| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Composition of two standard series |
| Difficulty | Standard +0.8 This requires composing two Taylor series (e^x and e^u) and carefully tracking terms through x^3, which demands systematic algebraic manipulation beyond routine recall. However, it's a standard Further Maths technique with clear scaffolding from part (a), making it moderately challenging but not requiring deep insight. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}\) or \(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}\) | B1 | Correct series, ignore terms beyond \(x^3\) |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Version 1: \(e^{(e^x-1)} = e^{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots-1}\) or Version 2: \(e^{(e^x-1)} = 1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}+\ldots\) | M1 | Correctly applies the exponential Maclaurin expansion at least once, either to the base exponent or in the index. Allow 2 for 2! and 6 for 3! in cube term. Follow through on series in (a) |
| \(= 1+\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)+\frac{1}{2}\left(x+\frac{x^2}{2!}+\ldots\right)^2+\frac{1}{6}(x+\ldots)^3+\ldots\) | M1 | A complete attempt to use the exponential Maclaurin series to produce a cubic expression in terms of \(x\) only. Polynomial in \(x\) must have been achieved. Condone a slip with one term. Follow through on series in (a) |
| \(= 1+x+\left(\frac{1}{2}+\frac{1}{2}\right)x^2+\left(\frac{1}{6}+\frac{1}{2}\times 2\times\frac{1}{2}+\frac{1}{6}\right)x^3+\ldots\) | dM1 | Dependent on previous M mark. Expands brackets and gathers terms (not necessarily fully simplified, but should have single term for each power) |
| \(= 1+x+x^2+\frac{5}{6}x^3+\ldots\) | A1 | Any two correct coefficients of \(x\), \(x^2\) and \(x^3\), need not be simplified |
| A1 | Fully correct answer with simplified terms | |
| (5) | NB: Use of differentiation scores no mark |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$ or $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$ | **B1** | Correct series, ignore terms beyond $x^3$ |
| | **(1)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Version 1:** $e^{(e^x-1)} = e^{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots-1}$ **or Version 2:** $e^{(e^x-1)} = 1+(e^x-1)+\frac{(e^x-1)^2}{2!}+\frac{(e^x-1)^3}{3!}+\ldots$ | **M1** | Correctly applies the exponential Maclaurin expansion at least once, either to the base exponent or in the index. Allow 2 for 2! and 6 for 3! in cube term. Follow through on series in (a) |
| $= 1+\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)+\frac{1}{2}\left(x+\frac{x^2}{2!}+\ldots\right)^2+\frac{1}{6}(x+\ldots)^3+\ldots$ | **M1** | A complete attempt to use the exponential Maclaurin series to produce a cubic expression in terms of $x$ only. Polynomial in $x$ must have been achieved. Condone a slip with one term. Follow through on series in (a) |
| $= 1+x+\left(\frac{1}{2}+\frac{1}{2}\right)x^2+\left(\frac{1}{6}+\frac{1}{2}\times 2\times\frac{1}{2}+\frac{1}{6}\right)x^3+\ldots$ | **dM1** | Dependent on previous M mark. Expands brackets and gathers terms (not necessarily fully simplified, but should have single term for each power) |
| $= 1+x+x^2+\frac{5}{6}x^3+\ldots$ | **A1** | Any two correct coefficients of $x$, $x^2$ and $x^3$, need not be simplified |
| | **A1** | Fully correct answer with simplified terms |
| | **(5)** | NB: Use of differentiation scores no mark |
---\begin{enumerate}
\item (a) Write down the Maclaurin series of $\mathrm { e } ^ { x }$, in ascending power of $x$, up to and including the term in $x ^ { 3 }$\\
(b) Hence, without differentiating, determine the Maclaurin series of
\end{enumerate}
$$\mathrm { e } ^ { \left( \mathrm { e } ^ { x } - 1 \right) }$$
in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each coefficient in simplest form.
\hfill \mbox{\textit{Edexcel CP2 2023 Q2 [6]}}