| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions in differential equations |
| Difficulty | Standard +0.8 This is a multi-step Further Maths question requiring partial fractions decomposition, separation of variables, integration of both 1/tan(y) and the partial fractions, and applying boundary conditions to find a constant. While the techniques are standard for P4/FM, the combination of steps and the need to manipulate the final answer into the specified form (sin^n y = f(x)) elevates it above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{(1+3x)(1-x)}=\frac{A}{1+3x}+\frac{B}{1-x} \Rightarrow 1=A(1-x)+B(1+3x)\) | B1 | Correct suitable identity without fractions, seen or implied |
| When \(x=1\): \(1=4B\Rightarrow B=...\) or when \(x=-\frac{1}{3}\): \(1=\frac{4}{3}A\Rightarrow A=...\) | M1 | Attempts to find one constant by substitution or equating coefficients |
| \(\frac{3}{4(1+3x)}+\frac{1}{4(1-x)}\) | A1 | Allow values for \(A\) and \(B\) stated following correct partial fraction form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int\cot y\,dy=\int...\,dx \Rightarrow "\ln\sin y"=\int...\,dx\) | M1 | Attempts to separate variables to form \(\cot y\frac{dy}{dx}=g(x)\); accept any changed function for attempt but must be attempting to integrate \(\cot y\) |
| \(...=\int\left(\frac{"3"}{4(1+3x)}+\frac{"1"}{4(1-x)}\right)dx=...\ln(1+3x)\pm...\ln(1-x)\ (+c)\) | M1 | Attempts to integrate partial fractions; award for \(\frac{...}{(1+3x)}\rightarrow...\ln(1+3x)\) or \(...\ln(4+12x)\) and \(\frac{...}{(1-x)}\rightarrow...\ln(1-x)\) or \(...\ln(4-4x)\) |
| \(\ln\sin y=\frac{1}{4}\ln(1+3x)-\frac{1}{4}\ln(1-x)\ (+c)\) oe | A1ft | Correct expression (any equivalent); follow through on constants for partial fractions; condone absence of constant of integration |
| \(\ln\sin(\frac{\pi}{2})=\frac{1}{4}\ln(1+3\times\frac{1}{2})-\frac{1}{4}\ln(1-\frac{1}{2})+c\Rightarrow c=...\left(=-\frac{1}{4}\ln 5\right)\) | dM1 | Depends on second M; uses initial conditions in equation with constant of integration; may integrate between limits |
| \(k\ln\sin y=m\ln(...)\Rightarrow\sin^k y=...^m\) or \(k\ln\sin y=...\Rightarrow\sin^k y=\exp(...)\) | M1 | Attempts to rearrange using log work to reach \(\sin^n y=f(x)\); must have had \(k\ln\sin y=...\); not dependent |
| \(\sin^4 y=\frac{1+3x}{5(1-x)}\) | A1 | oe in correct form |
# Question 2:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{(1+3x)(1-x)}=\frac{A}{1+3x}+\frac{B}{1-x} \Rightarrow 1=A(1-x)+B(1+3x)$ | B1 | Correct suitable identity without fractions, seen or implied |
| When $x=1$: $1=4B\Rightarrow B=...$ or when $x=-\frac{1}{3}$: $1=\frac{4}{3}A\Rightarrow A=...$ | M1 | Attempts to find one constant by substitution or equating coefficients |
| $\frac{3}{4(1+3x)}+\frac{1}{4(1-x)}$ | A1 | Allow values for $A$ and $B$ stated following correct partial fraction form |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\cot y\,dy=\int...\,dx \Rightarrow "\ln\sin y"=\int...\,dx$ | M1 | Attempts to separate variables to form $\cot y\frac{dy}{dx}=g(x)$; accept any changed function for attempt but must be attempting to integrate $\cot y$ |
| $...=\int\left(\frac{"3"}{4(1+3x)}+\frac{"1"}{4(1-x)}\right)dx=...\ln(1+3x)\pm...\ln(1-x)\ (+c)$ | M1 | Attempts to integrate partial fractions; award for $\frac{...}{(1+3x)}\rightarrow...\ln(1+3x)$ or $...\ln(4+12x)$ and $\frac{...}{(1-x)}\rightarrow...\ln(1-x)$ or $...\ln(4-4x)$ |
| $\ln\sin y=\frac{1}{4}\ln(1+3x)-\frac{1}{4}\ln(1-x)\ (+c)$ oe | A1ft | Correct expression (any equivalent); follow through on constants for partial fractions; condone absence of constant of integration |
| $\ln\sin(\frac{\pi}{2})=\frac{1}{4}\ln(1+3\times\frac{1}{2})-\frac{1}{4}\ln(1-\frac{1}{2})+c\Rightarrow c=...\left(=-\frac{1}{4}\ln 5\right)$ | dM1 | Depends on second M; uses initial conditions in equation with constant of integration; may integrate between limits |
| $k\ln\sin y=m\ln(...)\Rightarrow\sin^k y=...^m$ or $k\ln\sin y=...\Rightarrow\sin^k y=\exp(...)$ | M1 | Attempts to rearrange using log work to reach $\sin^n y=f(x)$; must have had $k\ln\sin y=...$; not dependent |
| $\sin^4 y=\frac{1+3x}{5(1-x)}$ | A1 | oe in correct form |
---\begin{enumerate}
\item (a) Express $\frac { 1 } { ( 1 + 3 x ) ( 1 - x ) }$ in partial fractions.\\
(b) Hence find the solution of the differential equation
\end{enumerate}
$$( 1 + 3 x ) ( 1 - x ) \frac { \mathrm { d } y } { \mathrm {~d} x } = \tan y \quad - \frac { 1 } { 3 } < x \leqslant \frac { 1 } { 2 }$$
for which $x = \frac { 1 } { 2 }$ when $y = \frac { \pi } { 2 }$\\
Give your answer in the form $\sin ^ { n } y = \mathrm { f } ( x )$ where $n$ is an integer to be found.
\hfill \mbox{\textit{Edexcel P4 2022 Q2 [9]}}