| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| 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 structured Further Maths question on Taylor series with guided steps. Part (a) requires differentiation using the chain rule (routine for FM students), part (b) is straightforward differentiation of the result from (a), and part (c) applies the standard Taylor series formula with evaluation at π/3. While it involves multiple derivatives and FM content, the question provides significant scaffolding and follows a predictable pattern, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.07b Gradient as rate of change: dy/dx notation4.08a Maclaurin series: find series for function |
| VIIIV SIHI NI J14M 10N OC | VIIN SIHI NI III HM ION OO | VERV SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \cot x\) | Given | |
| \(\frac{dy}{dx} = -\text{cosec}^2 x\) | ||
| \(\frac{d^2y}{dx^2} = (-2\text{cosec}\,x)(-\text{cosec}\,x\cot x)\) | M1 A1 | M1: Differentiates using chain rule or product/quotient rule; A1: Correct derivative |
| \(= 2\text{cosec}^2 x\cot x = 2\cot x + 2\cot^3 x\) * | A1cso* | A1: Correct completion to printed answer. \(1 + \cot^2 x = \text{cosec}^2 x\) or \(\cos^2 x + \sin^2 x = 1\) must be used. Full working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{\cos x}{\sin x} \rightarrow \frac{dy}{dx} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}\) | ||
| \(\frac{d^2y}{dx^2} = -(-2\sin^{-3}x\cos x) = ...\) | M1 A1 | |
| Correct completion to printed answer (see above) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d^3y}{dx^3} = -2\text{cosec}^2 x - 6\cot^2 x\,\text{cosec}^2 x\) | B1 | Correct third derivative |
| \(= -2(1 + \cot^2 x) - 6\cot^2 x(1 + \cot^2 x)\) | M1 | Uses \(1 + \cot^2 x = \text{cosec}^2 x\) |
| \(= -6\cot^4 x - 8\cot^2 x - 2\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f\!\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}\), \(f'\!\left(\frac{\pi}{3}\right) = -\frac{4}{3}\), \(f''\!\left(\frac{\pi}{3}\right) = \frac{8}{3\sqrt{3}}\), \(f'''\!\left(\frac{\pi}{3}\right) = -\frac{16}{3}\) | M1 | Attempts all 4 values at \(\frac{\pi}{3}\). No working need be shown |
| \((y =)\frac{1}{\sqrt{3}} - \frac{4}{3}\left(x - \frac{\pi}{3}\right) + \frac{4}{3\sqrt{3}}\left(x - \frac{\pi}{3}\right)^2 - \frac{8}{9}\left(x - \frac{\pi}{3}\right)^3\) | M1 A1 | M1: Correct application of Taylor using their values, must be up to and including \(\left(x - \frac{\pi}{3}\right)^3\); A1: Correct expression. Must start \(y = ...\) or \(\cot x\). Decimal equivalents allowed (min 3 sf): 0.577, 0.578, 1.33, 0.770 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \cot x$ | | Given |
| $\frac{dy}{dx} = -\text{cosec}^2 x$ | | |
| $\frac{d^2y}{dx^2} = (-2\text{cosec}\,x)(-\text{cosec}\,x\cot x)$ | M1 A1 | M1: Differentiates using chain rule or product/quotient rule; A1: Correct derivative |
| $= 2\text{cosec}^2 x\cot x = 2\cot x + 2\cot^3 x$ * | A1cso* | A1: Correct completion to printed answer. $1 + \cot^2 x = \text{cosec}^2 x$ or $\cos^2 x + \sin^2 x = 1$ must be used. Full working must be shown |
**Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{\cos x}{\sin x} \rightarrow \frac{dy}{dx} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}$ | | |
| $\frac{d^2y}{dx^2} = -(-2\sin^{-3}x\cos x) = ...$ | M1 A1 | |
| Correct completion to printed answer (see above) | A1 | |
**Total: (3)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d^3y}{dx^3} = -2\text{cosec}^2 x - 6\cot^2 x\,\text{cosec}^2 x$ | B1 | Correct third derivative |
| $= -2(1 + \cot^2 x) - 6\cot^2 x(1 + \cot^2 x)$ | M1 | Uses $1 + \cot^2 x = \text{cosec}^2 x$ |
| $= -6\cot^4 x - 8\cot^2 x - 2$ | A1 | cso |
**Total: (3)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$, $f'\!\left(\frac{\pi}{3}\right) = -\frac{4}{3}$, $f''\!\left(\frac{\pi}{3}\right) = \frac{8}{3\sqrt{3}}$, $f'''\!\left(\frac{\pi}{3}\right) = -\frac{16}{3}$ | M1 | Attempts all 4 values at $\frac{\pi}{3}$. No working need be shown |
| $(y =)\frac{1}{\sqrt{3}} - \frac{4}{3}\left(x - \frac{\pi}{3}\right) + \frac{4}{3\sqrt{3}}\left(x - \frac{\pi}{3}\right)^2 - \frac{8}{9}\left(x - \frac{\pi}{3}\right)^3$ | M1 A1 | M1: Correct application of Taylor using their values, must be up to and including $\left(x - \frac{\pi}{3}\right)^3$; A1: Correct expression. Must start $y = ...$ or $\cot x$. Decimal equivalents allowed (min 3 sf): 0.577, 0.578, 1.33, 0.770 |
**Total: (3)**\begin{enumerate}
\item Given that $y = \cot x$,\\
(a) show that
\end{enumerate}
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 2 \cot x + 2 \cot ^ { 3 } x$$
(b) Hence show that
$$\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = p \cot ^ { 4 } x + q \cot ^ { 2 } x + r$$
where $p , q$ and $r$ are integers to be found.\\
(c) Find the Taylor series expansion of $\cot x$ in ascending powers of $\left( x - \frac { \pi } { 3 } \right)$ up to and including the term in $\left( x - \frac { \pi } { 3 } \right) ^ { 3 }$.\\
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VIIIV SIHI NI J14M 10N OC & VIIN SIHI NI III HM ION OO & VERV SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel F2 2018 Q5 [9]}}