| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Exponential or trigonometric base functions |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on Maclaurin series with clear scaffolding. Part (a) is routine differentiation using the product rule. Parts (b) and (c) involve standard Maclaurin series techniques that are well-practiced at this level. Part (d) requires using the series to solve an equation, which adds some problem-solving but follows directly from the previous work. While it's a multi-part question worth 16 marks total, each component is methodical and the scaffolding guides students through. It's moderately above average difficulty due to the Further Maths content and length, but not exceptionally challenging. |
| Spec | 4.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = e^x\cos x - e^x\sin x\) | B1 | AO2 |
| \(f''(x) = e^x\cos x - e^x\sin x - e^x\sin x - e^x\cos x = -2e^x\sin x\) | B1 | AO2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'''(x) = -2e^x\sin x - 2e^x\cos x\) | B1 | AO1 |
| \(f^{(4)}(x) = -2e^x\sin x - 2e^x\cos x - 2e^x\cos x + 2e^x\sin x\) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0) = 1, f'(0) = 1, f''(0) = 0\) | B1 | AO1 |
| \(f'''(0) = -2, f^{(4)}(0) = -4\) | B1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^x\cos x = 1 + x - \frac{2x^3}{6} - \frac{4x^4}{24} + ...\) | M1 | AO1 |
| \(= 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + ...\) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^x\cos x - e^x\sin x = 1 - x^2 - \frac{2x^3}{3} + ...\) | M1 | AO1 |
| \(e^x\sin x = 1 + x - \frac{x^3}{3} - 1 + x^2 + \frac{2x^3}{3} + ...\) | A1 | AO1 |
| \(= x + x^2 + \frac{x^3}{3} + ...\) | A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(10\left(x + x^2 + \frac{x^3}{3}\right) - 11x = 0\) | M1 | AO3 |
| \(10x^3 + 30x^2 - 3x = 0\) | A1 | AO3 |
| \(x = \frac{-30 + \sqrt{900+120}}{20}\) | m1 | AO3 |
| \(= 0.097\) | A1 | AO3 |
## Part (a)
$f'(x) = e^x\cos x - e^x\sin x$ | B1 | AO2
$f''(x) = e^x\cos x - e^x\sin x - e^x\sin x - e^x\cos x = -2e^x\sin x$ | B1 | AO2
## Part (b)
$f'''(x) = -2e^x\sin x - 2e^x\cos x$ | B1 | AO1
$f^{(4)}(x) = -2e^x\sin x - 2e^x\cos x - 2e^x\cos x + 2e^x\sin x$ | B1 | AO1
$(= -4e^x\cos x)$
$f(0) = 1, f'(0) = 1, f''(0) = 0$ | B1 | AO1
$f'''(0) = -2, f^{(4)}(0) = -4$ | B1 | AO1
The Maclaurin series is:
$e^x\cos x = 1 + x - \frac{2x^3}{6} - \frac{4x^4}{24} + ...$ | M1 | AO1
$= 1 + x - \frac{x^3}{3} - \frac{x^4}{6} + ...$ | A1 | AO1
## Part (c)
Valid attempt at differentiating both sides:
$e^x\cos x - e^x\sin x = 1 - x^2 - \frac{2x^3}{3} + ...$ | M1 | AO1
$e^x\sin x = 1 + x - \frac{x^3}{3} - 1 + x^2 + \frac{2x^3}{3} + ...$ | A1 | AO1
$= x + x^2 + \frac{x^3}{3} + ...$ | A1 | AO1
## Part (d)
Replacing $e^x\sin x$ by its series:
$10\left(x + x^2 + \frac{x^3}{3}\right) - 11x = 0$ | M1 | AO3
$10x^3 + 30x^2 - 3x = 0$ | A1 | AO3
$x = \frac{-30 + \sqrt{900+120}}{20}$ | m1 | AO3
$= 0.097$ | A1 | AO3
**Total: [16]**The function $f$ is given by
$$f(x) = e^x \cos x.$$
\begin{enumerate}[label=(\alph*)]
\item Show that $f''(x) = -2e^x \sin x$. [2]
\item Determine the Maclaurin series for $f(x)$ as far as the $x^4$ term. [6]
\item Hence, by differentiating your series, determine the Maclaurin series for $e^x \sin x$ as far as the $x^3$ term. [4]
\item The equation
$$10e^x \sin x - 11x = 0$$
has a small positive root. Determine its approximate value, giving your answer correct to three decimal places. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q12 [16]}}