| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(trigonometric expressions) |
| Difficulty | Challenging +1.2 This is a structured Further Maths question requiring systematic application of standard techniques: recalling ln(1+x) series, computing successive derivatives of ln(cos x), applying Maclaurin's theorem, and using series to evaluate a limit. While it involves multiple steps and FP3 content, each part follows predictable patterns with clear scaffolding, making it moderately above average difficulty but not requiring significant novel insight. |
| 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 |
|---|---|---|
| (a)(i) \(\ln(1+2x) = 2x - 2x^2 + \frac{8}{3}x^3 \ldots\) | M1, A1 | Total: 2 |
| (ii) \(-\frac{1}{2} < x \leq \frac{1}{2}\) | B1 | Total: 1 |
| (b)(i) \(y=\ln\cos x \Rightarrow y'(x) = \frac{1}{\cos x}(-\sin x)\) | M1 | |
| \(y''(x) = -\sec^2x\) | A1 | |
| \(y'''(x) = -2\sec x(\sec x\tan x)\) | M1 | |
| \([y'''(x) = -2\tan x(\sec^2x)]\) | A1/ | Total: 4 |
| (ii) \(y''''(x) = -2[\sec^2x(\sec^2x) + \tan x(2\sec x(\sec x\tan x))]\) | M1, A1 | |
| \(y''''(0) = -2[1^2+0] = -2\) | A1/ | Total: 3 |
| (iii) \(\ln\cos x=0+0+ \frac{x^2}{2}(-1) + 0 + \frac{x^4}{4!}(-2)\) | M1 | |
| \(\simeq -\frac{x^2}{2} - \frac{x^4}{12}\) | A1 | Total: 2 |
| (c) \(\text{Limit} = \lim_{x \to 0}\frac{x\ln(1+2x)}{x^2 - \ln\cos x}\) | M1 | |
| \(= \lim_{x \to 0}\frac{x(2x - 2x^2 + \ldots)}{x^2 - (-\frac{x^2}{2} - \frac{x^4}{12} \ldots)}\) | ||
| \(\text{Limit} = \lim_{x \to 0}\frac{2x^2 - o(x^3)}{1.5x^2 + o(x^4)}\) | M1 | |
| \(= \lim_{x \to 0}\frac{2 - o(x)}{1.5 + o(x^2)} = \frac{4}{3}\) | M1, A1 | Total: 3 |
**(a)(i)** $\ln(1+2x) = 2x - 2x^2 + \frac{8}{3}x^3 \ldots$ | M1, A1 | Total: 2 | Use of expansion of ln(1+x); Simplified 'numerators'
**(ii)** $-\frac{1}{2} < x \leq \frac{1}{2}$ | B1 | Total: 1 |
**(b)(i)** $y=\ln\cos x \Rightarrow y'(x) = \frac{1}{\cos x}(-\sin x)$ | M1 | |
$y''(x) = -\sec^2x$ | A1 | | ACF
$y'''(x) = -2\sec x(\sec x\tan x)$ | M1 | | Chain rule OE
$[y'''(x) = -2\tan x(\sec^2x)]$ | A1/ | Total: 4 | Ft a slip...accept unsimplified
**(ii)** $y''''(x) = -2[\sec^2x(\sec^2x) + \tan x(2\sec x(\sec x\tan x))]$ | M1, A1 | | Product rule OE; ACF
$y''''(0) = -2[1^2+0] = -2$ | A1/ | Total: 3 | Ft a slip
**(iii)** $\ln\cos x=0+0+ \frac{x^2}{2}(-1) + 0 + \frac{x^4}{4!}(-2)$ | M1 | |
$\simeq -\frac{x^2}{2} - \frac{x^4}{12}$ | A1 | Total: 2 | CSO throughout part (b). AG
**(c)** $\text{Limit} = \lim_{x \to 0}\frac{x\ln(1+2x)}{x^2 - \ln\cos x}$ | M1 | |
$= \lim_{x \to 0}\frac{x(2x - 2x^2 + \ldots)}{x^2 - (-\frac{x^2}{2} - \frac{x^4}{12} \ldots)}$ | | |
$\text{Limit} = \lim_{x \to 0}\frac{2x^2 - o(x^3)}{1.5x^2 + o(x^4)}$ | M1 | | Using earlier expansions
$= \lim_{x \to 0}\frac{2 - o(x)}{1.5 + o(x^2)} = \frac{4}{3}$ | M1, A1 | Total: 3 | Need to see stage, division by $x^2$; The notation o(x^n) can be replaced by a term of the form $kx^p$7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the expansion of $\ln ( 1 + 2 x )$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$.
\item State the range of values of $x$ for which this expansion is valid.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \ln \cos x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$.
\item Find the value of $\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }$ when $x = 0$.
\item Hence, by using Maclaurin's theorem, show that the first two non-zero terms in the expansion, in ascending powers of $x$, of $\ln \cos x$ are
$$- \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 4 } } { 12 }$$
\end{enumerate}\item Find
$$\lim _ { x \rightarrow 0 } \left[ \frac { x \ln ( 1 + 2 x ) } { x ^ { 2 } - \ln \cos x } \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q7 [15]}}