| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions then differentiate |
| Difficulty | Standard +0.3 This is a straightforward two-part question: (a) requires standard partial fractions decomposition with cover-up method, yielding A=3, B=4, C=5; (b) asks to prove f is decreasing by showing f'(x)<0, which follows directly from differentiating the partial fractions form. Both parts are routine A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07i Differentiate x^n: for rational n and sums1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1+11x-6x^2 \equiv A(1-2x)(x-3)+B(1-2x)+C(x-3)\), finds \(B=\ldots, C=\ldots\) | M1 | AO2.1 — Uses correct identity in complete method |
| \(A = 3\) | B1 | AO1.1b |
| Uses substitution or compares terms to find either \(B\) or \(C\) | M1 | AO1.1b |
| \(B=4\) and \(C=-2\) found using correct identity | A1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Long division gives \(\frac{1+11x-6x^2}{(x-3)(1-2x)} \equiv 3 + \frac{-10x+10}{(x-3)(1-2x)}\), then \(-10x+10 \equiv B(1-2x)+C(x-3)\) | M1 | AO2.1 |
| \(A=3\) | B1 | AO1.1b |
| Uses substitution or compares terms to find either \(B\) or \(C\) | M1 | AO1.1b |
| \(B=4\) and \(C=-2\) from \(-10x+10 \equiv B(1-2x)+C(x-3)\) | A1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = -4(x-3)^{-2} - 4(1-2x)^{-2}\) i.e. \(\left\{=\pm\lambda(x-3)^{-2} \pm \mu(1-2x)^{-2};\ \lambda,\mu\neq 0\right\}\) | M1 | AO2.1 |
| \(f'(x) = -4(x-3)^{-2} - 4(1-2x)^{-2}\), simplified or unsimplified | A1ft | AO1.1b — ft on their \(B\) and \(C\) |
| Correct \(f'(x)\) and since \((x-3)^2>0\) and \((1-2x)^2>0\), then \(f'(x) = -(+\text{ve})-(+\text{ve}) < 0\), so \(f(x)\) is a decreasing function | A1 | AO2.4 — Requires correct explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{20}{(x-3)(1-2x)} \equiv \frac{D}{(x-3)} + \frac{E}{(1-2x)} \Rightarrow 20 \equiv D(1-2x) + E(x-3) \Rightarrow D=-4, E=-8\) | — | Alternative via dividing by \((x-3)\) first |
| \(\Rightarrow 3 + \frac{4}{(x-3)} - \frac{2}{(1-2x)}\); \(A=3, B=4, C=-2\) | — | |
| \(\frac{5}{(x-3)(1-2x)} \equiv \frac{D}{(x-3)} + \frac{E}{(1-2x)} \Rightarrow D=-1, E=-2\) | — | Alternative via dividing by \((1-2x)\) first |
| \(\Rightarrow 3 + \frac{4}{(x-3)} - \frac{2}{(1-2x)}\); \(A=3, B=4, C=-2\) | — |
# Question 11:
## Part (a) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1+11x-6x^2 \equiv A(1-2x)(x-3)+B(1-2x)+C(x-3)$, finds $B=\ldots, C=\ldots$ | M1 | AO2.1 — Uses correct identity in complete method |
| $A = 3$ | B1 | AO1.1b |
| Uses substitution or compares terms to find either $B$ or $C$ | M1 | AO1.1b |
| $B=4$ and $C=-2$ found using correct identity | A1 | AO1.1b |
## Part (a) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Long division gives $\frac{1+11x-6x^2}{(x-3)(1-2x)} \equiv 3 + \frac{-10x+10}{(x-3)(1-2x)}$, then $-10x+10 \equiv B(1-2x)+C(x-3)$ | M1 | AO2.1 |
| $A=3$ | B1 | AO1.1b |
| Uses substitution or compares terms to find either $B$ or $C$ | M1 | AO1.1b |
| $B=4$ and $C=-2$ from $-10x+10 \equiv B(1-2x)+C(x-3)$ | A1 | AO1.1b |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = -4(x-3)^{-2} - 4(1-2x)^{-2}$ i.e. $\left\{=\pm\lambda(x-3)^{-2} \pm \mu(1-2x)^{-2};\ \lambda,\mu\neq 0\right\}$ | M1 | AO2.1 |
| $f'(x) = -4(x-3)^{-2} - 4(1-2x)^{-2}$, simplified or unsimplified | A1ft | AO1.1b — ft on their $B$ and $C$ |
| Correct $f'(x)$ **and** since $(x-3)^2>0$ and $(1-2x)^2>0$, then $f'(x) = -(+\text{ve})-(+\text{ve}) < 0$, so $f(x)$ is a decreasing function | A1 | AO2.4 — Requires correct explanation |
# Question 11 (Alternatives):
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{20}{(x-3)(1-2x)} \equiv \frac{D}{(x-3)} + \frac{E}{(1-2x)} \Rightarrow 20 \equiv D(1-2x) + E(x-3) \Rightarrow D=-4, E=-8$ | — | Alternative via dividing by $(x-3)$ first |
| $\Rightarrow 3 + \frac{4}{(x-3)} - \frac{2}{(1-2x)}$; $A=3, B=4, C=-2$ | — | |
| $\frac{5}{(x-3)(1-2x)} \equiv \frac{D}{(x-3)} + \frac{E}{(1-2x)} \Rightarrow D=-1, E=-2$ | — | Alternative via dividing by $(1-2x)$ first |
| $\Rightarrow 3 + \frac{4}{(x-3)} - \frac{2}{(1-2x)}$; $A=3, B=4, C=-2$ | — | |
---11.
$$\frac { 1 + 11 x - 6 x ^ { 2 } } { ( x - 3 ) ( 1 - 2 x ) } \equiv A + \frac { B } { ( x - 3 ) } + \frac { C } { ( 1 - 2 x ) }$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constants $A , B$ and $C$.
$$f ( x ) = \frac { 1 + 11 x - 6 x ^ { 2 } } { ( x - 3 ) ( 1 - 2 x ) } \quad x > 3$$
\item Prove that $\mathrm { f } ( x )$ is a decreasing function.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q11 [7]}}