| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Deduce related series from given series |
| Difficulty | Standard +0.3 This is a structured Further Maths question that guides students through standard Maclaurin series techniques with clear scaffolding. While it involves multiple derivatives and series manipulation, each part follows directly from the previous one with no novel insight required. The final part (subtracting series to find ln of a quotient) is a standard textbook exercise. Being Further Maths raises the baseline slightly, but the heavy scaffolding keeps it below average difficulty. |
| Spec | 1.07l Derivative of ln(x): and related functions4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \ln(5+3x) \Rightarrow \frac{dy}{dx} = \frac{3}{5+3x}\) | B1 | Correct first derivative |
| \(\frac{d^2y}{dx^2} = -\frac{9}{(5+3x)^2} \Rightarrow \frac{d^3y}{dx^3} = \frac{54}{(5+3x)^3}\) | M1 | Continues differentiating to reach \(\frac{d^3y}{dx^3} = \frac{k}{(5+3x)^3}\) |
| Correct simplified third derivative | A1 | Allow \(\frac{54}{(5+3x)^3}\) or \(54(5+3x)^{-3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_0 = \ln 5,\ y_0' = \frac{3}{5},\ y_0'' = -\frac{9}{25},\ y_0''' = \frac{54}{125}\) | M1 | Attempts all values at \(x=0\) and applies Maclaurin's theorem; evidence from at least 2 terms; correct factorial form |
| \(\ln(5+3x) \approx \ln 5 + \frac{3}{5}x - \frac{9}{50}x^2 + \frac{9}{125}x^3 + \ldots\) | A1 | Correct expansion; "\(\ln(5+3x) =\)" not required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln(5-3x) \approx \ln 5 - \frac{3}{5}x - \frac{9}{50}x^2 - \frac{9}{125}x^3 + \ldots\) | B1ft | Correct expansion or correct follow-through with signs changed on odd power coefficients only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln\frac{(5+3x)}{(5-3x)} = \ln(5+3x) - \ln(5-3x)\) | M1 | Subtracts their 2 different series to obtain at least 2 non-zero terms |
| \(= \frac{6}{5}x + \frac{18}{125}x^3 + \ldots\) | A1 | Correct terms; allow \(0 + \frac{6}{5}x + 0x^2 + \frac{18}{125}x^3 + \ldots\) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln(5+3x) \Rightarrow \frac{dy}{dx} = \frac{3}{5+3x}$ | B1 | Correct first derivative |
| $\frac{d^2y}{dx^2} = -\frac{9}{(5+3x)^2} \Rightarrow \frac{d^3y}{dx^3} = \frac{54}{(5+3x)^3}$ | M1 | Continues differentiating to reach $\frac{d^3y}{dx^3} = \frac{k}{(5+3x)^3}$ |
| Correct simplified third derivative | A1 | Allow $\frac{54}{(5+3x)^3}$ or $54(5+3x)^{-3}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_0 = \ln 5,\ y_0' = \frac{3}{5},\ y_0'' = -\frac{9}{25},\ y_0''' = \frac{54}{125}$ | M1 | Attempts all values at $x=0$ and applies Maclaurin's theorem; evidence from at least 2 terms; correct factorial form |
| $\ln(5+3x) \approx \ln 5 + \frac{3}{5}x - \frac{9}{50}x^2 + \frac{9}{125}x^3 + \ldots$ | A1 | Correct expansion; "$\ln(5+3x) =$" not required |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(5-3x) \approx \ln 5 - \frac{3}{5}x - \frac{9}{50}x^2 - \frac{9}{125}x^3 + \ldots$ | B1ft | Correct expansion or correct follow-through with signs changed on odd power coefficients only |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln\frac{(5+3x)}{(5-3x)} = \ln(5+3x) - \ln(5-3x)$ | M1 | Subtracts their 2 different series to obtain at least 2 non-zero terms |
| $= \frac{6}{5}x + \frac{18}{125}x^3 + \ldots$ | A1 | Correct terms; allow $0 + \frac{6}{5}x + 0x^2 + \frac{18}{125}x^3 + \ldots$ |
---\begin{enumerate}
\item Given that $y = \ln ( 5 + 3 x )$\\
(a) determine, in simplest form, $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$\\
(b) Hence determine the Maclaurin series expansion of $\ln ( 5 + 3 x )$, in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, giving each coefficient in simplest form.\\
(c) Hence write down the Maclaurin series expansion of $\ln ( 5 - 3 x )$, in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, giving each coefficient in simplest form.\\
(d) Use the answers to parts (b) and (c) to determine the first 2 non-zero terms, in ascending powers of $x$, of the Maclaurin series expansion of
\end{enumerate}
$$\ln \left( \frac { 5 + 3 x } { 5 - 3 x } \right)$$
\hfill \mbox{\textit{Edexcel F2 2023 Q1 [8]}}