OCR MEI FP2 2011 January — Question 1 19 marks

Exam BoardOCR MEI
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJanuary
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyStandard +0.3 This is a standard FP2 question covering polar coordinates and Maclaurin series. Part (a) involves routine conversion between polar and cartesian forms, recognizing a circle, and applying the standard polar area formula. Part (b) requires differentiating arctan and using binomial expansion, then integrating term-by-term—all textbook techniques. While Further Maths content is inherently more advanced, these are straightforward applications without novel problem-solving, placing it slightly above average difficulty.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.08a Maclaurin series: find series for function4.09a Polar coordinates: convert to/from cartesian
  1. A curve has polar equation \(r = 2(\cos \theta + \sin \theta)\) for \(-\frac{1}{4}\pi \leq \theta \leq \frac{3}{4}\pi\).
    1. Show that a cartesian equation of the curve is \(x^2 + y^2 = 2x + 2y\). Hence or otherwise sketch the curve. [5]
    2. Find, by integration, the area of the region bounded by the curve and the lines \(\theta = 0\) and \(\theta = \frac{1}{2}\pi\). Give your answer in terms of \(\pi\). [7]
    1. Given that \(f(x) = \arctan(\frac{1}{2}x)\), find \(f'(x)\). [2]
    2. Expand \(f'(x)\) in ascending powers of \(x\) as far as the term in \(x^4\). Hence obtain an expression for \(f(x)\) in ascending powers of \(x\) as far as the term in \(x^5\). [5]
(a)(i)
\(x = r\cos\theta, y = r\sin\theta, x^2 + y^2 = r^2\)
\(r = 2(\cos\theta + \sin\theta)\)
\(\Rightarrow r^2 = 2r(\cos\theta + \sin\theta)\)
\(\Rightarrow x^2 + y^2 = 2x + 2y\)
\(\Rightarrow x^2 - 2x + y^2 - 2y = 0\)
\(\Rightarrow (x-1)^2 + (y-1)^2 = 2\)
which is a circle centre \((1,1)\) radius \(\sqrt{2}\)
AnswerMarks
M1Using at least one of these
A1 (ag)Working must be convincing
M1Recognise as circle or appropriate algebra leading to \((x-a)^2 + (y-b)^2 = r^2\)
G1Attempt at complete circle with centre in first quadrant
G1A circle with centre and radius indicated, or centre \((1,1)\) indicated and passing through \((0,0)\), or \((2,0)\) and \((0,2)\) indicated and passing through \((0,0)\)
Total5
(a)(ii)
Area \(= \frac{1}{2}\int_0^{\pi} r^2 d\theta\)
\(= 2\int_0^{\pi} (\cos\theta + \sin\theta)^2 d\theta\)
\(= 2\int_0^{\pi} (\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta) d\theta\)
\(= 2\int_0^{\pi} (1 + 2\sin\theta\cos\theta) d\theta\)
\(= 2\left[\theta - \frac{1}{2}\cos 2\theta\right]_0^{\pi}\) or \(2\left[\theta + \sin^2\theta\right]_0^0\) etc.
\(= 2\left(\left(\frac{\pi}{2} + \frac{1}{2}\right) - \left(0 - \frac{1}{2}\right)\right)\)
\(= \pi + 2\)
AnswerMarks
M1Integral expression involving \(r^2\) in terms of \(\theta\)
M1Multiplying out
A1\(\cos^2\theta + \sin^2\theta = 1\) used
A2Correct result of integration with correct limits. Give A1 for one error
M1Substituting limits. Dep. on both M1s
A1Mark final answer
Total7
(b)(i)
\(f'(x) = \frac{1}{2} \cdot \frac{1}{(1 + \frac{1}{4}x^2)} \cdot \frac{2}{4 + x^2}\)
AnswerMarks
M1Using Chain Rule
A1Correct derivative in any form
Total2
(b)(ii)
\(f'(x) = \frac{1}{2}(1 + x^2)^{-1} = \frac{1}{2}(1 - \frac{1}{4}x^2 + \frac{1}{16}x^4 - ...)\)
\(= \frac{1}{2} - \frac{1}{8}x^2 + \frac{1}{32}x^4 - ...\)
\(\Rightarrow f(x) = \frac{1}{2}x - \frac{1}{24}x^3 + \frac{1}{160}x^5 - ... + c\)
But \(c = 0\) because \(\arctan(0) = 0\)
AnswerMarks
M1Correctly using binomial expansion
A1Correct expansion
M1Integrating at least two terms
A1
A1Independent
Total5 
## (a)(i)

$x = r\cos\theta, y = r\sin\theta, x^2 + y^2 = r^2$

$r = 2(\cos\theta + \sin\theta)$

$\Rightarrow r^2 = 2r(\cos\theta + \sin\theta)$

$\Rightarrow x^2 + y^2 = 2x + 2y$

$\Rightarrow x^2 - 2x + y^2 - 2y = 0$

$\Rightarrow (x-1)^2 + (y-1)^2 = 2$

which is a circle centre $(1,1)$ radius $\sqrt{2}$

| | M1 | Using at least one of these |
| | A1 (ag) | Working must be convincing |
| | M1 | Recognise as circle or appropriate algebra leading to $(x-a)^2 + (y-b)^2 = r^2$ |
| | G1 | Attempt at complete circle with centre in first quadrant |
| | G1 | A circle with centre and radius indicated, or centre $(1,1)$ indicated and passing through $(0,0)$, or $(2,0)$ and $(0,2)$ indicated and passing through $(0,0)$ |
| **Total** | **5** | |

## (a)(ii)

Area $= \frac{1}{2}\int_0^{\pi} r^2 d\theta$

$= 2\int_0^{\pi} (\cos\theta + \sin\theta)^2 d\theta$

$= 2\int_0^{\pi} (\cos^2\theta + 2\sin\theta\cos\theta + \sin^2\theta) d\theta$

$= 2\int_0^{\pi} (1 + 2\sin\theta\cos\theta) d\theta$

$= 2\left[\theta - \frac{1}{2}\cos 2\theta\right]_0^{\pi}$ or $2\left[\theta + \sin^2\theta\right]_0^0$ etc.

$= 2\left(\left(\frac{\pi}{2} + \frac{1}{2}\right) - \left(0 - \frac{1}{2}\right)\right)$

$= \pi + 2$

| | M1 | Integral expression involving $r^2$ in terms of $\theta$ |
| | M1 | Multiplying out |
| | A1 | $\cos^2\theta + \sin^2\theta = 1$ used |
| | A2 | Correct result of integration with correct limits. Give A1 for one error |
| | M1 | Substituting limits. Dep. on both M1s |
| | A1 | Mark final answer |
| **Total** | **7** | |

## (b)(i)

$f'(x) = \frac{1}{2} \cdot \frac{1}{(1 + \frac{1}{4}x^2)} \cdot \frac{2}{4 + x^2}$

| | M1 | Using Chain Rule |
| | A1 | Correct derivative in any form |
| **Total** | **2** | |

## (b)(ii)

$f'(x) = \frac{1}{2}(1 + x^2)^{-1} = \frac{1}{2}(1 - \frac{1}{4}x^2 + \frac{1}{16}x^4 - ...)$

$= \frac{1}{2} - \frac{1}{8}x^2 + \frac{1}{32}x^4 - ...$

$\Rightarrow f(x) = \frac{1}{2}x - \frac{1}{24}x^3 + \frac{1}{160}x^5 - ... + c$

But $c = 0$ because $\arctan(0) = 0$

| | M1 | Correctly using binomial expansion |
| | A1 | Correct expansion |
| | M1 | Integrating at least two terms |
| | A1 | |
| | A1 | Independent |
| **Total** | **5** | |

---
\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = 2(\cos \theta + \sin \theta)$ for $-\frac{1}{4}\pi \leq \theta \leq \frac{3}{4}\pi$.
\begin{enumerate}[label=(\roman*)]
\item Show that a cartesian equation of the curve is $x^2 + y^2 = 2x + 2y$. Hence or otherwise sketch the curve. [5]
\item Find, by integration, the area of the region bounded by the curve and the lines $\theta = 0$ and $\theta = \frac{1}{2}\pi$. Give your answer in terms of $\pi$. [7]
\end{enumerate}
\item \begin{enumerate}[label=(\roman*)]
\item Given that $f(x) = \arctan(\frac{1}{2}x)$, find $f'(x)$. [2]
\item Expand $f'(x)$ in ascending powers of $x$ as far as the term in $x^4$.

Hence obtain an expression for $f(x)$ in ascending powers of $x$ as far as the term in $x^5$. [5]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI FP2 2011 Q1 [19]}}
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Q1 19 Q2 19 Q3 16 Q4 18 Q5 18