| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for inverse trigonometric functions |
| Difficulty | Standard +0.3 This is a standard FP2 question covering routine techniques: binomial expansion to derive Taylor series (bookwork), a straightforward trigonometric substitution integral, basic polar curve sketching and area calculation, and standard complex number manipulations using De Moivre's theorem. All parts follow well-established methods with no novel insights required, making it slightly easier than average for A-level. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.07d Differentiate/integrate: hyperbolic functions4.08a 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 cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \frac{1}{1+x^2}\) | B1 | |
| Binomial expansion gives \(f'(x) = 1 - x^2 + x^4 - \cdots\) | M1 | Three terms from \((1+x^2)^{-1}\) |
| Correct series | A1 [3] | Give full marks for correct series; ignore higher powers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to obtain \(f(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots (+c)\) | M1, A1 | Must use \(f'(x)\); just answer without \(+c\) is M0 |
| Use \(\arctan(0) = 0\) to find \(c = 0\) | A1 [3] | Can be earned after M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{3/4} \frac{1}{\sqrt{\frac{3}{4}-x^2}}\,dx\) | M1 | For arcsin (or arccos); or any sine (or cosine) substitution |
| \(\frac{1}{2}\left[\arcsin\frac{2x}{\sqrt{3}}\right]_0^{3/4}\) | A1 | For \(\arcsin\frac{2x}{\sqrt{3}}\) (o.e.); or \(2x = \sqrt{3}\sin u\) |
| \(\frac{1}{2}\left(\arcsin\frac{\sqrt{3}}{2} - \arcsin 0\right)\) | A1 | For \(\frac{1}{2}\); or \(\left[\frac{1}{2}u\right]\) |
| \(\frac{\pi}{6}\) | A1, A1 [5] | For \(\arcsin\frac{\sqrt{3}}{2} = \frac{\pi}{3}\) soi; e.g. new limit is \(\frac{\pi}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Spiral shape | G1 | Overall spiral shape (lenient) |
| Correct limits for \(\theta\) | G1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r\) tends to infinity as \(\theta\) tends to zero | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= \frac{1}{2}\int_{\pi/4}^{2\pi} \frac{a^2}{\theta}\,d\theta\) | M1 | For integral of \(\left(\frac{a}{\sqrt{\theta}}\right)^2\) |
| \(\frac{1}{2}\left[a^2\ln\theta\right]_{\pi/4}^{2\pi}\) | A1 | |
| \(\frac{1}{2}\left(a^2\ln 2\pi - a^2\ln\frac{\pi}{4}\right)\) | A1 | For \(\ln 2\pi - \ln\frac{\pi}{4}\) o.e. |
| Simplify to \(\frac{3}{2}a^2\ln 2\) | A1 [4] [18] | Or \(\frac{1}{2}a^2\ln 8\) or \(a^2\ln(\sqrt{8})\) etc |
# Question 1:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \frac{1}{1+x^2}$ | B1 | |
| Binomial expansion gives $f'(x) = 1 - x^2 + x^4 - \cdots$ | M1 | Three terms from $(1+x^2)^{-1}$ |
| Correct series | A1 [3] | Give full marks for correct series; ignore higher powers |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain $f(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots (+c)$ | M1, A1 | Must use $f'(x)$; just answer without $+c$ is M0 |
| Use $\arctan(0) = 0$ to find $c = 0$ | A1 [3] | Can be earned after M1A0 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{3/4} \frac{1}{\sqrt{\frac{3}{4}-x^2}}\,dx$ | M1 | For arcsin (or arccos); or any sine (or cosine) substitution |
| $\frac{1}{2}\left[\arcsin\frac{2x}{\sqrt{3}}\right]_0^{3/4}$ | A1 | For $\arcsin\frac{2x}{\sqrt{3}}$ (o.e.); or $2x = \sqrt{3}\sin u$ |
| $\frac{1}{2}\left(\arcsin\frac{\sqrt{3}}{2} - \arcsin 0\right)$ | A1 | For $\frac{1}{2}$; or $\left[\frac{1}{2}u\right]$ |
| $\frac{\pi}{6}$ | A1, A1 [5] | For $\arcsin\frac{\sqrt{3}}{2} = \frac{\pi}{3}$ soi; e.g. new limit is $\frac{\pi}{3}$ |
## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Spiral shape | G1 | Overall spiral shape (lenient) |
| Correct limits for $\theta$ | G1 [2] | |
## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r$ tends to infinity as $\theta$ tends to zero | B1 [1] | |
## Part (c)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \frac{1}{2}\int_{\pi/4}^{2\pi} \frac{a^2}{\theta}\,d\theta$ | M1 | For integral of $\left(\frac{a}{\sqrt{\theta}}\right)^2$ |
| $\frac{1}{2}\left[a^2\ln\theta\right]_{\pi/4}^{2\pi}$ | A1 | |
| $\frac{1}{2}\left(a^2\ln 2\pi - a^2\ln\frac{\pi}{4}\right)$ | A1 | For $\ln 2\pi - \ln\frac{\pi}{4}$ o.e. |
| Simplify to $\frac{3}{2}a^2\ln 2$ | A1 [4] [18] | Or $\frac{1}{2}a^2\ln 8$ or $a^2\ln(\sqrt{8})$ etc |
---
1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { f } ( x ) = \arctan x$, write down an expression for $\mathrm { f } ^ { \prime } ( x )$. Assuming that $x$ is small, use a binomial expansion to express $\mathrm { f } ^ { \prime } ( x )$ in ascending powers of $x$ as far as the term in $x ^ { 4 }$.
\item Hence express $\arctan x$ in ascending powers of $x$ as far as the term in $x ^ { 5 }$.
\end{enumerate}\item Find, in exact form, the value of the following integral.
$$\int _ { 0 } ^ { \frac { 3 } { 4 } } \frac { 1 } { \sqrt { 3 - 4 x ^ { 2 } } } \mathrm {~d} x$$
\item A curve has polar equation $r = \frac { a } { \sqrt { \theta } }$ where $a > 0$.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve for $\frac { \pi } { 4 } \leqslant \theta \leqslant 2 \pi$.
\item State what happens to $r$ as $\theta$ tends to zero.
\item Find the area of the region enclosed by the part of the curve sketched in part (i) and the lines $\theta = \frac { \pi } { 4 }$ and $\theta = 2 \pi$. Give your answer in an exact simplified form.\\
(a) (i) Express $2 \sin \frac { 1 } { 2 } \theta \left( \sin \frac { 1 } { 2 } \theta - \mathrm { j } \cos \frac { 1 } { 2 } \theta \right)$ in terms of $z$ where $z = \cos \theta + \mathrm { j } \sin \theta$.\\
(ii) The series $C$ and $S$ are defined as follows.
$$\begin{aligned}
C & = 1 - \binom { n } { 1 } \cos \theta + \binom { n } { 2 } \cos 2 \theta - \ldots + ( - 1 ) ^ { n } \binom { n } { n } \cos n \theta \\
S & = - \binom { n } { 1 } \sin \theta + \binom { n } { 2 } \sin 2 \theta - \ldots + ( - 1 ) ^ { n } \binom { n } { n } \sin n \theta
\end{aligned}$$
Show that
$$C + \mathrm { j } S = \left\{ - 2 \mathrm { j } \sin \frac { 1 } { 2 } \theta \left( \cos \frac { 1 } { 2 } \theta + \mathrm { j } \sin \frac { 1 } { 2 } \theta \right) \right\} ^ { n } .$$
Hence show that, for even values of $n$,
$$\frac { C } { S } = \cot \left( \frac { 1 } { 2 } n \theta \right)$$
(b) Write the complex number $z = \sqrt { 6 } + \mathrm { j } \sqrt { 2 }$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$, expressing $r$ and $\theta$ as simply as possible. Hence find the cube roots of $z$ in the form $r \mathrm { e } ^ { \mathrm { j } \theta }$.
Show the points representing $z$ and its cube roots on an Argand diagram.
\begin{enumerate}[label=(\roman*)]
\item Find the eigenvalues and eigenvectors of the matrix $\mathbf { M }$, where
$$\mathbf { M } = \left( \begin{array} { l l }
\frac { 1 } { 2 } & \frac { 1 } { 2 } \\
\frac { 2 } { 3 } & \frac { 1 } { 3 }
\end{array} \right)$$
Hence express $\mathbf { M }$ in the form $\mathbf { P D P } ^ { - 1 }$ where $\mathbf { D }$ is a diagonal matrix.
\item Write down an equation for $\mathbf { M } ^ { n }$ in terms of the matrices $\mathbf { P }$ and $\mathbf { D }$.
Hence obtain expressions for the elements of $\mathbf { M } ^ { n }$.\\
Show that $\mathbf { M } ^ { n }$ tends to a limit as $n$ tends to infinity. Find that limit.
\item Express $\mathbf { M } ^ { - 1 }$ in terms of the matrices $\mathbf { P }$ and $\mathbf { D }$. Hence determine whether or not $\left( \mathbf { M } ^ { - 1 } \right) ^ { n }$ tends to a limit as $n$ tends to infinity.
Section B (18 marks)\\
(i) Given that $y = \cosh x$, use the definition of $\cosh x$ in terms of exponential functions to prove that
$$x = \pm \ln \left( y + \sqrt { y ^ { 2 } - 1 } \right) .$$
(ii) Solve the equation
$$\cosh x + \cosh 2 x = 5$$
giving the roots in an exact logarithmic form.\\
(iii) Sketch the curve with equation $y = \cosh x + \cosh 2 x$. Show on your sketch the line $y = 5$.
Find the area of the finite region bounded by the curve and the line $y = 5$. Give your answer in an exact form that does not involve hyperbolic functions.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2016 Q1}}