| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about π/4 |
| Difficulty | Challenging +1.2 This is a structured Taylor series question with clear scaffolding through parts (a) and (b). While it requires multiple derivatives of trigonometric functions and evaluation at x=π/3, the steps are guided and the techniques are standard for FP2. The algebraic manipulation is moderately involved but follows predictable patterns for this topic. |
| 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{dy}{dx} = 2\tan x\sec^2 x\) | B1 | |
| OR \(\frac{dy}{dx} = 2\tan x(1+\tan^2 x)\) | ||
| \(\frac{d^2y}{dx^2} = 2\sec^4 x + 4\tan^2 x\sec^2 x\) OR \(2\sec^2 x + 6\tan^2 x\sec^2 x\) | M1 A1 | Attempt second derivative using product rule or \(\sec^2\theta = \tan^2\theta+1\); must start from result in (a); A1 correct second derivative in any form |
| \(= 6\sec^4 x - 4\sec^2 x\) * | A1cso (4) | Correct result; \(\sec^2\theta = \tan^2\theta+1\) must be seen or used |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{d^3y}{dx^3} = 24\sec^3 x\sec x\tan x - 8\sec^2 x\tan x\) | M1A1 | Attempt third derivative including chain rule; A1 correct derivative |
| \(= 8\sec^2 x\tan x(3\sec^2 x - 1)\) | A1cso (3) | Completely correct final result |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y_{\pi/3} = (\sqrt{3})^2 = 3\) and \(\left(\frac{dy}{dx}\right)_{\pi/3} = 2\sqrt{3}\times\left(\frac{2}{1}\right)^2 = 8\sqrt{3}\) | B1(both) | Both values correct |
| \(\left(\frac{d^2y}{dx^2}\right)_{\pi/3} = 6\times 2^4 - 4\times 2^2 = 80\) | M1(attempt both) | Obtaining values for 2nd and 3rd derivatives at \(\frac{\pi}{3}\) (need not be correct but must come from their derivatives) |
| \(\left(\frac{d^3y}{dx^3}\right)_{\pi/3} = 8\times 4\times\sqrt{3}(3\times 2^2-1) = 352\sqrt{3}\) | ||
| \(\tan^2 x = y_{\pi/3} + \left(x-\frac{\pi}{3}\right)\left(\frac{dy}{dx}\right)_{\pi/3} + \frac{1}{2!}\left(x-\frac{\pi}{3}\right)^2\left(\frac{d^2y}{dx^2}\right)_{\pi/3} + \frac{1}{3!}\left(x-\frac{\pi}{3}\right)^3\left(\frac{d^3y}{dx^3}\right)_{\pi/3}\) | M1 | Using correct Taylor expansion with \(\left(x-\frac{\pi}{3}\right)\); 2! or 2, 3! or 6 must be seen or implied |
| \(= 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + 40\left(x-\frac{\pi}{3}\right)^2 + \frac{176\sqrt{3}}{3}\left(x-\frac{\pi}{3}\right)^3\) | M1A1 (4)[11] | A1 correct final answer; must start \(\tan^2 x = \ldots\) or \(y = \ldots f(x)\); no factorials in final answer |
## Question 7:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{dy}{dx} = 2\tan x\sec^2 x$ | B1 | |
| OR $\frac{dy}{dx} = 2\tan x(1+\tan^2 x)$ | | |
| $\frac{d^2y}{dx^2} = 2\sec^4 x + 4\tan^2 x\sec^2 x$ OR $2\sec^2 x + 6\tan^2 x\sec^2 x$ | M1 A1 | Attempt second derivative using product rule or $\sec^2\theta = \tan^2\theta+1$; must start from result in (a); A1 correct second derivative in any form |
| $= 6\sec^4 x - 4\sec^2 x$ * | A1cso (4) | Correct result; $\sec^2\theta = \tan^2\theta+1$ must be seen or used |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{d^3y}{dx^3} = 24\sec^3 x\sec x\tan x - 8\sec^2 x\tan x$ | M1A1 | Attempt third derivative including chain rule; A1 correct derivative |
| $= 8\sec^2 x\tan x(3\sec^2 x - 1)$ | A1cso (3) | Completely correct final result |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $y_{\pi/3} = (\sqrt{3})^2 = 3$ and $\left(\frac{dy}{dx}\right)_{\pi/3} = 2\sqrt{3}\times\left(\frac{2}{1}\right)^2 = 8\sqrt{3}$ | B1(both) | Both values correct |
| $\left(\frac{d^2y}{dx^2}\right)_{\pi/3} = 6\times 2^4 - 4\times 2^2 = 80$ | M1(attempt both) | Obtaining values for 2nd and 3rd derivatives at $\frac{\pi}{3}$ (need not be correct but must come from their derivatives) |
| $\left(\frac{d^3y}{dx^3}\right)_{\pi/3} = 8\times 4\times\sqrt{3}(3\times 2^2-1) = 352\sqrt{3}$ | | |
| $\tan^2 x = y_{\pi/3} + \left(x-\frac{\pi}{3}\right)\left(\frac{dy}{dx}\right)_{\pi/3} + \frac{1}{2!}\left(x-\frac{\pi}{3}\right)^2\left(\frac{d^2y}{dx^2}\right)_{\pi/3} + \frac{1}{3!}\left(x-\frac{\pi}{3}\right)^3\left(\frac{d^3y}{dx^3}\right)_{\pi/3}$ | M1 | Using correct Taylor expansion with $\left(x-\frac{\pi}{3}\right)$; 2! or 2, 3! or 6 must be seen or implied |
| $= 3 + 8\sqrt{3}\left(x-\frac{\pi}{3}\right) + 40\left(x-\frac{\pi}{3}\right)^2 + \frac{176\sqrt{3}}{3}\left(x-\frac{\pi}{3}\right)^3$ | M1A1 (4)[11] | A1 correct final answer; must start $\tan^2 x = \ldots$ or $y = \ldots f(x)$; no factorials in final answer |7.
$$y = \tan ^ { 2 } x , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 6 \sec ^ { 4 } x - 4 \sec ^ { 2 } x$
\item Hence show that $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } } = 8 \sec ^ { 2 } x \tan x \left( A \sec ^ { 2 } x + B \right)$, where $A$ and $B$ are constants to be found.
\item Find the Taylor series expansion of $\tan ^ { 2 } x$, in ascending powers of $\left( x - \frac { \pi } { 3 } \right)$, up to and including the term in $\left( x - \frac { \pi } { 3 } \right) ^ { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2015 Q7 [11]}}