| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Deduce related series from given series |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring Taylor series manipulation and limit evaluation using series expansions. Part (a) is routine substitution, part (b) requires computing derivatives of a composite exponential function using Maclaurin's theorem (moderately technical), and part (c) is a challenging limit requiring careful combination of multiple series expansions and algebraic manipulation. The final limit calculation demands sophisticated understanding of dominant terms and is non-routine, placing this above average difficulty for Further Maths. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{\sin x} = 1+\left(x-\frac{x^3}{3!}\cdots\right)+\frac{1}{2!}\left(x-\frac{x^3}{3!}\cdots\right)^2+\frac{1}{3!}\left(x-\frac{x^3}{3!}\cdots\right)^3\cdots\) | M1 | For 1st 3 terms ignoring higher powers than those shown |
| \(= 1+x-\frac{x^3}{6}+\frac{1}{2}(x^2-\cdots)+\frac{1}{6}(x^3-\cdots)\) | ||
| \(= 1+x+\frac{1}{2}x^2\) | A1 | be convinced….ignore any powers of \(x\) above power 2 |
| Coefficient of \(x^3\): \(-\frac{x^3}{6}+\frac{1}{6}x^3 = 0\) | A1 | be convinced….ignore any powers of \(x\) above power 3 |
**Question 4 (SC extra notes):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{\sin x} = 1+\left(x-\frac{x^3}{3!}\cdots\right)+\frac{1}{2!}\left(x-\frac{x^3}{3!}\cdots\right)^2+\frac{1}{3!}\left(x-\frac{x^3}{3!}\cdots\right)^3\cdots$ | M1 | For 1st 3 terms ignoring higher powers than those shown |
| $= 1+x-\frac{x^3}{6}+\frac{1}{2}(x^2-\cdots)+\frac{1}{6}(x^3-\cdots)$ | | |
| $= 1+x+\frac{1}{2}x^2$ | A1 | **be convinced….ignore any powers of $x$ above power 2** |
| Coefficient of $x^3$: $-\frac{x^3}{6}+\frac{1}{6}x^3 = 0$ | A1 | **be convinced….ignore any powers of $x$ above power 3** |4
\begin{enumerate}[label=(\alph*)]
\item Use the series expansion
$$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 } + \ldots$$
to write down the first four terms in the expansion, in ascending powers of $x$, of $\ln ( 1 - x )$.
\item The function f is defined by
$$\mathrm { f } ( x ) = \mathrm { e } ^ { \sin x }$$
Use Maclaurin's theorem to show that when $\mathrm { f } ( x )$ is expanded in ascending powers of $x$ :
\begin{enumerate}[label=(\roman*)]
\item the first three terms are
$$1 + x + \frac { 1 } { 2 } x ^ { 2 }$$
\item the coefficient of $x ^ { 3 }$ is zero.
\end{enumerate}\item Find
$$\lim _ { x \rightarrow 0 } \frac { \mathrm { e } ^ { \sin x } - 1 + \ln ( 1 - x ) } { x ^ { 2 } \sin x }$$
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2006 Q4 [14]}}