| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about π/3 or π/6 |
| Difficulty | Standard +0.8 This is a Further Maths F2 question requiring multiple derivatives of a trigonometric function, pattern recognition to express the third derivative in a specific form, then application of Taylor series about a non-zero point (π/3). While the techniques are standard for FM students, the algebraic manipulation of trigonometric derivatives and evaluation at π/3 requires careful work across multiple steps, placing it moderately above average difficulty. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \tan^2 x \Rightarrow \frac{dy}{dx} = 2\tan x \sec^2 x\) | B1 | Correct first derivative any correct form |
| \(\frac{dy}{dx} = 2\tan x \sec^2 x \Rightarrow \frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x \tan^2 x\) | M1A1 | M1: Correct application of product rule and chain rule. A1: Correct expression |
| \(\frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x\tan^2 x \Rightarrow \frac{d^3y}{dx^3} = 8\sec^4 x\tan x + 8\sec^2 x\tan^3 x + 8\sec^4 x\tan x\) | M1 | M1: Attempt to differentiate using product and chain rule. At least one term to be correct |
| Or: \(\frac{d^2y}{dx^2} = 6\sec^4 x - 4\sec^2 x \Rightarrow \frac{d^3y}{dx^3} = 24\sec^4 x\tan x - 8\sec^2 x\tan x\) | ||
| \(= 8\sec^4 x\tan x + 8\sec^2 x\tan x(\sec^2 x - 1) + 8\sec^4 x\tan x\) | ||
| \(= 24\sec^4 x\tan x - 8\sec^2 x\tan x = 8\sec^2 x\tan x(3\sec^2 x - 1)\) | A1 | Fully correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y)_{\frac{\pi}{3}} = 3,\ (y')_{\frac{\pi}{3}} = 8\sqrt{3},\ (y'')_{\frac{\pi}{3}} = 80,\ (y''')_{\frac{\pi}{3}} = 352\sqrt{3}\) | M1 | Attempts the values up to the third derivative when \(x = \frac{\pi}{3}\) |
| \(y = 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + \frac{80}{2!}\left(x-\frac{\pi}{3}\right)^2 + \frac{352\sqrt{3}}{3!}\left(x-\frac{\pi}{3}\right)^3 + \ldots\) | M1 | Correct application of the Taylor series \(2!\) or \(2\), \(3!\) or \(6\) |
| \(y = 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + 40\left(x-\frac{\pi}{3}\right)^2 + \frac{176\sqrt{3}}{3}\left(x-\frac{\pi}{3}\right)^3 + \ldots\) | A1 | Correct expansion. Must start \(y = \ldots\) or \(\tan^2 x = \ldots\) \(f(x)\) only accepted if \(f(x)\) has been defined to be \(\tan^2 x\) |
## Question 5:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \tan^2 x \Rightarrow \frac{dy}{dx} = 2\tan x \sec^2 x$ | B1 | Correct first derivative any correct form |
| $\frac{dy}{dx} = 2\tan x \sec^2 x \Rightarrow \frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x \tan^2 x$ | M1A1 | M1: Correct application of product rule and chain rule. A1: Correct expression |
| $\frac{d^2y}{dx^2} = 2\sec^4 x + 4\sec^2 x\tan^2 x \Rightarrow \frac{d^3y}{dx^3} = 8\sec^4 x\tan x + 8\sec^2 x\tan^3 x + 8\sec^4 x\tan x$ | M1 | M1: Attempt to differentiate using product and chain rule. At least one term to be correct |
| Or: $\frac{d^2y}{dx^2} = 6\sec^4 x - 4\sec^2 x \Rightarrow \frac{d^3y}{dx^3} = 24\sec^4 x\tan x - 8\sec^2 x\tan x$ | | |
| $= 8\sec^4 x\tan x + 8\sec^2 x\tan x(\sec^2 x - 1) + 8\sec^4 x\tan x$ | | |
| $= 24\sec^4 x\tan x - 8\sec^2 x\tan x = 8\sec^2 x\tan x(3\sec^2 x - 1)$ | A1 | Fully correct expression |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y)_{\frac{\pi}{3}} = 3,\ (y')_{\frac{\pi}{3}} = 8\sqrt{3},\ (y'')_{\frac{\pi}{3}} = 80,\ (y''')_{\frac{\pi}{3}} = 352\sqrt{3}$ | M1 | Attempts the values up to the third derivative when $x = \frac{\pi}{3}$ |
| $y = 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + \frac{80}{2!}\left(x-\frac{\pi}{3}\right)^2 + \frac{352\sqrt{3}}{3!}\left(x-\frac{\pi}{3}\right)^3 + \ldots$ | M1 | Correct application of the Taylor series $2!$ or $2$, $3!$ or $6$ |
| $y = 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + 40\left(x-\frac{\pi}{3}\right)^2 + \frac{176\sqrt{3}}{3}\left(x-\frac{\pi}{3}\right)^3 + \ldots$ | A1 | Correct expansion. Must start $y = \ldots$ or $\tan^2 x = \ldots$ $f(x)$ only accepted if $f(x)$ has been defined to be $\tan^2 x$ |
---5. Given that $y = \tan ^ { 2 } x$
\begin{enumerate}[label=(\alph*)]
\item show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = 8 \tan x \sec ^ { 2 } x \left( p \sec ^ { 2 } x + q \right)$$
where $p$ and $q$ are integers to be determined.
\item Hence determine the Taylor series expansion about $\frac { \pi } { 3 }$ of $\tan ^ { 2 } x$ in ascending powers of $\left( x - \frac { \pi } { 3 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 3 } \right) ^ { 3 }$, giving each coefficient in simplest form.\\
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-19_33_407_306_258}\\
\includegraphics[max width=\textwidth, alt={}, center]{8fa1e7da-009f-4b7f-9fa8-21a1768bfd73-19_58_458_2752_150}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2021 Q5 [8]}}