| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for composite exponential/root functions |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP3 Maclaurin series question requiring chain rule differentiation of a composite function, evaluation at x=0, and a limit using the series. While it involves multiple steps and careful algebra, it follows a predictable template with no novel insights required. The limit in part (b) is a routine application of the series just derived. Slightly above average difficulty due to the algebraic manipulation involved. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07l Derivative of ln(x): and related functions4.08a 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 |
| \(f(x) = (9 + \tan x)^{\frac{1}{2}}\) | ||
| \(f'(x) = \frac{1}{2}(9 + \tan x)^{-\frac{1}{2}} \sec^2 x\) | M1, A1 | Chain rule |
| \(f''(x) = -\frac{1}{4}(9 + \tan x)^{-\frac{3}{2}} \sec^4 x + \frac{1}{2}(9 + \tan x)^{-\frac{1}{2}}(2\sec^2 x \tan x)\) | M1, A1 | Product rule, OE; ACF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(0) = 3\) | B1 | |
| \(f'(0) = \frac{1}{2}(9)^{-\frac{1}{2}} = \frac{1}{6}\); \(f''(0) = -\frac{1}{4}(9)^{-\frac{3}{2}} = -\frac{1}{108}\) | M1 | Both attempted and at least one correct ft on \(f'(x)\) and \(f''(x)\) |
| \((9 + \tan x)^{\frac{1}{2}} \approx 3 + \frac{x}{6} - \frac{x^2}{216}\) | A1 | CSO, AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{f(x)-3}{\sin 3x} \approx \frac{\frac{x}{6} - \frac{x^2}{216}\cdots}{3x - \frac{(3x)^3}{3!}\cdots}\) | M1 | Using series expansions |
| \(\approx \dfrac{\frac{1}{6} - \frac{x}{216}\cdots}{3 - \cdots}\) | m1 | Dividing numerator and denominator by \(x\) to get constant term in each |
| \(\lim_{x \to 0}\left[\frac{f(x)-3}{\sin 3x}\right] = \frac{1}{18}\) | A1 |
# Question 6:
## Part 6(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = (9 + \tan x)^{\frac{1}{2}}$ | | |
| $f'(x) = \frac{1}{2}(9 + \tan x)^{-\frac{1}{2}} \sec^2 x$ | M1, A1 | Chain rule |
| $f''(x) = -\frac{1}{4}(9 + \tan x)^{-\frac{3}{2}} \sec^4 x + \frac{1}{2}(9 + \tan x)^{-\frac{1}{2}}(2\sec^2 x \tan x)$ | M1, A1 | Product rule, OE; ACF | Total: 4 marks |
## Part 6(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(0) = 3$ | B1 | |
| $f'(0) = \frac{1}{2}(9)^{-\frac{1}{2}} = \frac{1}{6}$; $f''(0) = -\frac{1}{4}(9)^{-\frac{3}{2}} = -\frac{1}{108}$ | M1 | Both attempted and at least one correct ft on $f'(x)$ and $f''(x)$ |
| $(9 + \tan x)^{\frac{1}{2}} \approx 3 + \frac{x}{6} - \frac{x^2}{216}$ | A1 | CSO, AG | Total: 3 marks |
## Part 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{f(x)-3}{\sin 3x} \approx \frac{\frac{x}{6} - \frac{x^2}{216}\cdots}{3x - \frac{(3x)^3}{3!}\cdots}$ | M1 | Using series expansions |
| $\approx \dfrac{\frac{1}{6} - \frac{x}{216}\cdots}{3 - \cdots}$ | m1 | Dividing numerator and denominator by $x$ to get constant term in each |
| $\lim_{x \to 0}\left[\frac{f(x)-3}{\sin 3x}\right] = \frac{1}{18}$ | A1 | | Total: 3 marks |
---6 The function f is defined by
$$\mathrm { f } ( x ) = ( 9 + \tan x ) ^ { \frac { 1 } { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find $f ^ { \prime \prime } ( x )$.
\item By using Maclaurin's theorem, show that, for small values of $x$,
$$( 9 + \tan x ) ^ { \frac { 1 } { 2 } } \approx 3 + \frac { x } { 6 } - \frac { x ^ { 2 } } { 216 }$$
\end{enumerate}\item Find
$$\lim _ { x \rightarrow 0 } \left[ \frac { f ( x ) - 3 } { \sin 3 x } \right]$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2009 Q6 [10]}}