Edexcel C4 2011 January — Question 3 12 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2011
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypePartial fractions in differential equations
DifficultyStandard +0.3 This is a standard C4 partial fractions question with three routine parts: decomposition into partial fractions, integration, and solving a separable differential equation. All steps follow textbook procedures with no novel insight required, though it does require careful execution across multiple techniques. Slightly easier than average due to the straightforward structure and linear factors.
Spec1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions1.08k Separable differential equations: dy/dx = f(x)g(y)
3. (a) Express \(\frac { 5 } { ( x - 1 ) ( 3 x + 2 ) }\) in partial fractions.
(b) Hence find \(\int \frac { 5 } { ( x - 1 ) ( 3 x + 2 ) } \mathrm { d } x\), where \(x > 1\).
(c) Find the particular solution of the differential equation $$( x - 1 ) ( 3 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 5 y , \quad x > 1$$ for which \(y = 8\) at \(x = 2\). Give your answer in the form \(y = \mathrm { f } ( x )\).
Question 3:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\frac{5}{(x-1)(3x+2)} = \frac{A}{x-1} + \frac{B}{3x+2}\)
\(5 = A(3x+2) + B(x-1)\)
\(x\to 1\): \(5 = 5A \Rightarrow A=1\)M1 A1
\(x\to -\frac{2}{3}\): \(5 = -\frac{5}{3}B \Rightarrow B=-3\)A1 (3)
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\int\frac{5}{(x-1)(3x+2)}\,dx = \int\left(\frac{1}{x-1} - \frac{3}{3x+2}\right)dx\)
\(= \ln(x-1) - \ln(3x+2) \quad (+C)\)M1 A1ft A1ft ft constants (3)
Part (c):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\int\frac{5}{(x-1)(3x+2)}\,dx = \int\left(\frac{1}{y}\right)dy\)M1
\(\ln(x-1) - \ln(3x+2) = \ln y \quad (+C)\)M1 A1
\(y = \frac{K(x-1)}{3x+2}\)M1 dep depends on first two Ms in (c)
Using \((2,8)\): \(8 = \frac{K}{8}\)M1 dep depends on first two Ms in (c)
\(y = \frac{64(x-1)}{3x+2}\)A1 (6)
Total[12] 
# Question 3:

## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{5}{(x-1)(3x+2)} = \frac{A}{x-1} + \frac{B}{3x+2}$ | | |
| $5 = A(3x+2) + B(x-1)$ | | |
| $x\to 1$: $5 = 5A \Rightarrow A=1$ | M1 A1 | |
| $x\to -\frac{2}{3}$: $5 = -\frac{5}{3}B \Rightarrow B=-3$ | A1 | **(3)** |

## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int\frac{5}{(x-1)(3x+2)}\,dx = \int\left(\frac{1}{x-1} - \frac{3}{3x+2}\right)dx$ | | |
| $= \ln(x-1) - \ln(3x+2) \quad (+C)$ | M1 A1ft A1ft | ft constants **(3)** |

## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int\frac{5}{(x-1)(3x+2)}\,dx = \int\left(\frac{1}{y}\right)dy$ | M1 | |
| $\ln(x-1) - \ln(3x+2) = \ln y \quad (+C)$ | M1 A1 | |
| $y = \frac{K(x-1)}{3x+2}$ | M1 dep | depends on first two Ms in (c) |
| Using $(2,8)$: $8 = \frac{K}{8}$ | M1 dep | depends on first two Ms in (c) |
| $y = \frac{64(x-1)}{3x+2}$ | A1 | **(6)** |
| **Total** | **[12]** | |

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3. (a) Express $\frac { 5 } { ( x - 1 ) ( 3 x + 2 ) }$ in partial fractions.\\
(b) Hence find $\int \frac { 5 } { ( x - 1 ) ( 3 x + 2 ) } \mathrm { d } x$, where $x > 1$.\\
(c) Find the particular solution of the differential equation

$$( x - 1 ) ( 3 x + 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = 5 y , \quad x > 1$$

for which $y = 8$ at $x = 2$. Give your answer in the form $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{Edexcel C4 2011 Q3 [12]}}
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