OCR C4 2014 June — Question 9 9 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2014
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeRepeated linear factor with distinct linear factor – decompose and integrate
DifficultyStandard +0.3 This is a standard C4 partial fractions question with a repeated linear factor, requiring decomposition into A/(1+2x) + B/(1-x) + C/(1-x)², followed by routine integration. While it involves multiple steps (finding constants, integrating, evaluating limits), the techniques are entirely procedural with no novel insight required, making it slightly easier than average.
Spec1.02y Partial fractions: decompose rational functions1.08j Integration using partial fractions

9 Express \(\frac { 2 + x ^ { 2 } } { ( 1 + 2 x ) ( 1 - x ) ^ { 2 } }\) in partial fractions and hence show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } } \frac { 2 + x ^ { 2 } } { ( 1 + 2 x ) ( 1 - x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln \frac { 3 } { 2 } + \frac { 1 } { 3 }\).

Question 9:
AnswerMarks Guidance
AnswerMarks Guidance
\(\dfrac{A}{1+2x} + \dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}\)B1 or \(\dfrac{A}{1+2x} + \dfrac{Bx+C}{(1-x)^2}\); may be seen in later work
\(2 + x^2 \equiv A(1-x)^2 + B(1+2x)(1-x) + C(1+2x)\)M1 or \(A(1-x)^2 + (Bx+C)(1+2x)\)
\(A=1\), \(B=0\) and \(C=1\)A1A1A1
\(\int\left(\dfrac{1}{1+2x} + \dfrac{1}{(1-x)^2}\right)dx = a\ln(1+2x) + b(1-x)^{-1}\)M1* \(a\) and \(b\) are non-zero constants
\(F(x) = \frac{1}{2}\ln(1+2x) + (1-x)^{-1}\)A1
their \(\dfrac{1}{2}\ln\dfrac{3}{2} + \dfrac{4}{3} - \left(\dfrac{1}{2}\ln 1 + 1\right)\)M1dep*
\(\dfrac{1}{2}\ln\dfrac{3}{2} + \dfrac{4}{3} - 0 - 1\)A1 and completion to given result
[9]
# Question 9:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{A}{1+2x} + \dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}$ | B1 | or $\dfrac{A}{1+2x} + \dfrac{Bx+C}{(1-x)^2}$; may be seen in later work | if B0M0, SC1 for $\dfrac{1}{1+2x}$ seen |
| $2 + x^2 \equiv A(1-x)^2 + B(1+2x)(1-x) + C(1+2x)$ | M1 | or $A(1-x)^2 + (Bx+C)(1+2x)$ | allow only sign errors, not algebraic errors |
| $A=1$, $B=0$ and $C=1$ | A1A1A1 | | |
| $\int\left(\dfrac{1}{1+2x} + \dfrac{1}{(1-x)^2}\right)dx = a\ln(1+2x) + b(1-x)^{-1}$ | M1* | $a$ and $b$ are non-zero constants | ignore extra terms |
| $F(x) = \frac{1}{2}\ln(1+2x) + (1-x)^{-1}$ | A1 | | |
| their $\dfrac{1}{2}\ln\dfrac{3}{2} + \dfrac{4}{3} - \left(\dfrac{1}{2}\ln 1 + 1\right)$ | M1dep* | | |
| $\dfrac{1}{2}\ln\dfrac{3}{2} + \dfrac{4}{3} - 0 - 1$ | A1 | and completion to given result | NB $\dfrac{1}{2}\ln\dfrac{3}{2} + \dfrac{1}{3}$ |
| **[9]** | | |

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9 Express $\frac { 2 + x ^ { 2 } } { ( 1 + 2 x ) ( 1 - x ) ^ { 2 } }$ in partial fractions and hence show that $\int _ { 0 } ^ { \frac { 1 } { 4 } } \frac { 2 + x ^ { 2 } } { ( 1 + 2 x ) ( 1 - x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 2 } \ln \frac { 3 } { 2 } + \frac { 1 } { 3 }$.

\hfill \mbox{\textit{OCR C4 2014 Q9 [9]}}