| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Use binomial with exponential series |
| Difficulty | Standard +0.8 This is a substantial Further Maths question requiring multiple techniques: binomial series with fractional exponent, exponential series, product of series (requiring careful collection of terms to specified order), Maclaurin's theorem with second derivatives, and a limit using series approximations. While each component is standard FP3 material, the multi-part structure, the algebraic manipulation in part (a)(ii), and the synthesis required in part (c) place this above average difficulty for A-level, though not exceptionally hard for Further Maths students who have practiced these techniques. |
| 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \ldots\) | M1 | Clear use of \(x\to 2x\) in expansion of \(e^x\) |
| A1 | Total: 2; ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\{f(x)\} = e^{2x}(1+3x)^{-\frac{2}{3}}\) | ||
| \((1+3x)^{-\frac{2}{3}} = 1 + \left(-\frac{2}{3}\right)(3x) + \frac{\left(-\frac{2}{3}\right)\!\left(-\frac{5}{3}\right)(3x)^2}{2} - \frac{40}{3}x^3\) | M1 | First three terms as \(1+\left(-\frac{2}{3}\right)(3x) + kx^2\) OE |
| \(= 1 - 2x + 5x^2 - \frac{40}{3}x^3\) | A1 | |
| \(\{f(x)\} \approx\) \(1+2x+2x^2+\frac{4x^3}{3} - 2x - 4x^2 - 4x^3 + 5x^2 + 10x^3 - \frac{40x^3}{3}\) | m1, A1ft | Dep on both prev MS; condone one sign or numerical slip in mult. |
| \(= 1 + 3x^2 - 6x^3\) | A1 | Total: 5; CSO AG; A0 if binomial series not used |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = \ln(1+2\sin x) \Rightarrow \frac{dy}{dx} = \frac{1}{1+2\sin x}\times 2\cos x\) | M1, A1 | Chain rule |
| \(\frac{d^2y}{dx^2} = \frac{(1+2\sin x)(-2\sin x) - 2\cos x(2\cos x)}{(1+2\sin x)^2} = \frac{-2(\sin x + 2)}{(1+2\sin x)^2}\) | M1 | Quotient rule OE with \(u\) and \(v\) non constant |
| A1 | Total: 4; ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y(0) = 0,\quad y'(0) = 2,\quad y''(0) = -4\) | M1 | |
| McLaurin Thm.: \(\{\ln(1+2\sin x)\} \approx 0 + 2x - 4\!\left(\frac{x^2}{2}\right) + \ldots \approx 2x - 2x^2\) | A1 | Total: 2; CSO AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\lim_{x\to 0}\frac{1-f(x)}{x\ln(1+2\sin x)} = \lim_{x\to 0}\frac{-3x^2+6x^3}{2x^2-2x^3}\) | M1 | Using expansions |
| \(= \lim_{x\to 0}\frac{-3+6x}{2-2x}\) | m1 | Division by \(x^2\) stage before taking limit |
| \(= -\frac{3}{2}\) | A1 | Total: 3; CSO |
## Question 6:
### Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|---------|
| $e^{2x} = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \ldots$ | M1 | Clear use of $x\to 2x$ in expansion of $e^x$ |
| | A1 | **Total: 2**; ACF |
### Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|---------|
| $\{f(x)\} = e^{2x}(1+3x)^{-\frac{2}{3}}$ | | |
| $(1+3x)^{-\frac{2}{3}} = 1 + \left(-\frac{2}{3}\right)(3x) + \frac{\left(-\frac{2}{3}\right)\!\left(-\frac{5}{3}\right)(3x)^2}{2} - \frac{40}{3}x^3$ | M1 | First three terms as $1+\left(-\frac{2}{3}\right)(3x) + kx^2$ OE |
| $= 1 - 2x + 5x^2 - \frac{40}{3}x^3$ | A1 | |
| $\{f(x)\} \approx$ $1+2x+2x^2+\frac{4x^3}{3} - 2x - 4x^2 - 4x^3 + 5x^2 + 10x^3 - \frac{40x^3}{3}$ | m1, A1ft | Dep on both prev MS; condone one sign or numerical slip in mult. |
| $= 1 + 3x^2 - 6x^3$ | A1 | **Total: 5**; CSO AG; A0 if binomial series not used |
### Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = \ln(1+2\sin x) \Rightarrow \frac{dy}{dx} = \frac{1}{1+2\sin x}\times 2\cos x$ | M1, A1 | Chain rule |
| $\frac{d^2y}{dx^2} = \frac{(1+2\sin x)(-2\sin x) - 2\cos x(2\cos x)}{(1+2\sin x)^2} = \frac{-2(\sin x + 2)}{(1+2\sin x)^2}$ | M1 | Quotient rule OE with $u$ and $v$ non constant |
| | A1 | **Total: 4**; ACF |
### Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|---------|
| $y(0) = 0,\quad y'(0) = 2,\quad y''(0) = -4$ | M1 | |
| McLaurin Thm.: $\{\ln(1+2\sin x)\} \approx 0 + 2x - 4\!\left(\frac{x^2}{2}\right) + \ldots \approx 2x - 2x^2$ | A1 | **Total: 2**; CSO AG |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|---------|
| $\lim_{x\to 0}\frac{1-f(x)}{x\ln(1+2\sin x)} = \lim_{x\to 0}\frac{-3x^2+6x^3}{2x^2-2x^3}$ | M1 | Using expansions |
| $= \lim_{x\to 0}\frac{-3+6x}{2-2x}$ | m1 | Division by $x^2$ stage before taking limit |
| $= -\frac{3}{2}$ | A1 | **Total: 3**; CSO |6 The function f is defined by $\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } ( 1 + 3 x ) ^ { - \frac { 2 } { 3 } }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use the series expansion for $\mathrm { e } ^ { x }$ to write down the first four terms in the series expansion of $\mathrm { e } ^ { 2 x }$.
\item Use the binomial series expansion of $( 1 + 3 x ) ^ { - \frac { 2 } { 3 } }$ and your answer to part (a)(i) to show that the first three non-zero terms in the series expansion of $\mathrm { f } ( x )$ are $1 + 3 x ^ { 2 } - 6 x ^ { 3 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \ln ( 1 + 2 \sin x )$, find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item By using Maclaurin's theorem, show that, for small values of $x$,
$$\ln ( 1 + 2 \sin x ) \approx 2 x - 2 x ^ { 2 }$$
\end{enumerate}\item Find
$$\lim _ { x \rightarrow 0 } \frac { 1 - \mathrm { f } ( x ) } { x \ln ( 1 + 2 \sin x ) }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q6 [16]}}