| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Limit using series expansion |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring Taylor series manipulation and L'Hôpital-type limit evaluation. Part (a) involves standard series recall and algebraic combination, while part (b) requires recognizing an indeterminate form and using series expansions to evaluate it—a technique beyond standard A-level but routine for FP3 students who have practiced such problems. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln(1+2x) = 2x - 2x^2 + \frac{8x^3}{3} - 4x^4 + \cdots\) | B1 | Correct expansion to \(x^4\) term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln[(1+2x)(1-2x)] = \ln(1+2x) + \ln(1-2x)\) | M1 | Using log law |
| \(= -4x^2 - \frac{8x^4}{3} - \cdots\) (first two non-zero terms) | A1 | Correct terms |
| Valid for \(-\frac{1}{2} < x < \frac{1}{2}\) | A1 | Correct range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Numerator: \(3x - x\sqrt{9+x} = x(3 - \sqrt{9+x})\); expand \(\sqrt{9+x} = 3(1+\frac{x}{9})^{1/2} \approx 3 + \frac{x}{6} - \frac{x^2}{216}+\cdots\) | M1 | Series expansion of numerator |
| Numerator \(\approx 3x - x(3 + \frac{x}{6}) = -\frac{x^2}{6}\) | A1 | Correct leading term |
| Denominator \(\approx -4x^2\) (from (a)(ii)) | A1 | Using result from (a)(ii) |
| Limit \(= \frac{-x^2/6}{-4x^2} = \frac{1}{24}\) | A1 | Correct final answer |
# Question 3:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(1+2x) = 2x - 2x^2 + \frac{8x^3}{3} - 4x^4 + \cdots$ | B1 | Correct expansion to $x^4$ term |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln[(1+2x)(1-2x)] = \ln(1+2x) + \ln(1-2x)$ | M1 | Using log law |
| $= -4x^2 - \frac{8x^4}{3} - \cdots$ (first two non-zero terms) | A1 | Correct terms |
| Valid for $-\frac{1}{2} < x < \frac{1}{2}$ | A1 | Correct range |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Numerator: $3x - x\sqrt{9+x} = x(3 - \sqrt{9+x})$; expand $\sqrt{9+x} = 3(1+\frac{x}{9})^{1/2} \approx 3 + \frac{x}{6} - \frac{x^2}{216}+\cdots$ | M1 | Series expansion of numerator |
| Numerator $\approx 3x - x(3 + \frac{x}{6}) = -\frac{x^2}{6}$ | A1 | Correct leading term |
| Denominator $\approx -4x^2$ (from (a)(ii)) | A1 | Using result from (a)(ii) |
| Limit $= \frac{-x^2/6}{-4x^2} = \frac{1}{24}$ | A1 | Correct final answer |3
\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 $x ^ { 4 }$.
\item Hence, or otherwise, find the first two non-zero terms in the expansion of
$$\ln [ ( 1 + 2 x ) ( 1 - 2 x ) ]$$
in ascending powers of $x$ and state the range of values of $x$ for which the expansion is valid.
\end{enumerate}\item Find $\lim _ { x \rightarrow 0 } \left[ \frac { 3 x - x \sqrt { 9 + x } } { \ln [ ( 1 + 2 x ) ( 1 - 2 x ) ] } \right]$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2015 Q3 [8]}}