| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Repeated linear factor with distinct linear factor – decompose only |
| Difficulty | Moderate -0.3 This is a slightly below-average A-level question. Part (a) involves identifying an error in partial fractions work (accessible error-spotting), while part (b) is a standard partial fractions decomposition with a repeated linear factor—a routine technique covered in C4/FP1. The question requires methodical application of the cover-up rule and equation solving, but no novel insight or complex multi-step reasoning. |
| Spec | 1.02y Partial fractions: decompose rational functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(x=0 \Rightarrow LHS = 0\); \(RHS = \frac{1}{1} - \frac{6}{4} = -\frac{1}{2} \neq 0\) | M1 (AO 2.2a) | Deduces appropriate value for \(x\) and substitutes into at least one side; any value of \(x \neq -2,-1\) |
| \(\therefore\) Chloe's answer must be incorrect | R1 (AO 2.1) | Shows LHS \(\neq\) RHS and concludes Chloe's answer must be incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Chloe should have included \(\dfrac{c}{x+2}\) | E1 (AO 2.3) | Explains Chloe should have included additional term with \((x+2)\) in denominator, or that numerator for \((x+2)^2\) should be \((Bx+C)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{2x^2+x}{(x+1)(x+2)^2} \equiv \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}\) | M1 (AO 1.1a) | Writes identity of correct form; condone use of equals signs |
| \(2x^2 + x \equiv A(x+2)^2 + B(x+1)(x+2) + C(x+1)\) | M1 (AO 3.1a) | Uses suitable method to obtain all three constants; only award if identity used from correctly removing fractions |
| \(x=-1 \Rightarrow A=1\); \(x=-2 \Rightarrow C=-6\); \(x^2: A+B=2 \Rightarrow B=1\) | A1 (AO 1.1b) | Any two correct constants; if \(Bx+C\) used then \(B=1\) and \(C=-4\) |
| \(\dfrac{2x^2+x}{(x+1)(x+2)^2} \equiv \dfrac{1}{x+1} + \dfrac{1}{x+2} - \dfrac{6}{(x+2)^2}\) | A1 (AO 1.1b) | All three correct values |
## Question 9(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $x=0 \Rightarrow LHS = 0$; $RHS = \frac{1}{1} - \frac{6}{4} = -\frac{1}{2} \neq 0$ | M1 (AO 2.2a) | Deduces appropriate value for $x$ and substitutes into at least one side; any value of $x \neq -2,-1$ |
| $\therefore$ Chloe's answer must be incorrect | R1 (AO 2.1) | Shows LHS $\neq$ RHS and concludes Chloe's answer must be incorrect |
---
## Question 9(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Chloe should have included $\dfrac{c}{x+2}$ | E1 (AO 2.3) | Explains Chloe should have included additional term with $(x+2)$ in denominator, or that numerator for $(x+2)^2$ should be $(Bx+C)$ |
---
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{2x^2+x}{(x+1)(x+2)^2} \equiv \dfrac{A}{x+1} + \dfrac{B}{x+2} + \dfrac{C}{(x+2)^2}$ | M1 (AO 1.1a) | Writes identity of correct form; condone use of equals signs |
| $2x^2 + x \equiv A(x+2)^2 + B(x+1)(x+2) + C(x+1)$ | M1 (AO 3.1a) | Uses suitable method to obtain all three constants; only award if identity used from correctly removing fractions |
| $x=-1 \Rightarrow A=1$; $x=-2 \Rightarrow C=-6$; $x^2: A+B=2 \Rightarrow B=1$ | A1 (AO 1.1b) | Any two correct constants; if $Bx+C$ used then $B=1$ and $C=-4$ |
| $\dfrac{2x^2+x}{(x+1)(x+2)^2} \equiv \dfrac{1}{x+1} + \dfrac{1}{x+2} - \dfrac{6}{(x+2)^2}$ | A1 (AO 1.1b) | All three correct values |
---9
\begin{enumerate}[label=(\alph*)]
\item (i) By using a counter example, show that the answer obtained by Chloe cannot be correct.\\
9 (a) (ii) Explain her mistake in Step 1.\\
9
\item Write $\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } }$ as partial fractions, with constant numerators.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2020 Q9 [7]}}