms with same denominator |
| \(= \frac{x+k}{(x-2k)}\) | A1* | Proceeds without any errors (including bracketing) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Applies \(\frac{vu'-uv'}{v^2}\) to \(y = \frac{x+k}{x-2k}\) with \(u=x+k\) and \(v=x-2k\) | M1, A1 | |
| \(f'(x) = \frac{(x-2k)\times 1-(x+k)\times 1}{(x-2k)^2}\) | ||
| \(f'(x) = \frac{-3k}{(x-2k)^2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If \(f'(x) = \frac{-Ck}{(x-2k)^2} \Rightarrow f(x)\) is an increasing function as \(f'(x) > 0\) | M1 | |
| \(f'(x) = \frac{-3k}{(x-2k)^2} > 0\) for all values of \(x\) as \(\frac{\text{negative}\times\text{negative}}{\text{positive}} = \text{positive}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Applies \(\frac{vu'-uv'}{v^2}\) to \(y = \frac{x+k}{x-2k}\) with \(u = x+k\) and \(v = x-2k\) | M1 | If the rule is stated it must be correct. Can be implied by \(u = x+k\) and \(v = x-2k\) with their \(u'\), \(v'\) and \(\frac{vu'-uv'}{v^2}\). If neither stated nor implied only accept expressions of the form \(f'(x) = \frac{x-2k-x\pm k}{(x-2k)^2}\). Mark can also be scored for applying product rule to \(y=(x+k)(x-2k)^{-1}\). Alternatively writes \(y = 1 + \frac{3k}{x-2k}\) and differentiates to \(\frac{dy}{dx} = \frac{A}{(x-2k)^2}\) |
| Any correct form (unsimplified) of \(f'(x)\), e.g. \(f'(x) = \frac{(x-2k)\times 1-(x+k)\times 1}{(x-2k)^2}\) by quotient rule, or \(f'(x) = (x-2k)^{-1}-(x+k)(x-2k)^{-2}\) by product rule, or \(f'(x) = \frac{-3k}{(x-2k)^2}\) by third method | A1 | |
| cao \(f'(x) = \frac{-3k}{(x-2k)^2}\). Allow \(f'(x) = \frac{-3k}{x^2-4kx+4k^2}\) | A1 | As this answer is not given, candidates may recover from missing brackets |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Note that this is B1 B1 on e-pen. We are scoring it M1 A1 | — | |
| If in part (b) \(f'(x) = \frac{-Ck}{(x-2k)^2}\), look for \(f(x)\) is an increasing function as \(f'(x)/\text{gradient} > 0\). Accept a version that states as \(k < 0 \Rightarrow -Ck > 0\) hence increasing. If in part (b) \(f'(x) = \frac{(+)Ck}{(x-2k)^2}\), look for \(f(x)\) is a decreasing function as \(f'(x)/\text{gradient} < 0\). Similarly accept a version that states as \(k < 0 \Rightarrow (+)Ck < 0\) hence decreasing | M1 | |
| Must have \(f'(x) = \frac{-3k}{(x-2k)^2}\) and give a reason that links the gradient with its sign. There must have been reference to the sign of both numerator and denominator to justify the overall positive sign. | A1 |
# Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 3kx + 2k^2 = (x-2k)(x-k)$ | B1 | Seeing this factorisation anywhere in solution |
| $2 - \frac{(x-5k)(x-k)}{(x-2k)(x-k)} = 2 - \frac{(x-5k)}{(x-2k)} = \frac{2(x-2k)-(x-5k)}{(x-2k)}$ | M1 | Writing as single term or two terms with same denominator |
| $= \frac{x+k}{(x-2k)}$ | A1* | Proceeds without any errors (including bracketing) |
**(3 marks)**
---
# Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies $\frac{vu'-uv'}{v^2}$ to $y = \frac{x+k}{x-2k}$ with $u=x+k$ and $v=x-2k$ | M1, A1 | |
| $f'(x) = \frac{(x-2k)\times 1-(x+k)\times 1}{(x-2k)^2}$ | | |
| $f'(x) = \frac{-3k}{(x-2k)^2}$ | A1 | |
**(3 marks)**
---
# Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| If $f'(x) = \frac{-Ck}{(x-2k)^2} \Rightarrow f(x)$ is an increasing function as $f'(x) > 0$ | M1 | |
| $f'(x) = \frac{-3k}{(x-2k)^2} > 0$ for all values of $x$ as $\frac{\text{negative}\times\text{negative}}{\text{positive}} = \text{positive}$ | A1 | |
**(2 marks)**
## Question (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Applies $\frac{vu'-uv'}{v^2}$ to $y = \frac{x+k}{x-2k}$ with $u = x+k$ and $v = x-2k$ | M1 | If the rule is stated it must be correct. Can be implied by $u = x+k$ and $v = x-2k$ with their $u'$, $v'$ and $\frac{vu'-uv'}{v^2}$. If neither stated nor implied only accept expressions of the form $f'(x) = \frac{x-2k-x\pm k}{(x-2k)^2}$. Mark can also be scored for applying product rule to $y=(x+k)(x-2k)^{-1}$. Alternatively writes $y = 1 + \frac{3k}{x-2k}$ and differentiates to $\frac{dy}{dx} = \frac{A}{(x-2k)^2}$ |
| Any correct form (unsimplified) of $f'(x)$, e.g. $f'(x) = \frac{(x-2k)\times 1-(x+k)\times 1}{(x-2k)^2}$ by quotient rule, or $f'(x) = (x-2k)^{-1}-(x+k)(x-2k)^{-2}$ by product rule, or $f'(x) = \frac{-3k}{(x-2k)^2}$ by third method | A1 | |
| cao $f'(x) = \frac{-3k}{(x-2k)^2}$. Allow $f'(x) = \frac{-3k}{x^2-4kx+4k^2}$ | A1 | As this answer is not given, candidates may recover from missing brackets |
---
## Question (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Note that this is B1 B1 on e-pen. We are scoring it M1 A1 | — | |
| If in part (b) $f'(x) = \frac{-Ck}{(x-2k)^2}$, look for $f(x)$ is an increasing function as $f'(x)/\text{gradient} > 0$. Accept a version that states as $k < 0 \Rightarrow -Ck > 0$ hence increasing. If in part (b) $f'(x) = \frac{(+)Ck}{(x-2k)^2}$, look for $f(x)$ is a decreasing function as $f'(x)/\text{gradient} < 0$. Similarly accept a version that states as $k < 0 \Rightarrow (+)Ck < 0$ hence decreasing | M1 | |
| Must have $f'(x) = \frac{-3k}{(x-2k)^2}$ and give a reason that links the gradient with its sign. There must have been reference to the sign of both numerator and denominator to justify the overall positive sign. | A1 | |9. Given that $k$ is a negative constant and that the function $\mathrm { f } ( x )$ is defined by
$$f ( x ) = 2 - \frac { ( x - 5 k ) ( x - k ) } { x ^ { 2 } - 3 k x + 2 k ^ { 2 } } , \quad x \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item show that $\mathrm { f } ( x ) = \frac { x + k } { x - 2 k }$
\item Hence find $\mathrm { f } ^ { \prime } ( x )$, giving your answer in its simplest form.
\item State, with a reason, whether $\mathrm { f } ( x )$ is an increasing or a decreasing function.
Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2015 Q9 [8]}}