| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Limit using series expansion |
| Difficulty | Standard +0.8 This is a Further Maths question requiring Taylor series expansion and limit evaluation. Part (a) is straightforward recall, but part (b) requires recognizing the indeterminate form, expanding both sin 3x and cos 2x to sufficient order (x³), careful algebraic manipulation, and cancellation. The multi-step nature and need to determine the correct order of expansion elevates this above average difficulty. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin 3x = 3x - \frac{1}{3!}(3x)^3 = 3x - 4.5x^3 + \ldots\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(= -\frac{3}{10}\) | B1, M1, m1, A1 | Using expansions; Division by \(x^3\) stage to reach relevant form of quotient before taking limit; CSO OE; 4 marks total |
**(a)**
$\sin 3x = 3x - \frac{1}{3!}(3x)^3 = 3x - 4.5x^3 + \ldots$ | B1 | 1 mark total
**(b)**
$\cos 2x = 1 - \frac{1}{2!}(2x)^2 + \ldots$
$\lim_{x \to 0}\frac{3x\cos 2x - \sin 3x}{5x^3} =$
$\lim_{x \to 0}\frac{3x - 6x^3 - 3x + 4.5x^3 + \ldots}{5x^3}$
$= \lim_{x \to 0}\frac{-1.5 + o(x^2)\ldots}{5}$
$= -\frac{3}{10}$ | B1, M1, m1, A1 | Using expansions; Division by $x^3$ stage to reach relevant form of quotient before taking limit; CSO OE; 4 marks total
**Total: 5 marks**
---4
\begin{enumerate}[label=(\alph*)]
\item Write down the expansion of $\sin 3 x$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$.
\item Find
$$\lim _ { x \rightarrow 0 } \left[ \frac { 3 x \cos 2 x - \sin 3 x } { 5 x ^ { 3 } } \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2010 Q4 [5]}}