| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable with partial fractions |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation requiring standard partial fractions (part a guides the student) followed by routine separation and integration. The question scaffolds the solution by providing the partial fractions step and the final form to verify, making it slightly easier than a typical A-level DE question but still requiring competent execution of multiple techniques. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(4x+1)(x+1)} = \frac{A}{4x+1} + \frac{B}{x+1}\); \(1 = A(x+1) + B(4x+1)\) | M1 | 1.1a — Method mark implied by correct answer |
| \(x = -1 \Rightarrow 1 = -3B \Rightarrow B = -\frac{1}{3}\) | A1 | 1.1 |
| \(x = -\frac{1}{4} \Rightarrow 1 = \frac{3}{4}A \Rightarrow A = \frac{4}{3}\) | ||
| \(\frac{1}{(4x+1)(x+1)} = \frac{4}{3(4x+1)} - \frac{1}{3(x+1)}\) | A1 | 1.1 — Final solution needed for A1. Can be recovered in 10(b) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{y}\,dy = \int \frac{1}{(4x+1)(x+1)}\,dx\) | M1 | 3.1a — Separation of variables — both sides seen. Integral signs, \(dy\) and \(dx\) needed |
| \(\int \frac{1}{y}\,dy = \int \left\{\frac{4}{3(4x+1)} - \frac{1}{3(x+1)}\right\}dx\) | M1 | 2.2a — Use of their (a). RHS only needed. Condone no \(dx\) |
| \(\ln | y | = \frac{1}{3}\ln |
| When \(x = 0\), \(y = 2 \Rightarrow c = \ln 2\) | B1 | 2.2a — Finding constant (FT their (a)) |
| \(\ln | y | = \ln 2\left(\frac{ |
| \(y = 2\left(\frac{4x+1}{x+1}\right)^{\frac{1}{3}}\) | A1 | 2.1 — Answer with correct values of \(A\) and \(B\) |
| [6] |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(4x+1)(x+1)} = \frac{A}{4x+1} + \frac{B}{x+1}$; $1 = A(x+1) + B(4x+1)$ | M1 | 1.1a — Method mark implied by correct answer |
| $x = -1 \Rightarrow 1 = -3B \Rightarrow B = -\frac{1}{3}$ | A1 | 1.1 |
| $x = -\frac{1}{4} \Rightarrow 1 = \frac{3}{4}A \Rightarrow A = \frac{4}{3}$ | | |
| $\frac{1}{(4x+1)(x+1)} = \frac{4}{3(4x+1)} - \frac{1}{3(x+1)}$ | A1 | 1.1 — Final solution needed for A1. Can be recovered in 10(b) |
| | **[3]** | |
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## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{y}\,dy = \int \frac{1}{(4x+1)(x+1)}\,dx$ | M1 | 3.1a — Separation of variables — both sides seen. Integral signs, $dy$ and $dx$ needed |
| $\int \frac{1}{y}\,dy = \int \left\{\frac{4}{3(4x+1)} - \frac{1}{3(x+1)}\right\}dx$ | M1 | 2.2a — Use of their (a). RHS only needed. Condone no $dx$ |
| $\ln|y| = \frac{1}{3}\ln|4x+1| - \frac{1}{3}\ln|x+1| + c$ | M1 | 1.1 — Integration. One correct $x$ term (ft their partial fractions). Condone missing modulus signs |
| When $x = 0$, $y = 2 \Rightarrow c = \ln 2$ | B1 | 2.2a — Finding constant (FT their (a)) |
| $\ln|y| = \ln 2\left(\frac{|4x+1|}{|x+1|}\right)^{\frac{1}{3}}$ | M1 | 1.1 — For use of $a\ln m = \ln m^a$ and $\ln m - \ln n = \ln\frac{m}{n}$. Not dep on $c$. Condone missing modulus signs throughout |
| $y = 2\left(\frac{4x+1}{x+1}\right)^{\frac{1}{3}}$ | A1 | 2.1 — Answer with correct values of $A$ and $B$ |
| | **[6]** | |
---10
\begin{enumerate}[label=(\alph*)]
\item Express $\frac { 1 } { ( 4 x + 1 ) ( x + 1 ) }$ in partial fractions.
\item A curve passes through the point $( 0,2 )$ and satisfies the differential equation $\frac { d y } { d x } = \frac { y } { ( 4 x + 1 ) ( x + 1 ) }$,\\
for $x > - \frac { 1 } { 4 }$.\\
Show by integration that $\mathrm { y } = \mathrm { A } \left( \frac { 4 \mathrm { x } + 1 } { \mathrm { x } + 1 } \right) ^ { \mathrm { B } }$ where $A$ and $B$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q10 [9]}}