| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Deduce related series from given series |
| Difficulty | Moderate -0.3 Part (a) is direct recall of a standard Maclaurin series. Part (b) requires manipulating logarithms using log laws, then combining two standard series expansions and collecting terms—straightforward algebraic manipulation with no novel insight required. This is slightly easier than average due to being mostly procedural, though the algebraic work in part (b) requires care. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks |
|---|---|
| Use of formula booklet expansion with \(-x\) | M1 |
| \(\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: | ||
| \(-2\ln\left(\frac{1-x}{(1+x)^2}\right) = -2[\ln(1-x) - 2\ln(1+x)]\) | M1A1 | |
| \(= -2 \times \left[-x - \frac{x^2}{2} - \frac{x^3}{3} - 2\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)\right]\) | A1 | FT their (a) |
| \(= -2 \times \left[-3x + \frac{x^2}{2} - x^3\right]\) | A1 | cao Mark final answer |
| \(= 6x - x^2 + 2x^3\) | ||
| METHOD 2: Let \(f(x) = -2\ln\left(\frac{1-x}{(1+x)^2}\right)\) {\(= -2\ln(1-x) + 4\ln(1+x)\)} | (M1) | |
| \(f(0) = 0\) | ||
| \(f'(x) = \frac{-2}{1-x} + \frac{4}{1+x}\) | ||
| \(f''(x) = 2(1-x)^{-2} - 4(1+x)^{-2}\) | ||
| \(f'''(x) = 4(1-x)^{-3} + 8(1+x)^{-3}\) | ||
| \(f'(0) = 6\) | ||
| \(f''(0) = -2\) | ||
| \(f'''(0) = 12\) | ||
| Therefore, | ||
| \(-2\ln\left(\frac{1-x}{(1+x)^2}\right) = 0 + x(6) + x^2\left(\frac{-2}{2!}\right) + x^3\left(\frac{12}{3!}\right)\) | (A3) | cao A1 each correct term Mark final answer |
| \(-2\ln\left(\frac{1-x}{(1+x)^2}\right) = 6x - x^2 + 2x^3\) |
## Part a)
Use of formula booklet expansion with $-x$ | M1 |
$\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$ | A1 |
## Part b)
METHOD 1: | |
$-2\ln\left(\frac{1-x}{(1+x)^2}\right) = -2[\ln(1-x) - 2\ln(1+x)]$ | M1A1 |
$= -2 \times \left[-x - \frac{x^2}{2} - \frac{x^3}{3} - 2\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)\right]$ | A1 | FT their (a)
$= -2 \times \left[-3x + \frac{x^2}{2} - x^3\right]$ | A1 | cao Mark final answer
$= 6x - x^2 + 2x^3$ | |
METHOD 2: Let $f(x) = -2\ln\left(\frac{1-x}{(1+x)^2}\right)$ {$= -2\ln(1-x) + 4\ln(1+x)$} | (M1) |
$f(0) = 0$ | |
$f'(x) = \frac{-2}{1-x} + \frac{4}{1+x}$ | |
$f''(x) = 2(1-x)^{-2} - 4(1+x)^{-2}$ | |
$f'''(x) = 4(1-x)^{-3} + 8(1+x)^{-3}$ | |
$f'(0) = 6$ | |
$f''(0) = -2$ | |
$f'''(0) = 12$ | |
Therefore, | |
$-2\ln\left(\frac{1-x}{(1+x)^2}\right) = 0 + x(6) + x^2\left(\frac{-2}{2!}\right) + x^3\left(\frac{12}{3!}\right)$ | (A3) | cao A1 each correct term Mark final answer
$-2\ln\left(\frac{1-x}{(1+x)^2}\right) = 6x - x^2 + 2x^3$ | |\begin{enumerate}[label=(\alph*)]
\item Write down the Maclaurin series expansion for $\ln(1 - x)$ as far as the term in $x^3$. [2]
\item Show that $-2\ln\left(\frac{1-x}{(1+x)^2}\right)$ can be expressed in the form $ax + bx^2 + cx^3 + \ldots$, where $a$, $b$, $c$ are integers whose values are to be determined. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q7 [6]}}