| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Use series to find error or validity |
| Difficulty | Standard +0.3 This is a straightforward application of Maclaurin series with standard substitution (part a), error analysis requiring knowledge of radius of convergence (part b), and a routine calculation with justification (part c). While it tests understanding of validity conditions for series, the concepts are standard Further Maths AS content with no novel problem-solving required. The error identification is conceptual but accessible, making this slightly easier than average. |
| 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 | Marks | Guidance |
| \(\ln(1+3x)=3x-\frac{(3x)^2}{2}+\frac{(3x)^3}{3}-\cdots \approx 3x-\frac{9}{2}x^2+9x^3\) | R1 | Substitutes \(3x\) for \(x\) in \(\ln(1+x)\) series and simplifies; condone missing LHS |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
The expansion is valid for \(-1<3x\leq1 \Rightarrow -\frac{1}{3}| M1 |
States \((-1<)\ 3x\leq1\); condone \(3x<1\) |
|
| \(x=1\) is not in the valid range, so Julia should not have substituted \(x=1\) | E1 | Explains that \(x=1\) is not a valid value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(x = -\frac{1}{6}\) into \(3x - \frac{9}{2}x^2 + 9x^3\) and \(\ln(1+3x)\), PI by \(-2 \times \left(3\left(-\frac{1}{6}\right) - \frac{9}{2}\left(-\frac{1}{6}\right)^2 + 9\left(-\frac{1}{6}\right)^3\right)\) | M1 (1.1a) | PI by correct substitution shown |
| Obtains \(\ln\frac{1}{2} \approx -\frac{2}{3}\); accept AWRT \(-0.667\); PI by \(\ln 4 \approx -2 \times \left(3\left(-\frac{1}{6}\right) - \frac{9}{2}\left(-\frac{1}{6}\right)^2 + 9\left(-\frac{1}{6}\right)^3\right)\) or \(\ln 4 \approx -2 \times \left(-\frac{2}{3}\right)\) | A1 (1.1b) | |
| Uses \(\ln 4 = -2\ln\frac{1}{2}\) | M1 (1.1a) | |
| Deduces \(\ln 4 \approx \frac{4}{3}\); accept AWRT 1.33 | R1 (2.2a) | Condone equals sign instead of approximation sign for all four marks |
## Question 15(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln(1+3x)=3x-\frac{(3x)^2}{2}+\frac{(3x)^3}{3}-\cdots \approx 3x-\frac{9}{2}x^2+9x^3$ | R1 | Substitutes $3x$ for $x$ in $\ln(1+x)$ series and simplifies; condone missing LHS |
---
## Question 15(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The expansion is valid for $-1<3x\leq1 \Rightarrow -\frac{1}{3}<x\leq\frac{1}{3}$ | M1 | States $(-1<)\ 3x\leq1$; condone $3x<1$ |
| $x=1$ is not in the valid range, so Julia should not have substituted $x=1$ | E1 | Explains that $x=1$ is not a valid value |
## Question 15(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $x = -\frac{1}{6}$ into $3x - \frac{9}{2}x^2 + 9x^3$ **and** $\ln(1+3x)$, PI by $-2 \times \left(3\left(-\frac{1}{6}\right) - \frac{9}{2}\left(-\frac{1}{6}\right)^2 + 9\left(-\frac{1}{6}\right)^3\right)$ | M1 (1.1a) | PI by correct substitution shown |
| Obtains $\ln\frac{1}{2} \approx -\frac{2}{3}$; accept AWRT $-0.667$; PI by $\ln 4 \approx -2 \times \left(3\left(-\frac{1}{6}\right) - \frac{9}{2}\left(-\frac{1}{6}\right)^2 + 9\left(-\frac{1}{6}\right)^3\right)$ or $\ln 4 \approx -2 \times \left(-\frac{2}{3}\right)$ | A1 (1.1b) | |
| Uses $\ln 4 = -2\ln\frac{1}{2}$ | M1 (1.1a) | |
| Deduces $\ln 4 \approx \frac{4}{3}$; accept AWRT 1.33 | R1 (2.2a) | Condone equals sign instead of approximation sign for all four marks |
---15
\begin{enumerate}[label=(\alph*)]
\item Use Maclaurin's series expansion for $\ln ( 1 + x )$ to show that the first three terms of the Maclaurin's series expansion of $\ln ( 1 + 3 x )$ are
$$3 x - \frac { 9 } { 2 } x ^ { 2 } + 9 x ^ { 3 }$$
15
\item Julia attempts to use the series expansion found in part (a) to find an approximation for $\ln 4$
Julia's incorrect working is shown below.
$$\begin{array} { r }
\text { Let } 1 + 3 x = 4 \\
3 x = 3 \\
x = 1
\end{array}$$
$$\text { So } \begin{aligned}
\ln 4 & \approx 3 \times 1 - \frac { 9 } { 2 } \times 1 ^ { 2 } + 9 \times 1 ^ { 3 } \\
& \approx 3 - 4.5 + 9 \\
& \approx 7.5
\end{aligned}$$
Explain the error in Julia's working.\\
15
\item Use $x = - \frac { 1 } { 6 }$ in the series expansion found in part (a) to find an approximation for $\ln 4$ Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q15 [7]}}