| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (i) sketching a cardioid in polar coordinates, (ii) computing a polar area integral involving (1-cos θ)² which requires double angle formulas, (iii) a trigonometric substitution integral with careful algebraic manipulation, and (iv) series expansion of arccos(2x) via differentiation and binomial expansion. While each component uses standard FP2 techniques, the combination of multiple advanced topics and the need for careful algebraic manipulation across several steps places this above average difficulty for A-level, though not exceptionally hard for Further Maths students. |
| Spec | 4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08h Integration: inverse trig/hyperbolic substitutions4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
**(a)(i) Sketch the curve** [2]
B1: Correct identification of cardioid shape
B1: Correct positioning and orientation
**(a)(ii) Find the area of the region** [6]
M1: Set up integral using $A = \frac{1}{2}\int_0^{\pi/2} r^2 \, d\theta$
M1: Substitute $r = a(1 + \cos\theta)$ to get $A = \frac{1}{2}\int_0^{\pi/2} a^2(1 + \cos\theta)^2 \, d\theta$
M1: Expand $(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta$
M1: Use $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ to obtain integrable form
A1: Correct integration of each term
A1: Correct final answer $\frac{5\pi a^2}{8}$
**(b) Trigonometric substitution integral** [4]
M1: Use substitution $x = 2\sin\theta$ (or $x = 2\cos\theta$), $dx = 2\cos\theta \, d\theta$
M1: Simplify $(4 - x^2)^3 = 8\cos^6\theta$ and reduce integral to $\int_0^{\pi/6} \frac{1}{4\cos^5\theta} \, d\theta$
M1: Integrate using appropriate trigonometric identities or reduction formula
A1: Correct final answer $\frac{1}{4\sqrt{3}}$ or equivalent form
**(c)(i) Find $f'(x)$ where $f(x) = \arccos(2x)$** [2]
M1: Use chain rule: $f'(x) = \frac{-2}{\sqrt{1-(2x)^2}}$
A1: $f'(x) = \frac{-2}{\sqrt{1-4x^2}}$
**(c)(ii) Series expansion** [4]
M1: Write $\sqrt{1-4x^2} = (1-4x^2)^{-1/2}$ and use binomial series
M1: Apply binomial expansion correctly with $n = -\frac{1}{2}$, $u = -4x^2$
M1: Integrate term by term from $0$ to $x$ to find $f(x)$
A1: Correct series up to $x^5$: $f(x) = \frac{\pi}{2} - 2x - \frac{2x^3}{3} - \frac{4x^5}{15} + \ldots$
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\begin{enumerate}[label=(\alph*)]
\item A curve has polar equation $r = a ( 1 - \cos \theta )$, where $a$ is a positive constant.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve.
\item Find the area of the region enclosed by the section of the curve for which $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$ and the line $\theta = \frac { 1 } { 2 } \pi$.
\end{enumerate}\item Use a trigonometric substitution to show that $\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 4 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x = \frac { 1 } { 4 \sqrt { 3 } }$.
\item In this part of the question, $\mathrm { f } ( x ) = \arccos ( 2 x )$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mathrm { f } ^ { \prime } ( x )$.
\item Use a standard series to expand $\mathrm { f } ^ { \prime } ( x )$, and hence find the series for $\mathrm { f } ( x )$ in ascending powers of $x$, up to the term in $x ^ { 5 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2007 Q1 [18]}}