OCR FP2 2016 June — Question 2 6 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration with Partial Fractions
TypePartial fractions with irreducible quadratic
DifficultyChallenging +1.2 This is a Further Maths FP2 question requiring partial fractions with an irreducible quadratic factor, followed by integration involving ln and arctan. While it requires multiple techniques (partial fractions decomposition, integrating 1/(x+1) and handling the x²+1 term), the structure is standard for FP2 with no novel insights needed. The irreducible quadratic and definite integral evaluation add moderate complexity above typical A-level questions, but this is a textbook-style FP2 exercise.
Spec1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators

2 It is given that \(\mathrm { f } ( x ) = \frac { x ( x - 1 ) } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) }\). Express \(\mathrm { f } ( x )\) in partial fractions and hence find the exact value of \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\).

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
\(f(x)=\frac{x(x-1)}{(x+1)(x^2+1)} \equiv \frac{A}{(x+1)}+\frac{Bx+C}{(x^2+1)}\)M1 Correct partial fractions
\(\Rightarrow A(x^2+1)+(Bx+C)(x+1) \equiv x(x-1)\); equate coefficients: \(A+B=1,\ B+C=-1,\ A+C=0\)M1 Dep on 1st M; or sub values of \(x\) or division
\(\Rightarrow A=1, B=0, C=-1\); \(\Rightarrow f(x)=\frac{1}{(x+1)}-\frac{1}{(x^2+1)}\)A1 Dep on both M marks
\(\Rightarrow \int_0^1 f(x)\,dx = \int_0^1\left(\frac{1}{(x+1)}-\frac{1}{(x^2+1)}\right)dx\)B1 ft for integrating 1st term correctly (\(A/(x+1)\), \(A\neq0\))
\(=\Big[\ln(1+x)-\tan^{-1}x\Big]_0^1 = \ln 2 - \frac{\pi}{4}\)B1 ft for subsequent term(s) correctly
B1In exact form, dep on both previous B marks
[6]
# Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x)=\frac{x(x-1)}{(x+1)(x^2+1)} \equiv \frac{A}{(x+1)}+\frac{Bx+C}{(x^2+1)}$ | M1 | Correct partial fractions |
| $\Rightarrow A(x^2+1)+(Bx+C)(x+1) \equiv x(x-1)$; equate coefficients: $A+B=1,\ B+C=-1,\ A+C=0$ | M1 | Dep on 1st M; or sub values of $x$ or division |
| $\Rightarrow A=1, B=0, C=-1$; $\Rightarrow f(x)=\frac{1}{(x+1)}-\frac{1}{(x^2+1)}$ | A1 | Dep on both M marks |
| $\Rightarrow \int_0^1 f(x)\,dx = \int_0^1\left(\frac{1}{(x+1)}-\frac{1}{(x^2+1)}\right)dx$ | B1 | ft for integrating 1st term correctly ($A/(x+1)$, $A\neq0$) |
| $=\Big[\ln(1+x)-\tan^{-1}x\Big]_0^1 = \ln 2 - \frac{\pi}{4}$ | B1 | ft for subsequent term(s) correctly |
| | B1 | In exact form, dep on both previous B marks |
| | **[6]** | |

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2 It is given that $\mathrm { f } ( x ) = \frac { x ( x - 1 ) } { ( x + 1 ) \left( x ^ { 2 } + 1 \right) }$. Express $\mathrm { f } ( x )$ in partial fractions and hence find the exact value of $\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$.

\hfill \mbox{\textit{OCR FP2 2016 Q2 [6]}}