| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for composite exponential/root functions |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard Maclaurin series techniques. Part (a) is routine recall, part (b) involves mechanical differentiation and substitution into Maclaurin's formula, and part (c) applies L'Hôpital's rule or uses the series from parts (a) and (b). While it requires multiple techniques, each step follows standard procedures without requiring novel insight or complex problem-solving. |
| 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 |
|---|---|---|
| \(\sin 2x \approx 2x - \frac{(2x)^3}{3!} + \ldots = 2x - \frac{4}{3}x^3 + \ldots\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{1}{2}(3+e^x)^{-\frac{1}{2}}(e^x)\) | M1, A1 | Chain rule |
| \(\frac{d^2y}{dx^2} = \frac{1}{2}e^x(3+e^x)^{-\frac{1}{2}} - \frac{1}{4}(3+e^x)^{-\frac{3}{2}}(e^{2x})\) | M1, A1 | Product rule OE |
| \(y'(0) = \frac{1}{4}\); \(y''(0) = \frac{1}{4} - \frac{1}{32} = \frac{7}{32}\) | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{3+e^x} \approx 2 + \frac{1}{4}x + \frac{7}{64}x^2\) | M1, A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\frac{\sqrt{3+e^x}-2}{\sin 2x}\right] = \left[\frac{2+\frac{1}{4}x+\frac{7}{64}x^2-2}{2x-\frac{4}{3}x^3}\right]\) | M1 | |
| \(= \left[\frac{\frac{1}{4}+\frac{7}{64}x+\ldots}{2-\frac{4}{3}x^2+\ldots}\right]\) | m1 | Dividing numerator and denominator by \(x\) to get constant term in each |
| \(\lim_{x \to 0}\left[\frac{\sqrt{3+e^x}-2}{\sin 2x}\right] = \frac{\frac{1}{4}}{2} = \frac{1}{8}\) | A1F | Total: 3 |
# Question 7:
## Part (a):
$\sin 2x \approx 2x - \frac{(2x)^3}{3!} + \ldots = 2x - \frac{4}{3}x^3 + \ldots$ | B1 | Total: 1 |
## Part (b)(i):
$\frac{dy}{dx} = \frac{1}{2}(3+e^x)^{-\frac{1}{2}}(e^x)$ | M1, A1 | Chain rule |
$\frac{d^2y}{dx^2} = \frac{1}{2}e^x(3+e^x)^{-\frac{1}{2}} - \frac{1}{4}(3+e^x)^{-\frac{3}{2}}(e^{2x})$ | M1, A1 | Product rule OE |
$y'(0) = \frac{1}{4}$; $y''(0) = \frac{1}{4} - \frac{1}{32} = \frac{7}{32}$ | A1 | Total: 5 | CSO |
## Part (b)(ii):
$y(0) = 2$; $y'(0) = \frac{1}{4}$; $y''(0) = \frac{1}{4} - \frac{1}{32} = \frac{7}{32}$
Maclaurin: $y(0) + x\,y'(0) + \frac{x^2}{2}\,y''(0)$
$\sqrt{3+e^x} \approx 2 + \frac{1}{4}x + \frac{7}{64}x^2$ | M1, A1 | Total: 2 | CSO; AG |
## Part (c):
$\left[\frac{\sqrt{3+e^x}-2}{\sin 2x}\right] = \left[\frac{2+\frac{1}{4}x+\frac{7}{64}x^2-2}{2x-\frac{4}{3}x^3}\right]$ | M1 | |
$= \left[\frac{\frac{1}{4}+\frac{7}{64}x+\ldots}{2-\frac{4}{3}x^2+\ldots}\right]$ | m1 | Dividing numerator and denominator by $x$ to get constant term in each |
$\lim_{x \to 0}\left[\frac{\sqrt{3+e^x}-2}{\sin 2x}\right] = \frac{\frac{1}{4}}{2} = \frac{1}{8}$ | A1F | Total: 3 | Ft on candidate's answer to (a) provided of the form $ax+bx^3$ |
---7
\begin{enumerate}[label=(\alph*)]
\item Write down the expansion of $\sin 2 x$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \sqrt { 3 + \mathrm { e } ^ { x } }$, find the values of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = 0$.
\item Using Maclaurin's theorem, show that, for small values of $x$,
$$\sqrt { 3 + \mathrm { e } ^ { x } } \approx 2 + \frac { 1 } { 4 } x + \frac { 7 } { 64 } x ^ { 2 }$$
\end{enumerate}\item Find
$$\lim _ { x \rightarrow 0 } \left[ \frac { \sqrt { 3 + \mathrm { e } ^ { x } } - 2 } { \sin 2 x } \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q7 [11]}}