Pre-U Pre-U 9795/1 2014 June — Question 11 9 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2014
SessionJune
Marks9
TopicPolar coordinates
TypeArea of region with line boundary
DifficultyChallenging +1.2 This is a multi-part polar coordinates question requiring standard techniques: identifying key points, sketching, and using Maclaurin series for area approximation. Parts (i)-(iii) are routine. Part (iv) requires deriving e^(sin θ) ≈ 1 + θ + θ²/2 and integrating r²/2, which is methodical rather than insightful. Slightly above average due to the series manipulation and multi-step nature, but follows standard Further Maths polar area procedures.
Spec4.08a Maclaurin series: find series for function4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
11 A curve has polar equation \(r = \mathrm { e } ^ { \sin \theta }\) for \(- \pi < \theta \leqslant \pi\).
  1. State the polar coordinates of the point where the curve crosses the initial line.
  2. State also the polar coordinates of the points where \(r\) takes its least and greatest values.
  3. Sketch the curve.
  4. By deriving a suitable Maclaurin series up to and including the term in \(\theta ^ { 2 }\), find an approximation, to 3 decimal places, for the area of the region enclosed by the curve, the initial line and the line \(\theta = 0.3\).
(i) \((r, \theta) = (1, 0)\) B1 [1]
Condone \((r, \theta)\) given as \((\theta, r)\) throughout
(ii) \(\mathrm{e}^{\sin\theta}\) min./max. when \(\sin\theta\) min./max. B1 B1 [2]
giving min. at \((r, \theta) = \left(\mathrm{e}^{-1}, -\frac{\pi}{2}\right)\) and max. at \((r, \theta) = \left(\mathrm{e}, \frac{\pi}{2}\right)\)
(iii) Symmetry in \(y\)-axis B1
Closed curve B1
Shape essentially correct – don't penalise kinks. ft from (ii) where suitable B1 [3]
(iv) \(A = \int_0^{0.3} \frac{1}{2}\mathrm{e}^{2\sin\theta}\ \mathrm{d}\theta\) Use of formula; correct M1 A1
\(\mathrm{f}(\theta) = \mathrm{e}^{2\sin\theta} \Rightarrow \mathrm{f}'(\theta) = \mathrm{e}^{2\sin\theta} \cdot 2\cos\theta\) and \(\mathrm{f}''(\theta) = \mathrm{e}^{2\sin\theta}(4\cos^2\theta - 2\sin\theta)\) B1 B1
\(\mathrm{f}(0) = 1\), \(\mathrm{f}'(0) = 2\), \(\mathrm{f}''(0) = 4\) B1
\(\Rightarrow \mathrm{f}(\theta) = 1 + 2\theta + 2\theta^2 \ldots\) M1 A1
\(A = \frac{1}{2}\left[\theta + \theta^2 + \frac{2}{3}\theta^3\right]_0^{0.3} = 0.204\) or \(\frac{51}{250}\) 1st A1 for correct \(\int\) of a 3-term quadratic A1
Accept 0.205 from correct \(\int\) of quartic \((1 + 2\theta + 2\theta^2 + \theta^3 + \frac{1}{4}\theta^4)\) A1 [9]
NB Correct answer is 0.204 98...
ALTERNATIVES (middle 5 marks)
Alt. I Ignoring terms in \(\theta^3\) and above, \(\sin\theta \approx \theta \ldots\) B1
\(\mathrm{e}^{\sin\theta} \approx 1 + \theta + \frac{1}{2}\theta^2 \ldots\) M1 A1
\(\Rightarrow (\mathrm{e}^{\sin\theta})^2 \approx 1 + 2\theta + \theta^2 \ldots + 2 \times \frac{1}{2}\theta^2 \ldots = 1 + 2\theta + 2\theta^2 \ldots\) M1 A1
Alt. II \(\mathrm{f}(\theta) = \mathrm{e}^{\sin\theta} \ldots\) \(\mathrm{f}'(\theta) = \cos\theta\,\mathrm{e}^{\sin\theta}\) and \(\mathrm{f}''(\theta) = (\cos^2\theta - \sin\theta)\,\mathrm{e}^{\sin\theta}\) B1 B1
\(\mathrm{f}(0) = \mathrm{f}'(0) = \mathrm{f}''(0) = 1\) B1
\(\Rightarrow \mathrm{f}(\theta) = 1 + \theta + \frac{1}{2}\theta^2 + \ldots\) Maclaurin attempt at as a function of \(\theta\) M1
\(\Rightarrow [\mathrm{f}(\theta)]^2 = 1 + 2\theta + 2\theta^2 + \ldots\) Ignore higher power terms A1
(i) $(r, \theta) = (1, 0)$ **B1** [1]

Condone $(r, \theta)$ given as $(\theta, r)$ throughout

(ii) $\mathrm{e}^{\sin\theta}$ min./max. when $\sin\theta$ min./max. **B1 B1** [2]

giving min. at $(r, \theta) = \left(\mathrm{e}^{-1}, -\frac{\pi}{2}\right)$ and max. at $(r, \theta) = \left(\mathrm{e}, \frac{\pi}{2}\right)$

(iii) Symmetry in $y$-axis **B1**

Closed curve **B1**

Shape essentially correct – don't penalise kinks. ft from (ii) where suitable **B1** [3]

(iv) $A = \int_0^{0.3} \frac{1}{2}\mathrm{e}^{2\sin\theta}\ \mathrm{d}\theta$ Use of formula; correct **M1 A1**

$\mathrm{f}(\theta) = \mathrm{e}^{2\sin\theta} \Rightarrow \mathrm{f}'(\theta) = \mathrm{e}^{2\sin\theta} \cdot 2\cos\theta$ and $\mathrm{f}''(\theta) = \mathrm{e}^{2\sin\theta}(4\cos^2\theta - 2\sin\theta)$ **B1 B1**

$\mathrm{f}(0) = 1$, $\mathrm{f}'(0) = 2$, $\mathrm{f}''(0) = 4$ **B1**

$\Rightarrow \mathrm{f}(\theta) = 1 + 2\theta + 2\theta^2 \ldots$ **M1 A1**

$A = \frac{1}{2}\left[\theta + \theta^2 + \frac{2}{3}\theta^3\right]_0^{0.3} = 0.204$ or $\frac{51}{250}$ 1st A1 for correct $\int$ of a 3-term quadratic **A1**

Accept 0.205 from correct $\int$ of quartic $(1 + 2\theta + 2\theta^2 + \theta^3 + \frac{1}{4}\theta^4)$ **A1** [9]

NB Correct answer is 0.204 98...

**ALTERNATIVES** (middle 5 marks)

**Alt. I** Ignoring terms in $\theta^3$ and above, $\sin\theta \approx \theta \ldots$ **B1**

$\mathrm{e}^{\sin\theta} \approx 1 + \theta + \frac{1}{2}\theta^2 \ldots$ **M1 A1**

$\Rightarrow (\mathrm{e}^{\sin\theta})^2 \approx 1 + 2\theta + \theta^2 \ldots + 2 \times \frac{1}{2}\theta^2 \ldots = 1 + 2\theta + 2\theta^2 \ldots$ **M1 A1**

**Alt. II** $\mathrm{f}(\theta) = \mathrm{e}^{\sin\theta} \ldots$ $\mathrm{f}'(\theta) = \cos\theta\,\mathrm{e}^{\sin\theta}$ and $\mathrm{f}''(\theta) = (\cos^2\theta - \sin\theta)\,\mathrm{e}^{\sin\theta}$ **B1 B1**

$\mathrm{f}(0) = \mathrm{f}'(0) = \mathrm{f}''(0) = 1$ **B1**

$\Rightarrow \mathrm{f}(\theta) = 1 + \theta + \frac{1}{2}\theta^2 + \ldots$ Maclaurin attempt at as a function of $\theta$ **M1**

$\Rightarrow [\mathrm{f}(\theta)]^2 = 1 + 2\theta + 2\theta^2 + \ldots$ Ignore higher power terms **A1**
11 A curve has polar equation $r = \mathrm { e } ^ { \sin \theta }$ for $- \pi < \theta \leqslant \pi$.\\
(i) State the polar coordinates of the point where the curve crosses the initial line.\\
(ii) State also the polar coordinates of the points where $r$ takes its least and greatest values.\\
(iii) Sketch the curve.\\
(iv) By deriving a suitable Maclaurin series up to and including the term in $\theta ^ { 2 }$, find an approximation, to 3 decimal places, for the area of the region enclosed by the curve, the initial line and the line $\theta = 0.3$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q11 [9]}}
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