| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about π/3 or π/6 |
| Difficulty | Challenging +1.2 This is a standard Further Maths Taylor series question requiring systematic differentiation of tan(3x/2), evaluation at π/6, and substitution. Part (b) is straightforward substitution. While it requires careful algebra with trigonometric values and multiple derivatives, it follows a well-practiced template with no novel insight needed. Slightly above average due to the algebraic manipulation involved and being Further Maths content. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^24.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = \tan\left(\frac{3x}{2}\right) \Rightarrow y' = \frac{3}{2}\sec^2\left(\frac{3x}{2}\right)\) | B1 | Any correct first derivative. Not implied by \(y'\!\left(\frac{\pi}{6}\right)=3\) |
| \(y'' = \frac{9}{2}\sec^2\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right)\) | M1 | Attempts second derivative achieving \(k\sec^2\!\left(\frac{3x}{2}\right)\tan\!\left(\frac{3x}{2}\right)\). Not implied by \(y''\!\left(\frac{\pi}{6}\right)=9\) |
| \(y''' = \frac{27}{4}\sec^4\!\left(\frac{3x}{2}\right) + \frac{27}{2}\sec^2\!\left(\frac{3x}{2}\right)\tan^2\!\left(\frac{3x}{2}\right)\) | dM1, A1 | Product rule achieving \(P\sec^4\!\left(\frac{3x}{2}\right)+Q\sec^2\!\left(\frac{3x}{2}\right)\tan^2\!\left(\frac{3x}{2}\right)\). Requires previous M. A1 correct, unsimplified accepted |
| \(y\!\left(\frac{\pi}{6}\right)=1,\ y'\!\left(\frac{\pi}{6}\right)=3,\ y''\!\left(\frac{\pi}{6}\right)=9,\ y'''\!\left(\frac{\pi}{6}\right)=54\) | M1 | Attempts values for \(y\) and its 3 derivatives at \(\frac{\pi}{6}\) |
| \(y = 1 + 3\!\left(x-\frac{\pi}{6}\right) + \frac{9}{2!}\!\left(x-\frac{\pi}{6}\right)^2 + \frac{54}{3!}\!\left(x-\frac{\pi}{6}\right)^3 + \ldots\) | dM1 | Applies Taylor's correctly about \(\frac{\pi}{6}\). Requires previous M |
| \(y = 1 + 3\!\left(x-\frac{\pi}{6}\right) + \frac{9}{2}\!\left(x-\frac{\pi}{6}\right)^2 + 9\!\left(x-\frac{\pi}{6}\right)^3 + \ldots\) | A1 | Correct expression with coefficients in simplest form. Score A0 if clear evidence of use of any wrong derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y\left(\frac{\pi}{4}\right) = 1 + 3\left(\frac{\pi}{4} - \frac{\pi}{6}\right) + \frac{9}{2}\left(\frac{\pi}{4} - \frac{\pi}{6}\right)^2 + 9\left(\frac{\pi}{4} - \frac{\pi}{6}\right)^3\) or \(1 + 3\left(\frac{\pi}{12}\right) + \frac{9}{2}\left(\frac{\pi}{12}\right)^2 + 9\left(\frac{\pi}{12}\right)^3\) | M1 | Substitutes \(\frac{\pi}{4}\) into expression for \(y\) of correct form with at least first three terms (series about \(\frac{\pi}{6}\)). Must have values not unevaluated trig expressions. If only decimal given must be correct awrt 2.26. If no working must obtain \(a + b\pi + c\pi^2\) with correct exact \(a\), \(b\), \(c\) for series or \(1 + \frac{\pi}{4} + c\pi^2\) with correct exact \(c\) |
| \(= 1 + \frac{\pi}{4} + \frac{\pi^2}{32} + \frac{\pi^3}{192}\) or \(1 + \frac{1}{4}\pi + \frac{1}{32}\pi^2 + \frac{1}{192}\pi^3\) | A1 | Correct answer or values for \(A\) (32) and \(B\) (192). Can be awarded if full marks not scored in (a) |
# Question 4:
## Part 4(a):
| Working | Mark | Notes |
|---------|------|-------|
| $y = \tan\left(\frac{3x}{2}\right) \Rightarrow y' = \frac{3}{2}\sec^2\left(\frac{3x}{2}\right)$ | B1 | Any correct first derivative. Not implied by $y'\!\left(\frac{\pi}{6}\right)=3$ |
| $y'' = \frac{9}{2}\sec^2\left(\frac{3x}{2}\right)\tan\left(\frac{3x}{2}\right)$ | M1 | Attempts second derivative achieving $k\sec^2\!\left(\frac{3x}{2}\right)\tan\!\left(\frac{3x}{2}\right)$. Not implied by $y''\!\left(\frac{\pi}{6}\right)=9$ |
| $y''' = \frac{27}{4}\sec^4\!\left(\frac{3x}{2}\right) + \frac{27}{2}\sec^2\!\left(\frac{3x}{2}\right)\tan^2\!\left(\frac{3x}{2}\right)$ | dM1, A1 | Product rule achieving $P\sec^4\!\left(\frac{3x}{2}\right)+Q\sec^2\!\left(\frac{3x}{2}\right)\tan^2\!\left(\frac{3x}{2}\right)$. Requires previous M. A1 correct, unsimplified accepted |
| $y\!\left(\frac{\pi}{6}\right)=1,\ y'\!\left(\frac{\pi}{6}\right)=3,\ y''\!\left(\frac{\pi}{6}\right)=9,\ y'''\!\left(\frac{\pi}{6}\right)=54$ | M1 | Attempts values for $y$ and its 3 derivatives at $\frac{\pi}{6}$ |
| $y = 1 + 3\!\left(x-\frac{\pi}{6}\right) + \frac{9}{2!}\!\left(x-\frac{\pi}{6}\right)^2 + \frac{54}{3!}\!\left(x-\frac{\pi}{6}\right)^3 + \ldots$ | dM1 | Applies Taylor's correctly about $\frac{\pi}{6}$. Requires previous M |
| $y = 1 + 3\!\left(x-\frac{\pi}{6}\right) + \frac{9}{2}\!\left(x-\frac{\pi}{6}\right)^2 + 9\!\left(x-\frac{\pi}{6}\right)^3 + \ldots$ | A1 | Correct expression with coefficients in simplest form. Score A0 if clear evidence of use of any wrong derivative |
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y\left(\frac{\pi}{4}\right) = 1 + 3\left(\frac{\pi}{4} - \frac{\pi}{6}\right) + \frac{9}{2}\left(\frac{\pi}{4} - \frac{\pi}{6}\right)^2 + 9\left(\frac{\pi}{4} - \frac{\pi}{6}\right)^3$ or $1 + 3\left(\frac{\pi}{12}\right) + \frac{9}{2}\left(\frac{\pi}{12}\right)^2 + 9\left(\frac{\pi}{12}\right)^3$ | M1 | Substitutes $\frac{\pi}{4}$ into expression for $y$ of correct form with at least first three terms (series about $\frac{\pi}{6}$). Must have values not unevaluated trig expressions. If only decimal given must be correct awrt 2.26. If no working must obtain $a + b\pi + c\pi^2$ with correct exact $a$, $b$, $c$ for series or $1 + \frac{\pi}{4} + c\pi^2$ with correct exact $c$ |
| $= 1 + \frac{\pi}{4} + \frac{\pi^2}{32} + \frac{\pi^3}{192}$ or $1 + \frac{1}{4}\pi + \frac{1}{32}\pi^2 + \frac{1}{192}\pi^3$ | A1 | Correct answer or values for $A$ (32) and $B$ (192). Can be awarded if full marks not scored in (a) |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(a) Determine, in ascending powers of $\left( x - \frac { \pi } { 6 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 6 } \right) ^ { 3 }$, the Taylor series expansion about $\frac { \pi } { 6 }$ of
$$y = \tan \left( \frac { 3 x } { 2 } \right)$$
giving each coefficient in simplest form.\\
(b) Hence show that
$$\tan \frac { 3 \pi } { 8 } \approx 1 + \frac { \pi } { 4 } + \frac { \pi ^ { 2 } } { A } + \frac { \pi ^ { 3 } } { B }$$
where $A$ and $B$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2024 Q4 [9]}}