| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions then differentiate |
| Difficulty | Standard +0.3 This is a straightforward partial fractions question with standard techniques: cover-up method or substitution for parts A, B, C, then differentiate the separated form and show derivative is negative. The algebra is routine and the proof in part (c) requires only checking the sign of f'(x), which is a standard textbook exercise slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6-5x-4x^2 = A(2-x)(1+2x)+B(1+2x)+C(2-x)\) | M1 | Write in correct form and attempt to find any constant |
| Coefficients of \(x^2 \Rightarrow A=2\) | B1 | For the \(2+\ldots\) OR \(A=2\) |
| Sub \(x=2 \Rightarrow -20=5B \Rightarrow B=-4\); \(x=-\frac{1}{2} \Rightarrow 7.5=2.5C \Rightarrow C=3\) | dM1A1 | Dependent on M1; correct method for \(B\) or \(C\). A1 for \(\frac{-4}{(2-x)}+\frac{3}{(1+2x)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x)=2-\frac{4}{(2-x)}+\frac{3}{(1+2x)} \Rightarrow f'(x)=-\frac{4}{(2-x)^2}-\frac{6}{(1+2x)^2}\) | M1A1ftA1 | M1: chain rule application: \(\frac{\ldots}{(2-x)}\rightarrow\frac{\ldots}{(2-x)^2}\) or \(\frac{\ldots}{(1+2x)}\rightarrow\frac{\ldots}{(1+2x)^2}\). Allow full marks even if \(A\) incorrect in (a). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| As \((2-x)^2>0\) and \((1+2x)^2>0 \Rightarrow f'(x)<0\) | B1 | Depends on fully correct derivative in (b); squares always positive. Attempts based on \(f(x)\) alone sent to review. |
# Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6-5x-4x^2 = A(2-x)(1+2x)+B(1+2x)+C(2-x)$ | M1 | Write in correct form and attempt to find any constant |
| Coefficients of $x^2 \Rightarrow A=2$ | B1 | For the $2+\ldots$ OR $A=2$ |
| Sub $x=2 \Rightarrow -20=5B \Rightarrow B=-4$; $x=-\frac{1}{2} \Rightarrow 7.5=2.5C \Rightarrow C=3$ | dM1A1 | Dependent on M1; correct method for $B$ or $C$. A1 for $\frac{-4}{(2-x)}+\frac{3}{(1+2x)}$ |
# Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x)=2-\frac{4}{(2-x)}+\frac{3}{(1+2x)} \Rightarrow f'(x)=-\frac{4}{(2-x)^2}-\frac{6}{(1+2x)^2}$ | M1A1ftA1 | M1: chain rule application: $\frac{\ldots}{(2-x)}\rightarrow\frac{\ldots}{(2-x)^2}$ or $\frac{\ldots}{(1+2x)}\rightarrow\frac{\ldots}{(1+2x)^2}$. **Allow full marks even if $A$ incorrect in (a).** |
# Question 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| As $(2-x)^2>0$ and $(1+2x)^2>0 \Rightarrow f'(x)<0$ | B1 | Depends on fully correct derivative in (b); squares always positive. Attempts based on $f(x)$ alone sent to review. |
---5.
$$\frac { 6 - 5 x - 4 x ^ { 2 } } { ( 2 - x ) ( 1 + 2 x ) } \equiv A + \frac { B } { ( 2 - x ) } + \frac { C } { ( 1 + 2 x ) }$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A , B$ and $C$.
$$f ( x ) = \frac { 6 - 5 x - 4 x ^ { 2 } } { ( 2 - x ) ( 1 + 2 x ) } \quad x > 2$$
\item Using part (a), find $\mathrm { f } ^ { \prime } ( x )$.
\item Prove that $\mathrm { f } ( x )$ is a decreasing function.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q5 [8]}}