Standard +0.3 This is a straightforward Maclaurin series question requiring factorization of ln(e+2ex) = ln(e(1+2x)) = 1+ln(1+2x), then applying the standard series for ln(1+u) with u=2x and collecting terms. The algebraic manipulation is routine for Further Maths students, making it slightly easier than average.
7 Show that the Maclaurin series for \(\ln ( \mathrm { e } + 2 \mathrm { e } x )\) is
$$1 + 2 x - 2 x ^ { 2 } + a x ^ { 3 } - \ldots$$
where \(a\) is to be determined.
Writes expression in form \(m+\ln(1+f(x))\) where \(m\) is non-zero constant (allow unsimplified). Or obtains first three derivatives of \(\ln(e+2ex)\) in form \(p(q+rx)^n\).
From expression of form \(\ln(1+f(x))\), substitutes \(f(x)\) into Maclaurin series. Condone sign errors and missing brackets.
\(= 1+2x-2x^2+\frac{8}{3}x^3 - \ldots\)
R1 (AO2.1)
Completes fully correct argument and correctly calculates \(a\). Do not condone missing brackets in working.
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln(e+2ex) = \ln(e(1+2x))$ | M1 (AO3.1a) | Writes expression in form $m+\ln(1+f(x))$ where $m$ is non-zero constant (allow unsimplified). Or obtains first three derivatives of $\ln(e+2ex)$ in form $p(q+rx)^n$. |
| $= \ln e + \ln(1+2x)$ $= 1 + 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \ldots$ | M1 (AO1.1a) | From expression of form $\ln(1+f(x))$, substitutes $f(x)$ into Maclaurin series. Condone sign errors and missing brackets. |
| $= 1+2x-2x^2+\frac{8}{3}x^3 - \ldots$ | R1 (AO2.1) | Completes fully correct argument and correctly calculates $a$. Do not condone missing brackets in working. |
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7 Show that the Maclaurin series for $\ln ( \mathrm { e } + 2 \mathrm { e } x )$ is
$$1 + 2 x - 2 x ^ { 2 } + a x ^ { 3 } - \ldots$$
where $a$ is to be determined.\\
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q7 [3]}}