| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(exponential expressions) |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring systematic application of Maclaurin series through differentiation, algebraic manipulation of series, and limit evaluation using series expansions. While each individual step follows standard techniques, the question demands careful execution across multiple parts with the final limit requiring insight to combine series and use the known expansion of sin(x). The computational burden and multi-step reasoning elevate this above average difficulty. |
| 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 |
| \(f(0) = \ln 2\) | B1 | |
| \(f'(x) = \frac{e^x}{1+e^x}\), \(f'(0) = \frac{1}{2}\) | M1, A1 | Chain rule |
| \(f''(x) = \frac{(1+e^x)e^x - e^x e^x}{(1+e^x)^2} = \frac{e^x}{(1+e^x)^2}\) | M1, A1 | Quotient rule OE |
| \(f''(0) = \frac{1}{4}\) | ||
| \(f(x) = \ln 2 + \frac{1}{2}x + \frac{1}{4}\cdot\frac{x^2}{2!} = \ln 2 + \frac{1}{2}x + \frac{1}{8}x^2\) | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'''(x) = \frac{(1+e^x)^2 e^x - e^x\left[2(1+e^x)e^x\right]}{(1+e^x)^4}\) | M1, A1ft | Chain rule with quotient/product rule; ft on \(f''(x) = ke^x(1+e^x)^n\) (integer \(n<0\)) |
| \(f'''(0) = \frac{4-4}{2^4} = 0\), so coefficient of \(x^3\) is zero | A1 | 3 marks total |
| SC for those not using Maclaurin's theorem: maximum of 4/9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}x + \frac{1}{8}x^2\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ln\left(1-\frac{x}{2}\right) = \left(-\frac{x}{2}\right) - \frac{1}{2}\left(-\frac{x}{2}\right)^2 + \frac{1}{3}\left(-\frac{x}{2}\right)^3 - \ldots\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ln\left(\frac{1+e^x}{2}\right) + \ln\left(1-\frac{x}{2}\right) = -\frac{x^3}{24} + \ldots\) | M1 | Uses previous expansions to obtain first non-zero term of the form \(kx^3\) |
| \(x - \sin x \approx x - \left[x - \frac{x^3}{3!} + \ldots\right] \approx \frac{x^3}{3!} + \ldots\) | B1 | |
| \(\left[\frac{\ln\left(\frac{1+e^x}{2}\right) + \ln\left(1-\frac{x}{2}\right)}{x - \sin x}\right] = \frac{-\frac{1}{24}x^3 + \ldots}{\frac{1}{6}x^3 + o(x^5)}\) | M1 | |
| \(= \frac{-\frac{1}{24}x^3 + \ldots}{x^3\left[\frac{1}{6} + o(x^2)\right]} = \frac{-\frac{1}{24} + \ldots}{\frac{1}{6} + o(x^2)}\) | ||
| \(\lim_{x \to 0} \ldots = -\frac{1}{4}\) | A1 | 4 marks total |
# Question 6:
## Part 6(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0) = \ln 2$ | B1 | |
| $f'(x) = \frac{e^x}{1+e^x}$, $f'(0) = \frac{1}{2}$ | M1, A1 | Chain rule |
| $f''(x) = \frac{(1+e^x)e^x - e^x e^x}{(1+e^x)^2} = \frac{e^x}{(1+e^x)^2}$ | M1, A1 | Quotient rule OE |
| $f''(0) = \frac{1}{4}$ | | |
| $f(x) = \ln 2 + \frac{1}{2}x + \frac{1}{4}\cdot\frac{x^2}{2!} = \ln 2 + \frac{1}{2}x + \frac{1}{8}x^2$ | A1 | 6 marks total | CSO; AG |
## Part 6(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'''(x) = \frac{(1+e^x)^2 e^x - e^x\left[2(1+e^x)e^x\right]}{(1+e^x)^4}$ | M1, A1ft | Chain rule with quotient/product rule; ft on $f''(x) = ke^x(1+e^x)^n$ (integer $n<0$) |
| $f'''(0) = \frac{4-4}{2^4} = 0$, so coefficient of $x^3$ is zero | A1 | 3 marks total | CSO; AG; all previous differentiation correct |
| **SC** for those not using Maclaurin's theorem: maximum of 4/9 | | |
## Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}x + \frac{1}{8}x^2$ | B1 | 1 mark total |
## Part 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln\left(1-\frac{x}{2}\right) = \left(-\frac{x}{2}\right) - \frac{1}{2}\left(-\frac{x}{2}\right)^2 + \frac{1}{3}\left(-\frac{x}{2}\right)^3 - \ldots$ | B1 | 1 mark total |
## Part 6(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln\left(\frac{1+e^x}{2}\right) + \ln\left(1-\frac{x}{2}\right) = -\frac{x^3}{24} + \ldots$ | M1 | Uses previous expansions to obtain first non-zero term of the form $kx^3$ |
| $x - \sin x \approx x - \left[x - \frac{x^3}{3!} + \ldots\right] \approx \frac{x^3}{3!} + \ldots$ | B1 | |
| $\left[\frac{\ln\left(\frac{1+e^x}{2}\right) + \ln\left(1-\frac{x}{2}\right)}{x - \sin x}\right] = \frac{-\frac{1}{24}x^3 + \ldots}{\frac{1}{6}x^3 + o(x^5)}$ | M1 | |
| $= \frac{-\frac{1}{24}x^3 + \ldots}{x^3\left[\frac{1}{6} + o(x^2)\right]} = \frac{-\frac{1}{24} + \ldots}{\frac{1}{6} + o(x^2)}$ | | |
| $\lim_{x \to 0} \ldots = -\frac{1}{4}$ | A1 | 4 marks total | CSO |
---6
\begin{enumerate}[label=(\alph*)]
\item The function f is defined by
$$\mathrm { f } ( x ) = \ln \left( 1 + \mathrm { e } ^ { x } \right)$$
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
$$\ln 2 + \frac { 1 } { 2 } x + \frac { 1 } { 8 } x ^ { 2 }$$
\item the coefficient of $x ^ { 3 }$ is zero.
\end{enumerate}\item Hence write down the first two non-zero terms in the expansion, in ascending powers of $x$, of $\ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right)$.
\item Use the series expansion
$$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \ldots$$
to write down the first three terms in the expansion, in ascending powers of $x$, of $\ln \left( 1 - \frac { x } { 2 } \right)$.
\item Use your answers to parts (b) and (c) to find
$$\lim _ { x \rightarrow 0 } \left[ \frac { \ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right) + \ln \left( 1 - \frac { x } { 2 } \right) } { x - \sin x } \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q6 [15]}}