| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Maclaurin series for ln(a+bx) |
| Difficulty | Standard +0.3 This is a straightforward Maclaurin series question requiring routine differentiation of a logarithmic function, substitution into the standard formula, and a simple numerical application. While it involves multiple steps and Further Maths content, it follows a completely standard template with no conceptual challenges or novel insights required—making it slightly easier than average for FP2 students. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08a 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 |
| \(y = \ln(1-2x)^{-1} = (\ln 1) - \ln(1-2x)\) | ||
| \(\frac{dy}{dx} = -\frac{1}{1-2x} \times -2 = \frac{2}{1-2x}\) | M1A1 | M1: Must use chain rule, \(\frac{k}{1-2x}\) with \(k \neq \pm 1\). Minus sign may be missing. A1: Correct derivative |
| OR \(\frac{dy}{dx} = (1-2x) \times -(1-2x)^{-2} \times -2 = \frac{2}{1-2x}\) | M1A1 | M1: Chain rule. Minus sign may be missing. A1: Correct derivative |
| \(\frac{d^2y}{dx^2} = -2 \times (1-2x)^{-2} \times -2 = \frac{4}{(1-2x)^2}\) | A1 | Correct second derivative from correct first derivative |
| \(\frac{d^3y}{dx^3} = -8 \times (1-2x)^{-3} \times -2 = \frac{16}{(1-2x)^3}\) | A1 (4) | Correct third derivative from correct first and second derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \ln\!\left(\frac{1}{1-2x}\right) \Rightarrow e^y = \frac{1}{1-2x} = (1-2x)^{-1}\) | ||
| \(e^y \frac{dy}{dx} = 2(1-2x)^{-2}\) | M1 | Differentiates using implicit differentiation and chain rule |
| \(\frac{dy}{dx} = 2e^{-y}(1-2x)^{-2}\) or \(\frac{2}{1-2x}\) | A1 | Correct derivative in either form. Equivalents accepted |
| If \(\frac{dy}{dx} = \frac{2}{1-2x}\) used from here, see main scheme for second and third derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y_0=0), y'_0=2, y''_0=4, y'''_0=16\) | M1 | Attempt values at \(x=0\); \(y_0=0\) need not be seen but other 3 values must be attempted |
| \((y=)(0)+2x+\frac{4x^2}{2!}+\frac{16x^3}{3!}\) | M1 | Uses values in correct Maclaurin series; must see \(x^3\) term; correct for their values with \(2!,3!\) or \(2\) and \(6\) |
| \(y=2x+2x^2+\frac{8}{3}x^3\) | A1cao | Must start \(y=...\) or \(\ln\left(\frac{1}{1-2x}\right)=...\); \(f(x)=...\) allowed only if \(f(x)\) defined as one of these |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=\ln\left(\frac{1}{1-2x}\right)=-\ln(1-2x)\) | M1 | Log power law applied correctly |
| \(=-\left((-2x)-\frac{(-2x)^2}{2}+\frac{(-2x)^3}{3}\right)\) | M1 | Replaces \(x\) with \(-2x\) in expansion for \(\ln(1+x)\) (in formula book) |
| \(y=2x+2x^2+\frac{8}{3}x^3\) | A1cao | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{1-2x}=\frac{3}{2}\Rightarrow x=\frac{1}{6}\) | B1 | Correct value for \(x\), seen explicitly or substituted in their expansion |
| \(\ln\left(\frac{3}{2}\right)\approx 2\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)^2+\frac{8}{3}\left(\frac{1}{6}\right)^3\) | M1 | Substitute their value of \(x\) into expansion; correct for their expansion and \(x\); calculator value for \(\ln\left(\frac{3}{2}\right)\) is \(0.405\) |
| \(=0.401\) | A1cso | Must come from correct work |
| NB: \(\ln 3-\ln 2\) or \(\ln 3+\ln\left(\frac{1}{2}\right)\) scores 0/3 as \( | x | \) must be \(<\frac{1}{2}\); Answer with no working scores 0/3 |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln(1-2x)^{-1} = (\ln 1) - \ln(1-2x)$ | | |
| $\frac{dy}{dx} = -\frac{1}{1-2x} \times -2 = \frac{2}{1-2x}$ | M1A1 | M1: Must use chain rule, $\frac{k}{1-2x}$ with $k \neq \pm 1$. Minus sign may be missing. A1: Correct derivative |
| **OR** $\frac{dy}{dx} = (1-2x) \times -(1-2x)^{-2} \times -2 = \frac{2}{1-2x}$ | M1A1 | M1: Chain rule. Minus sign may be missing. A1: Correct derivative |
| $\frac{d^2y}{dx^2} = -2 \times (1-2x)^{-2} \times -2 = \frac{4}{(1-2x)^2}$ | A1 | Correct second derivative from correct first derivative |
| $\frac{d^3y}{dx^3} = -8 \times (1-2x)^{-3} \times -2 = \frac{16}{(1-2x)^3}$ | A1 **(4)** | Correct third derivative from correct first and second derivatives |
**Alternative by implicit differentiation:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln\!\left(\frac{1}{1-2x}\right) \Rightarrow e^y = \frac{1}{1-2x} = (1-2x)^{-1}$ | | |
| $e^y \frac{dy}{dx} = 2(1-2x)^{-2}$ | M1 | Differentiates using implicit differentiation and chain rule |
| $\frac{dy}{dx} = 2e^{-y}(1-2x)^{-2}$ or $\frac{2}{1-2x}$ | A1 | Correct derivative in either form. Equivalents accepted |
| If $\frac{dy}{dx} = \frac{2}{1-2x}$ used from here, see main scheme for second and third derivatives | | |
# Question (b) [Maclaurin Series]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y_0=0), y'_0=2, y''_0=4, y'''_0=16$ | M1 | Attempt values at $x=0$; $y_0=0$ need not be seen but other 3 values must be attempted |
| $(y=)(0)+2x+\frac{4x^2}{2!}+\frac{16x^3}{3!}$ | M1 | Uses values in correct Maclaurin series; must see $x^3$ term; correct for their values with $2!,3!$ or $2$ and $6$ |
| $y=2x+2x^2+\frac{8}{3}x^3$ | A1cao | Must start $y=...$ or $\ln\left(\frac{1}{1-2x}\right)=...$; $f(x)=...$ allowed only if $f(x)$ defined as one of these |
**Alternative (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\ln\left(\frac{1}{1-2x}\right)=-\ln(1-2x)$ | M1 | Log power law applied correctly |
| $=-\left((-2x)-\frac{(-2x)^2}{2}+\frac{(-2x)^3}{3}\right)$ | M1 | Replaces $x$ with $-2x$ in expansion for $\ln(1+x)$ (in formula book) |
| $y=2x+2x^2+\frac{8}{3}x^3$ | A1cao | Correct expression |
# Question (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{1-2x}=\frac{3}{2}\Rightarrow x=\frac{1}{6}$ | B1 | Correct value for $x$, seen explicitly or substituted in their expansion |
| $\ln\left(\frac{3}{2}\right)\approx 2\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)^2+\frac{8}{3}\left(\frac{1}{6}\right)^3$ | M1 | Substitute their value of $x$ into expansion; correct for their expansion and $x$; calculator value for $\ln\left(\frac{3}{2}\right)$ is $0.405$ |
| $=0.401$ | A1cso | Must come from correct work |
NB: $\ln 3-\ln 2$ or $\ln 3+\ln\left(\frac{1}{2}\right)$ scores 0/3 as $|x|$ must be $<\frac{1}{2}$; Answer with no working scores 0/3
---4.
$$y = \ln \left( \frac { 1 } { 1 - 2 x } \right) , \quad | x | < \frac { 1 } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$
\item Hence, or otherwise, find the series expansion of $\ln \left( \frac { 1 } { 1 - 2 x } \right)$ about $x = 0$, in ascending powers of $x$, up to and including the term in $x ^ { 3 }$. Give each coefficient in its simplest form.
\item Use your expansion to find an approximate value for $\ln \left( \frac { 3 } { 2 } \right)$, giving your answer\\
to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2017 Q4 [10]}}