| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Partial fractions with linear factors – decompose and integrate (definite) |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard A-level techniques: identifying asymptotes from denominators (trivial), substituting a point to find constants (routine algebra), and integrating partial fractions to find area (standard method). While it requires multiple steps, each individual step is a textbook exercise with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) Explains \(2x - q = 0\) when \(x = 2\), hence \(q = 4\) | B1* | Must have words explaining why \(q=4\), referencing asymptote \(x=2\), curve undefined at \(x=2\), or denominator is 0 at \(x=2\) |
| (ii) Substitutes \(\left(3, \frac{1}{2}\right)\) into \(y = \frac{p-3x}{(2x-4)(x+3)}\) and solves | M1 | Alternatively substitutes into \(y = \frac{15-3x}{(2x-4)(x+3)}\) and shows \(\frac{1}{2} = \frac{6}{(2)\times(6)}\) |
| \(\frac{1}{2} = \frac{p-9}{(2)\times(6)} \Rightarrow p - 9 = 6 \Rightarrow p = 15\) | A1* | Full proof showing all necessary steps |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to write \(\frac{15-3x}{(2x-4)(x+3)}\) in partial fractions and integrates using ln, between 3 and another value of \(x\) | M1 | Overall attempt at PFs and integrating with lns, with sight of limits 3 and another value |
| \(\frac{15-3x}{(2x-4)(x+3)} = \frac{A}{(2x-4)} + \frac{B}{(x+3)}\) leading to \(A\) and \(B\) | M1 | |
| \(\frac{15-3x}{(2x-4)(x+3)} = \frac{1.8}{(2x-4)} - \frac{2.4}{(x+3)}\) or \(\frac{0.9}{(x-2)} - \frac{2.4}{(x+3)}\) | A1 | Must be written in PF form, not just correct \(A\) and \(B\) |
| \(I = \int \frac{15-3x}{(2x-4)(x+3)}dx = m\ln(2x-4) + n\ln(x+3) + (c)\) | M1 | |
| \(I = 0.9\ln(2x-4) - 2.4\ln(x+3)\) | A1ft | FT on their \(A\) and \(B\) |
| Deduces Area \(= \int_3^5 \frac{15-3x}{(2x-4)(x+3)}dx\) or \([\ldots]_3^5\) | B1 | Cannot be awarded just from 5 being marked on figure |
| Uses \(\ln 6 = \ln 2 + \ln 3\) or \(\ln 8 = 3\ln 2\): \([0.9\ln(6) - 2.4\ln(8)] - [0.9\ln(2) - 2.4\ln(6)] = 3.3\ln 6 - 7.2\ln 2 - 0.9\ln 2\) | dM1 | Dependent on correct limits and having achieved \(m\ln(2x-4)+n\ln(x+3)\) |
| \(= 3.3\ln 3 - 4.8\ln 2\) | A1 |
# Question 13:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) Explains $2x - q = 0$ when $x = 2$, hence $q = 4$ | B1* | Must have words explaining why $q=4$, referencing asymptote $x=2$, curve undefined at $x=2$, or denominator is 0 at $x=2$ |
| (ii) Substitutes $\left(3, \frac{1}{2}\right)$ into $y = \frac{p-3x}{(2x-4)(x+3)}$ and solves | M1 | Alternatively substitutes into $y = \frac{15-3x}{(2x-4)(x+3)}$ and shows $\frac{1}{2} = \frac{6}{(2)\times(6)}$ |
| $\frac{1}{2} = \frac{p-9}{(2)\times(6)} \Rightarrow p - 9 = 6 \Rightarrow p = 15$ | A1* | Full proof showing all necessary steps |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to write $\frac{15-3x}{(2x-4)(x+3)}$ in partial fractions and integrates using ln, between 3 and another value of $x$ | M1 | Overall attempt at PFs and integrating with lns, with sight of limits 3 and another value |
| $\frac{15-3x}{(2x-4)(x+3)} = \frac{A}{(2x-4)} + \frac{B}{(x+3)}$ leading to $A$ and $B$ | M1 | |
| $\frac{15-3x}{(2x-4)(x+3)} = \frac{1.8}{(2x-4)} - \frac{2.4}{(x+3)}$ or $\frac{0.9}{(x-2)} - \frac{2.4}{(x+3)}$ | A1 | Must be written in PF form, not just correct $A$ and $B$ |
| $I = \int \frac{15-3x}{(2x-4)(x+3)}dx = m\ln(2x-4) + n\ln(x+3) + (c)$ | M1 | |
| $I = 0.9\ln(2x-4) - 2.4\ln(x+3)$ | A1ft | FT on their $A$ and $B$ |
| Deduces Area $= \int_3^5 \frac{15-3x}{(2x-4)(x+3)}dx$ or $[\ldots]_3^5$ | B1 | Cannot be awarded just from 5 being marked on figure |
| Uses $\ln 6 = \ln 2 + \ln 3$ or $\ln 8 = 3\ln 2$: $[0.9\ln(6) - 2.4\ln(8)] - [0.9\ln(2) - 2.4\ln(6)] = 3.3\ln 6 - 7.2\ln 2 - 0.9\ln 2$ | dM1 | Dependent on correct limits and having achieved $m\ln(2x-4)+n\ln(x+3)$ |
| $= 3.3\ln 3 - 4.8\ln 2$ | A1 | |
---\begin{enumerate}
\item The curve $C$ with equation
\end{enumerate}
$$y = \frac { p - 3 x } { ( 2 x - q ) ( x + 3 ) } \quad x \in \mathbb { R } , x \neq - 3 , x \neq 2$$
where $p$ and $q$ are constants, passes through the point $\left( 3 , \frac { 1 } { 2 } \right)$ and has two vertical asymptotes\\
with equations $x = 2$ and $x = - 3$ with equations $x = 2$ and $x = - 3$\\
(a) (i) Explain why you can deduce that $q = 4$\\
(ii) Show that $p = 15$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{91a2f26a-add2-4b58-997d-2ae229548217-38_616_889_842_587}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of part of the curve $C$. The region $R$, shown shaded in Figure 4, is bounded by the curve $C$, the $x$-axis and the line with equation $x = 3$\\
(b) Show that the exact value of the area of $R$ is $a \ln 2 + b \ln 3$, where $a$ and $b$ are rational constants to be found.
\hfill \mbox{\textit{Edexcel Paper 1 2019 Q13 [11]}}