| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Find series for logarithmic function |
| Difficulty | Standard +0.3 This is a straightforward Maclaurin series question requiring standard differentiation of logarithmic and trigonometric functions, followed by routine evaluation at x=0. Part (b) is given as 'show that', making it easier. The derivatives are manageable and the series construction follows a standard algorithm with no novel insight required. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07d Second derivatives: d^2y/dx^2 notation1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = \frac{k\cos kx}{1 + \sin kx}\) | M1 A1 | |
| Subtotal | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f''(x) = \frac{-(1+\sin kx)k^2\sin kx - k\cos kx(k\cos kx)}{(1+\sin kx)^2}\) | M1 | |
| So \(f''(x) = \frac{-k^2\sin kx - k^2(\sin^2 kx + \cos^2 kx)}{(1+\sin kx)^2}\) and use \(\sin^2 kx + \cos^2 kx = 1\) | M1 | |
| \(f''(x) = \frac{-k^2(1+\sin kx)}{(1+\sin kx)^2} = \frac{-k^2}{1+\sin kx}\) | A1cao | |
| Subtotal | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'''(x) = \frac{k^3\cos kx}{(1+\sin kx)^2}\) | B1 | |
| Finds \(f(0), f'(0), f''(0)\) and \(f'''(0)\) | M1 | |
| Uses Maclaurin Expansion to obtain \(f(x) = 0 + kx - \frac{k^2}{2}x^2 + \frac{k^3}{6}x^3\ldots\) | M1 A1 | |
| Subtotal | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} + \ldots\) with \(y = \sin kx\) | B1 | |
| Use \(\sin kx = kx - \frac{k^3x^3}{6}\ldots\) in ln expansion | M1 | |
| Obtains \(0 + kx - \frac{k^2}{2}x^2 + \frac{k^3}{6}x^3\ldots\) | M1 A1 | |
| Subtotal | (4) |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = \frac{k\cos kx}{1 + \sin kx}$ | M1 A1 | |
| **Subtotal** | **(2)** | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f''(x) = \frac{-(1+\sin kx)k^2\sin kx - k\cos kx(k\cos kx)}{(1+\sin kx)^2}$ | M1 | |
| So $f''(x) = \frac{-k^2\sin kx - k^2(\sin^2 kx + \cos^2 kx)}{(1+\sin kx)^2}$ and use $\sin^2 kx + \cos^2 kx = 1$ | M1 | |
| $f''(x) = \frac{-k^2(1+\sin kx)}{(1+\sin kx)^2} = \frac{-k^2}{1+\sin kx}$ | A1cao | |
| **Subtotal** | **(3)** | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'''(x) = \frac{k^3\cos kx}{(1+\sin kx)^2}$ | B1 | |
| Finds $f(0), f'(0), f''(0)$ and $f'''(0)$ | M1 | |
| Uses Maclaurin Expansion to obtain $f(x) = 0 + kx - \frac{k^2}{2}x^2 + \frac{k^3}{6}x^3\ldots$ | M1 A1 | |
| **Subtotal** | **(4)** | |
### Part (c) Alternative Method
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} + \ldots$ with $y = \sin kx$ | B1 | |
| Use $\sin kx = kx - \frac{k^3x^3}{6}\ldots$ in ln expansion | M1 | |
| Obtains $0 + kx - \frac{k^2}{2}x^2 + \frac{k^3}{6}x^3\ldots$ | M1 A1 | |
| **Subtotal** | **(4)** | |
---3.
$$f ( x ) = \ln ( 1 + \sin k x )$$
where $k$ is a constant, $x \in \mathbb { R }$ and $- \frac { \pi } { 2 } < k x < \frac { 3 \pi } { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Find f ${ } ^ { \prime } ( x )$
\item Show that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { - k ^ { 2 } } { 1 + \sin k x }$
\item Find the Maclaurin series of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q3 [9]}}