| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Combining or manipulating standard series |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of Maclaurin series and the double-angle identity cos²x = (1+cos2x)/2. Part (a) is routine substitution into the standard sin x series. Part (b) requires recognizing the identity, integrating the series from (a), and careful algebraic manipulation—more conceptually demanding than standard A-level calculus but still a recognizable technique for Further Maths students. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| a) | METHOD 1: | |
| \(\sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!}\) | M1 | Use of formula booklet expansion of \(\sin x\) |
| \(\sin 2x = 2x - \frac{8}{6}x^3 + \frac{32}{120}x^5 + \ldots\) | ||
| \(\sin 2x = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5 + \ldots\) | A1 | |
| METHOD 2: | ||
| \(\sin 2x = \sin 0 + 2x\cos 0 + \frac{(2x)^2}{2!}(-\sin 0) + \frac{(2x)^3}{3!}(-\cos 0) + \frac{(2x)^4}{4!}(\sin 0) + \frac{(2x)^5}{5!}(\cos 0)\) | (M1) | Use of formula booklet expansion of \(f(x)\) with \(\sin x\) and \(2x\) |
| \(\sin 2x = 2x - \frac{8}{6}x^3 + \frac{32}{120}x^5 + \ldots\) | ||
| \(\sin 2x = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5 + \ldots\) | (A1) | |
| (2) | ||
| b) | Differentiating: \(2\cos 2x = 2 - 4x^2 + \frac{4}{3}x^4\) | M1, A1 |
| \(\cos 2x = 1 - 2x^2 + \frac{2}{3}x^4\) | ||
| \(\cos 2x = 2\cos^2 x - 1\) | ||
| Therefore: \(1 - 2x^2 + \frac{2}{3}x^4 = 2\cos^2 x - 1\) | m1, A1 | |
| \(2\cos^2 x = 2 - 2x^2 + \frac{2}{3}x^4\) | ||
| \(\cos^2 x = 1 - x^2 + \frac{1}{3}x^4\) | A1 | cao |
| (5) | ||
| [7] |
a) | **METHOD 1:** | | |
| $\sin 2x = 2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!}$ | M1 | Use of formula booklet expansion of $\sin x$ |
| $\sin 2x = 2x - \frac{8}{6}x^3 + \frac{32}{120}x^5 + \ldots$ | | |
| $\sin 2x = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5 + \ldots$ | A1 | |
| | | |
| **METHOD 2:** | | |
| $\sin 2x = \sin 0 + 2x\cos 0 + \frac{(2x)^2}{2!}(-\sin 0) + \frac{(2x)^3}{3!}(-\cos 0) + \frac{(2x)^4}{4!}(\sin 0) + \frac{(2x)^5}{5!}(\cos 0)$ | (M1) | Use of formula booklet expansion of $f(x)$ with $\sin x$ and $2x$ |
| $\sin 2x = 2x - \frac{8}{6}x^3 + \frac{32}{120}x^5 + \ldots$ | | |
| $\sin 2x = 2x - \frac{4}{3}x^3 + \frac{4}{15}x^5 + \ldots$ | (A1) | |
| | (2) | |
b) | Differentiating: $2\cos 2x = 2 - 4x^2 + \frac{4}{3}x^4$ | M1, A1 | FT (a); oe |
| | | |
| $\cos 2x = 1 - 2x^2 + \frac{2}{3}x^4$ | | |
| | | |
| $\cos 2x = 2\cos^2 x - 1$ | | |
| | | |
| Therefore: $1 - 2x^2 + \frac{2}{3}x^4 = 2\cos^2 x - 1$ | m1, A1 | |
| | | |
| $2\cos^2 x = 2 - 2x^2 + \frac{2}{3}x^4$ | | |
| | | |
| $\cos^2 x = 1 - x^2 + \frac{1}{3}x^4$ | A1 | cao |
| | (5) | |
| | [7] | |\begin{enumerate}[label=(\alph*)]
\item Write down and simplify the Maclaurin series for $\sin 2x$ as far as the term in $x^5$. [2]
\item Using your answer to part (a), determine the Maclaurin series for $\cos^2 x$ as far as the term in $x^4$. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q5 [7]}}