| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about x=1: direct function expansion |
| Difficulty | Standard +0.3 This is a straightforward application of Taylor series formula with routine differentiation of ln(x), followed by a simple limit evaluation using the series. Part (ii) is a standard L'Hôpital's rule application. All techniques are direct applications of given formulas with no novel insight required, making it slightly easier than average for Further Maths. |
| 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x)=\ln x \Rightarrow f(1)=0\); \(f'(x)=\frac{1}{x} \Rightarrow f'(1)=1\); \(f''= -\frac{1}{x^2}f'\) | M1, A1 | Differentiates \(f(x)=\ln x\) twice and finds \(f(1)\), \(f'(1)\), \(f''(1)\) |
| \(\ln x = (x-1)-\frac{1}{2}(x-1)^2+\ldots\) | M1, A1 | Correct expansion with simplified coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lim_{x\to1}\left(\frac{\ln x}{x-1}\right)=\lim_{x\to1}\left(\frac{(x-1)-\frac{1}{2}(x-1)^2+\ldots}{x-1}\right)\) | M1 | Substitutes Taylor series; cancels factor \((x-1)\). Must use result from (a) — no L'Hôpital |
| \(=\lim_{x\to1}\left(1-\frac{1}{2}(x-1)+\ldots\right)=1\) | A1* | Must see \(1-\frac{1}{2}(x-1)\); cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Writes as indeterminate form, e.g. \(\frac{\sin(2x)}{(x+3)\tan(6x)}\) or \(\frac{\sin(2x)\cos(6x)}{(x+3)\sin(6x)}\) | M1 | \(\frac{f(0)}{g(0)}=\frac{0}{0}\) or \(\frac{\infty}{\infty}\) |
| Differentiates numerator and denominator: \(\frac{2\cos(2x)}{\tan(6x)+6(x+3)\sec^2(6x)}\) or \(\frac{2\cos(2x)\cos(6x)-6\sin(2x)\sin(6x)}{\sin(6x)+6(x+3)\cos(6x)}\) | M1, A1 | Allow slips in coefficients; form correct. A1 depends on both M marks |
| \(\lim_{x\to0}\left(\frac{1}{(x+3)\tan(6x)\csc(2x)}\right)=\frac{2}{18}=\frac{1}{9}\) | A1cso | Deduces correct limit from fully correct work |
## Question 8:
### Part (i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=\ln x \Rightarrow f(1)=0$; $f'(x)=\frac{1}{x} \Rightarrow f'(1)=1$; $f''= -\frac{1}{x^2}f'$ | M1, A1 | Differentiates $f(x)=\ln x$ twice and finds $f(1)$, $f'(1)$, $f''(1)$ |
| $\ln x = (x-1)-\frac{1}{2}(x-1)^2+\ldots$ | M1, A1 | Correct expansion with simplified coefficients |
### Part (i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lim_{x\to1}\left(\frac{\ln x}{x-1}\right)=\lim_{x\to1}\left(\frac{(x-1)-\frac{1}{2}(x-1)^2+\ldots}{x-1}\right)$ | M1 | Substitutes Taylor series; cancels factor $(x-1)$. Must use result from (a) — no L'Hôpital |
| $=\lim_{x\to1}\left(1-\frac{1}{2}(x-1)+\ldots\right)=1$ | A1* | Must see $1-\frac{1}{2}(x-1)$; cso |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes as indeterminate form, e.g. $\frac{\sin(2x)}{(x+3)\tan(6x)}$ or $\frac{\sin(2x)\cos(6x)}{(x+3)\sin(6x)}$ | M1 | $\frac{f(0)}{g(0)}=\frac{0}{0}$ or $\frac{\infty}{\infty}$ |
| Differentiates numerator and denominator: $\frac{2\cos(2x)}{\tan(6x)+6(x+3)\sec^2(6x)}$ or $\frac{2\cos(2x)\cos(6x)-6\sin(2x)\sin(6x)}{\sin(6x)+6(x+3)\cos(6x)}$ | M1, A1 | Allow slips in coefficients; form correct. A1 depends on both M marks |
| $\lim_{x\to0}\left(\frac{1}{(x+3)\tan(6x)\csc(2x)}\right)=\frac{2}{18}=\frac{1}{9}$ | A1cso | Deduces correct limit from fully correct work |\begin{enumerate}
\item The Taylor series expansion of $f ( x )$ about $x = a$ is given by
\end{enumerate}
$$f ( x ) = f ( a ) + ( x - a ) f ^ { \prime } ( a ) + \frac { ( x - a ) ^ { 2 } } { 2 ! } f ^ { \prime \prime } ( a ) + \ldots + \frac { ( x - a ) ^ { r } } { r ! } f ^ { ( r ) } ( a ) + \ldots$$
(i) (a) Use differentiation to determine the Taylor series expansion of $\ln x$, in ascending powers of ( $x - 1$ ), up to and including the term in $( x - 1 ) ^ { 2 }$\\
(b) Hence prove that
$$\lim _ { x \rightarrow 1 } \left( \frac { \ln x } { x - 1 } \right) = 1$$
(ii) Use L'Hospital's rule to determine
$$\lim _ { x \rightarrow 0 } \left( \frac { 1 } { ( x + 3 ) \tan ( 6 x ) \operatorname { cosec } ( 2 x ) } \right)$$
(Solutions relying entirely on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel FP1 2022 Q8 [10]}}