| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Use binomial with exponential series |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring systematic application of Maclaurin series, product of series, and limit evaluation using series expansions. While each individual step follows standard techniques (differentiation, series multiplication, L'Hôpital alternative), the question requires careful algebraic manipulation across multiple parts and coordination of different series. The limit in part (d) requires recognizing that series expansion is the appropriate method. This is moderately challenging for FP3 level but follows predictable patterns once the approach is identified. |
| 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(f'(x) = \frac{1}{2}(1+2x)^{-\frac{1}{2}}(2) = (1+2x)^{-\frac{1}{2}}\) | M1A1 | |
| \(f''(x) = -(1+2x)^{-\frac{3}{2}}\) | A1F | ft a slip |
| \(f'''(x) = 3(1+2x)^{-\frac{5}{2}}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(f(x) = (1+2x)^{\frac{1}{2}} \Rightarrow f(0) = 1\) | B1 | |
| \(f'(0) = 1;\ f''(0) = -1;\ f'''(0) = 3\) | M1 | All three attempted |
| A1F | ft on \(k(1+2x)^n\) | |
| \(f(x) \approx 1 + x - \dfrac{x^2}{2} + \dfrac{x^3}{2}\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(e^x(1+2x)^{\frac{1}{2}} \approx \left(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}\right)\left(1+x-\dfrac{x^2}{2}+\dfrac{x^3}{2}\right)\) | M1 | Attempt to expand needed |
| \(\approx 1 + x(1+1) + x^2(-0.5+1+0.5) + x^3\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}+\frac{1}{6}\right)\) | A1 | |
| \(\approx 1 + 2x + x^2 + \dfrac{2}{3}x^3\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(e^{2x} = 1 + 2x + \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{6} + ...\) | B1 | |
| \(= 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + ...\) | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(1 - \cos x = \frac{1}{2}x^2 + \{o(x^4)\}\) | B1 | |
| \(\dfrac{e^x(1+2x)^{\frac{1}{2}} - e^{2x}}{1-\cos x} =\) | ||
| \(\dfrac{1+2x+x^2+\frac{2}{3}x^3 - \left[1+2x+2x^2+\frac{4}{3}x^3\right]}{\frac{1}{2}x^2 + \{o(x^4)\}}\) | M1 | Series used |
| \(\lim_{x\to 0} = \lim_{x\to 0} \dfrac{-x^2 + \{o(x^3)\}}{\frac{1}{2}x^2 + \{o(x^4)\}}\) | A1F | |
| \(\lim_{x\to 0} \dfrac{-1+o(x)}{\frac{1}{2}+o(x^2)} = -2\) | A1F | Total: 4 |
## Question 6:
### Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f'(x) = \frac{1}{2}(1+2x)^{-\frac{1}{2}}(2) = (1+2x)^{-\frac{1}{2}}$ | M1A1 | |
| $f''(x) = -(1+2x)^{-\frac{3}{2}}$ | A1F | ft a slip |
| $f'''(x) = 3(1+2x)^{-\frac{5}{2}}$ | A1 | Total: 4 | |
### Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $f(x) = (1+2x)^{\frac{1}{2}} \Rightarrow f(0) = 1$ | B1 | |
| $f'(0) = 1;\ f''(0) = -1;\ f'''(0) = 3$ | M1 | All three attempted |
| | A1F | ft on $k(1+2x)^n$ |
| $f(x) \approx 1 + x - \dfrac{x^2}{2} + \dfrac{x^3}{2}$ | A1 | Total: 4 | CSO **AG** |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $e^x(1+2x)^{\frac{1}{2}} \approx \left(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}\right)\left(1+x-\dfrac{x^2}{2}+\dfrac{x^3}{2}\right)$ | M1 | Attempt to expand needed |
| $\approx 1 + x(1+1) + x^2(-0.5+1+0.5) + x^3\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2}+\frac{1}{6}\right)$ | A1 | |
| $\approx 1 + 2x + x^2 + \dfrac{2}{3}x^3$ | A1 | Total: 3 | CSO |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $e^{2x} = 1 + 2x + \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{6} + ...$ | B1 | |
| $= 1 + 2x + 2x^2 + \dfrac{4}{3}x^3 + ...$ | | Total: 1 | |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $1 - \cos x = \frac{1}{2}x^2 + \{o(x^4)\}$ | B1 | |
| $\dfrac{e^x(1+2x)^{\frac{1}{2}} - e^{2x}}{1-\cos x} =$ | | |
| $\dfrac{1+2x+x^2+\frac{2}{3}x^3 - \left[1+2x+2x^2+\frac{4}{3}x^3\right]}{\frac{1}{2}x^2 + \{o(x^4)\}}$ | M1 | Series used |
| $\lim_{x\to 0} = \lim_{x\to 0} \dfrac{-x^2 + \{o(x^3)\}}{\frac{1}{2}x^2 + \{o(x^4)\}}$ | A1F | |
| $\lim_{x\to 0} \dfrac{-1+o(x)}{\frac{1}{2}+o(x^2)} = -2$ | A1F | Total: 4 | ft a slip but must see the intermediate stage |6 The function f is defined by $\mathrm { f } ( x ) = ( 1 + 2 x ) ^ { \frac { 1 } { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find f'''(x).
\item Using Maclaurin's theorem, show that, for small values of $x$,
$$\mathrm { f } ( x ) \approx 1 + x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 2 } x ^ { 3 }$$
\end{enumerate}\item Use the expansion of $\mathrm { e } ^ { x }$ together with the result in part (a)(ii) to show that, for small values of $x$,
$$\mathrm { e } ^ { x } ( 1 + 2 x ) ^ { \frac { 1 } { 2 } } \approx 1 + 2 x + x ^ { 2 } + k x ^ { 3 }$$
where $k$ is a rational number to be found.
\item Write down the first four terms in the expansion, in ascending powers of $x$, of $\mathrm { e } ^ { 2 x }$.
\item Find
$$\lim _ { x \rightarrow 0 } \frac { \mathrm { e } ^ { x } ( 1 + 2 x ) ^ { \frac { 1 } { 2 } } - \mathrm { e } ^ { 2 x } } { 1 - \cos x }$$
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2007 Q6 [16]}}