| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series of shifted function |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (1) differentiation of arcsin and verification of second derivative (routine), (2) Maclaurin expansion of a shifted function requiring chain rule and evaluation at t=1/2 (moderately challenging), (3) polar curve sketching and area calculation (standard FP2), and (4) integration using arctan (routine). The shifted Maclaurin series is the most demanding part, requiring careful application of the chain rule to derivatives and evaluation at a non-zero point, which is above average difficulty but well within FP2 scope. |
| Spec | 4.08a Maclaurin series: find series for function4.08g Derivatives: inverse trig and hyperbolic functions4.08h 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(t) = \frac{1}{\sqrt{1-t^2}} = (1-t^2)^{-\frac{1}{2}}\) | B1 | Any form |
| \(f''(t) = -\frac{1}{2}(1-t^2)^{-\frac{3}{2}} \times -2t\) | M1 | Using Chain Rule |
| \(= \frac{t}{(1-t^2)^{\frac{3}{2}}}\) | A1 (ag) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(0) = \arcsin(\frac{1}{2}) = \frac{\pi}{6}\) | B1 (ag) | \(\frac{\pi}{6}\) obtained clearly from \(f(0)\) |
| \(f'(0) = \left(1-\left(\frac{1}{2}\right)^2\right)^{-\frac{1}{2}} = \frac{2}{\sqrt{3}}\) | M1, A1 (ag) | Clear substitution of \(x=0\) or \(t=\frac{1}{2}\) |
| \(f''(0) = \frac{\frac{1}{2}}{\left(1-\left(\frac{1}{2}\right)^2\right)^{\frac{3}{2}}} = \frac{4\sqrt{3}}{9}\) | ||
| \(f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + \ldots\) | M1 | Evaluating \(f''(0)\) and dividing by 2 |
| Term in \(x^2\) is \(\frac{2\sqrt{3}}{9}x^2\) | A1 | Accept \(0.385x^2\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Spiral sketch | G1 | Complete spiral with \(r(2\pi) < r(0)\) |
| \(r(0)=a\), \(r(2\pi)=a/3\) indicated | G1 | \(r(0) > r(\pi/2) > r(\pi) > r(3\pi/2) > r(2\pi)\); Dep. on G1; Max G1 if not fully correct |
| Area \(= \int_0^{\pi} \frac{1}{2}r^2\,d\theta\) | ||
| \(= \int_0^{\pi} \frac{\pi^2 a^2}{2(\pi+\theta)^2}\,d\theta = \frac{\pi^2 a^2}{2}\int_0^{\pi}\frac{1}{(\pi+\theta)^2}\,d\theta\) | M1 | Integral expression involving \(r^2\) |
| \(= \frac{\pi^2 a^2}{2}\left[\frac{-1}{\pi+\theta}\right]_0^{\pi}\) | A1 | Correct result of integration with correct limits |
| \(= \frac{\pi^2 a^2}{2}\left(\frac{-1}{2\pi}+\frac{1}{\pi}\right)\) | M1 | Substituting limits into form \(\frac{k}{\pi+\theta}\); Dep. on M1 above |
| \(= \frac{1}{4}\pi a^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_0^{\frac{3}{2}} \frac{1}{9+4x^2}\,dx = \frac{1}{4}\int_0^{\frac{3}{2}}\frac{1}{\frac{9}{4}+x^2}\,dx = \frac{1}{4}\times\left[\frac{2}{3}\arctan\frac{2x}{3}\right]_0^{\frac{3}{2}}\) | M1 | arctan |
| A1A1 | \(\frac{1}{4}\times\frac{2}{3}\) and \(\frac{2x}{3}\) | |
| \(= \frac{1}{6}\arctan 1\) | M1 | Substituting limits; Dep. on M1 above |
| \(= \frac{\pi}{24}\) | A1 | Evaluated in terms of \(\pi\) |
# Question 1:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(t) = \frac{1}{\sqrt{1-t^2}} = (1-t^2)^{-\frac{1}{2}}$ | B1 | Any form |
| $f''(t) = -\frac{1}{2}(1-t^2)^{-\frac{3}{2}} \times -2t$ | M1 | Using Chain Rule |
| $= \frac{t}{(1-t^2)^{\frac{3}{2}}}$ | A1 (ag) | |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(0) = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$ | B1 (ag) | $\frac{\pi}{6}$ obtained clearly from $f(0)$ |
| $f'(0) = \left(1-\left(\frac{1}{2}\right)^2\right)^{-\frac{1}{2}} = \frac{2}{\sqrt{3}}$ | M1, A1 (ag) | Clear substitution of $x=0$ or $t=\frac{1}{2}$ |
| $f''(0) = \frac{\frac{1}{2}}{\left(1-\left(\frac{1}{2}\right)^2\right)^{\frac{3}{2}}} = \frac{4\sqrt{3}}{9}$ | | |
| $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + \ldots$ | M1 | Evaluating $f''(0)$ and dividing by 2 |
| Term in $x^2$ is $\frac{2\sqrt{3}}{9}x^2$ | A1 | Accept $0.385x^2$ or better |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Spiral sketch | G1 | Complete spiral with $r(2\pi) < r(0)$ |
| $r(0)=a$, $r(2\pi)=a/3$ indicated | G1 | $r(0) > r(\pi/2) > r(\pi) > r(3\pi/2) > r(2\pi)$; Dep. on G1; Max G1 if not fully correct |
| Area $= \int_0^{\pi} \frac{1}{2}r^2\,d\theta$ | | |
| $= \int_0^{\pi} \frac{\pi^2 a^2}{2(\pi+\theta)^2}\,d\theta = \frac{\pi^2 a^2}{2}\int_0^{\pi}\frac{1}{(\pi+\theta)^2}\,d\theta$ | M1 | Integral expression involving $r^2$ |
| $= \frac{\pi^2 a^2}{2}\left[\frac{-1}{\pi+\theta}\right]_0^{\pi}$ | A1 | Correct result of integration with correct limits |
| $= \frac{\pi^2 a^2}{2}\left(\frac{-1}{2\pi}+\frac{1}{\pi}\right)$ | M1 | Substituting limits into form $\frac{k}{\pi+\theta}$; Dep. on M1 above |
| $= \frac{1}{4}\pi a^2$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{3}{2}} \frac{1}{9+4x^2}\,dx = \frac{1}{4}\int_0^{\frac{3}{2}}\frac{1}{\frac{9}{4}+x^2}\,dx = \frac{1}{4}\times\left[\frac{2}{3}\arctan\frac{2x}{3}\right]_0^{\frac{3}{2}}$ | M1 | arctan |
| | A1A1 | $\frac{1}{4}\times\frac{2}{3}$ and $\frac{2x}{3}$ |
| $= \frac{1}{6}\arctan 1$ | M1 | Substituting limits; Dep. on M1 above |
| $= \frac{\pi}{24}$ | A1 | Evaluated in terms of $\pi$ |
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\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { f } ( t ) = \arcsin t$, write down an expression for $\mathrm { f } ^ { \prime } ( t )$ and show that
$$\mathrm { f } ^ { \prime \prime } ( t ) = \frac { t } { \left( 1 - t ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$$
\item Show that the Maclaurin expansion of the function $\arcsin \left( x + \frac { 1 } { 2 } \right)$ begins
$$\frac { \pi } { 6 } + \frac { 2 } { \sqrt { 3 } } x$$
and find the term in $x ^ { 2 }$.
\end{enumerate}\item Sketch the curve with polar equation $r = \frac { \pi a } { \pi + \theta }$, where $a > 0$, for $0 \leqslant \theta < 2 \pi$.
Find, in terms of $a$, the area of the region bounded by the part of the curve for which $0 \leqslant \theta \leqslant \pi$ and the lines $\theta = 0$ and $\theta = \pi$.
\item Find the exact value of the integral
$$\int _ { 0 } ^ { \frac { 3 } { 2 } } \frac { 1 } { 9 + 4 x ^ { 2 } } \mathrm {~d} x$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2010 Q1 [19]}}