| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for inverse trigonometric functions |
| Difficulty | Standard +0.8 This FP2 question combines standard inverse trig calculus (sketching arcsin, deriving its derivative, evaluating an integral) with a sophisticated complex number approach to summing trigonometric series. Part (a) is routine FP1/FP2 material (~0.0 difficulty), but part (b) requires recognizing that C + jS forms a geometric series in complex form, summing it, then separating real/imaginary parts—a multi-step problem requiring genuine insight into the connection between complex exponentials and trig series. The 11-mark allocation and non-standard series summation technique elevate this above typical A-level questions. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Graph: Correct basic shape (positive gradient, through (0, 0)) | G1 | |
| \(y = \arcsin x \Rightarrow \sin y = x\) | M1 | \(\sin y = x\) and attempt to diff. both sides |
| \(\Rightarrow \frac{dx}{dy} = \cos y\) | A1 | Or \(\cos y \frac{dx}{dy} = 1\) |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}\) | A1 | www. SC1 if quoted without working |
| Positive square root because gradient positive | B1 | Dep. on graph of an increasing function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{\sqrt{2-x^2}} dx = \left[\arcsin \frac{x}{\sqrt{2}}\right]\) | M1 | arcsin function alone, or any sine substitution |
| \(= \frac{\pi}{4}\) | A1 | Evaluated in terms of π |
| Answer | Marks | Guidance |
|---|---|---|
| \(C + jS = e^{j\theta} + \frac{1}{2}e^{j\theta} + \frac{1}{2}e^{j\theta} + \ldots\) | M1 | Forming C + jS as a series of powers. Identifying geometric series and attempting sum to infinity or to n terms |
| This is a geometric series | M1 | |
| with first term \(a = e^{j\theta}\), common ratio \(r = \frac{1}{2}e^{j\theta}\) | A1 | Correct a and r |
| Sum to infinity \(= \frac{a}{1-r} = \frac{e^{j\theta}}{1-\frac{1}{2}e^{j\theta}} = \frac{3e^{j\theta}}{3-e^{j\theta}}\) | A1 | Sum to infinity. Multiplying numerator and denominator by \(1-\frac{1}{2}e^{j\theta}\) o.e. Or writing in terms of trig functions and realising the denominator |
| \(= \frac{3e^{j\theta} \cdot 3-e^{-j\theta}}{3-e^{j\theta} \cdot 3-e^{-j\theta}}\) | M1* | Multiplying out numerator and denominator. Dep. on M1*. Valid attempt to express in terms of trig functions. If trig functions used from start, M1 for using the compound angle formulae and Pythagoras |
| \(= \frac{9e^{j\theta} - 3e^{-j\theta}}{9 - 3e^{j\theta} - 3e^{-j\theta} + 1}\) | M1 | |
| \(= \frac{9(\cos\theta + j\sin\theta) - 3(\cos\theta - j\sin\theta)}{10 - 3(\cos 2\theta + \sin 2\theta) - 3(\cos 2\theta + j\sin 2\theta)}\) | M1 | |
| \(= \frac{6\cos\theta + 12j\sin\theta}{10 - 6\cos 2\theta}\) | A1 | |
| \(\Rightarrow C = \frac{6\cos\theta}{10 - 6\cos 2\theta}\) | M1 | Equating real and imaginary parts. Dep. on M1* |
| \(= \frac{3\cos\theta}{5-3\cos 2\theta}\) | A1 (ag) | |
| \(S = \frac{6\sin\theta}{5-3\cos 2\theta}\) | A1 | o.e. |
## Part (a)(i)
**Graph**: Correct basic shape (positive gradient, through (0, 0)) | G1 |
$y = \arcsin x \Rightarrow \sin y = x$ | M1 | $\sin y = x$ and attempt to diff. both sides
$\Rightarrow \frac{dx}{dy} = \cos y$ | A1 | Or $\cos y \frac{dx}{dy} = 1$
$\Rightarrow \frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}}$ | A1 | www. SC1 if quoted without working
Positive square root because gradient positive | B1 | Dep. on graph of an increasing function | | 4 marks total
## Part (a)(ii)
$\int \frac{1}{\sqrt{2-x^2}} dx = \left[\arcsin \frac{x}{\sqrt{2}}\right]$ | M1 | arcsin function alone, or any sine substitution
$= \frac{\pi}{4}$ | A1 | Evaluated in terms of π | | 3 marks total
## Part (b)
$C + jS = e^{j\theta} + \frac{1}{2}e^{j\theta} + \frac{1}{2}e^{j\theta} + \ldots$ | M1 | Forming C + jS as a series of powers. Identifying geometric series and attempting sum to infinity or to n terms
This is a geometric series | M1 |
with first term $a = e^{j\theta}$, common ratio $r = \frac{1}{2}e^{j\theta}$ | A1 | Correct a and r
Sum to infinity $= \frac{a}{1-r} = \frac{e^{j\theta}}{1-\frac{1}{2}e^{j\theta}} = \frac{3e^{j\theta}}{3-e^{j\theta}}$ | A1 | Sum to infinity. Multiplying numerator and denominator by $1-\frac{1}{2}e^{j\theta}$ o.e. Or writing in terms of trig functions and realising the denominator
$= \frac{3e^{j\theta} \cdot 3-e^{-j\theta}}{3-e^{j\theta} \cdot 3-e^{-j\theta}}$ | M1* | Multiplying out numerator and denominator. Dep. on M1*. Valid attempt to express in terms of trig functions. If trig functions used from start, M1 for using the compound angle formulae and Pythagoras
$= \frac{9e^{j\theta} - 3e^{-j\theta}}{9 - 3e^{j\theta} - 3e^{-j\theta} + 1}$ | M1 |
$= \frac{9(\cos\theta + j\sin\theta) - 3(\cos\theta - j\sin\theta)}{10 - 3(\cos 2\theta + \sin 2\theta) - 3(\cos 2\theta + j\sin 2\theta)}$ | M1 |
$= \frac{6\cos\theta + 12j\sin\theta}{10 - 6\cos 2\theta}$ | A1 |
$\Rightarrow C = \frac{6\cos\theta}{10 - 6\cos 2\theta}$ | M1 | Equating real and imaginary parts. Dep. on M1*
$= \frac{3\cos\theta}{5-3\cos 2\theta}$ | A1 (ag) |
$S = \frac{6\sin\theta}{5-3\cos 2\theta}$ | A1 | o.e. | | 11 marks total, 19 marks for Question 3
---\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = \arcsin x$ for $-1 \leq x \leq 1$. [1]
Find $\frac{dy}{dx}$, justifying the sign of your answer by reference to your sketch. [4]
\item Find the exact value of the integral $\int_0^1 \frac{1}{\sqrt{2 - x^2}} dx$. [3]
\end{enumerate}
\item The infinite series $C$ and $S$ are defined as follows.
$$C = \cos \theta + \frac{1}{3}\cos 3\theta + \frac{1}{5}\cos 5\theta + \ldots$$
$$S = \sin \theta + \frac{1}{3}\sin 3\theta + \frac{1}{5}\sin 5\theta + \ldots$$
By considering $C + jS$, show that
$$C = \frac{3\cos \theta}{5 - 3\cos 2\theta},$$
and find a similar expression for $S$. [11]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q3 [19]}}