| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Taylor series about π/4 |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of sec²x and Taylor series expansion about a non-zero point. Part (a) is routine differentiation using chain rule, while part (b) requires evaluating derivatives at π/4 and applying the Taylor series formula. The calculations involve standard trig values at π/4, making this moderately challenging but still a standard FP2 exercise with clear methodology. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07d Second derivatives: d^2y/dx^2 notation4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = 2(\sec x)^1(\sec x\tan x) = 2\sec^2 x\tan x\) | B1 aef | Either \(2(\sec x)^1(\sec x\tan x)\) or \(2\sec^2 x\tan x\) |
| \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 4\sec^2 x\tan^2 x + 2\sec^4 x\) | M1 | Two terms added with one of either \(A\sec^2 x\tan^2 x\) or \(B\sec^4 x\) in correct form |
| A1 | Correct differentiation | |
| \(= 4\sec^2 x(\sec^2 x-1)+2\sec^4 x = 6\sec^4 x - 4\sec^2 x\) | A1 AG | Applies \(\tan^2 x = \sec^2 x-1\) leading to correct result (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y_{\frac{\pi}{4}} = (\sqrt{2})^2 = 2\), \(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_{\!\frac{\pi}{4}} = 2(\sqrt{2})^2(1) = 4\) | B1 | Both \(y_{\frac{\pi}{4}}=2\) and \(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_{\!\frac{\pi}{4}}=4\) |
| \(\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right)_{\!\frac{\pi}{4}} = 6(\sqrt{2})^4 - 4(\sqrt{2})^2 = 24-8 = 16\) | M1 | Attempts to substitute \(x=\frac{\pi}{4}\) into both terms of \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\) |
| \(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 24\sec^3 x(\sec x\tan x) - 8\sec x(\sec x\tan x) = 24\sec^4 x\tan x - 8\sec^2 x\tan x\) | M1 | Two terms differentiated with either \(24\sec^4 x\tan x\) or \(-8\sec^2 x\tan x\) being correct |
| \(\left(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\right)_{\!\frac{\pi}{4}} = 24(\sqrt{2})^4(1)-8(\sqrt{2})^2(1) = 96-16 = 80\) | B1 | \(\left(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\right)_{\!\frac{\pi}{4}}=80\) |
| \(\sec^2 x \approx 2+4\!\left(x-\frac{\pi}{4}\right)+\frac{16}{2}\!\left(x-\frac{\pi}{4}\right)^2+\frac{80}{6}\!\left(x-\frac{\pi}{4}\right)^3+\ldots\) | M1 | Applies Taylor expansion with at least 3 of 4 terms correctly |
| \(\sec^2 x \approx 2+4\!\left(x-\frac{\pi}{4}\right)+8\!\left(x-\frac{\pi}{4}\right)^2+\frac{40}{3}\!\left(x-\frac{\pi}{4}\right)^3+\ldots\) | A1 | Correct Taylor series expansion (6 marks) |
# Question 5:
## Part (a)
| $\frac{\mathrm{d}y}{\mathrm{d}x} = 2(\sec x)^1(\sec x\tan x) = 2\sec^2 x\tan x$ | B1 **aef** | Either $2(\sec x)^1(\sec x\tan x)$ or $2\sec^2 x\tan x$ |
|---|---|---|
| $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 4\sec^2 x\tan^2 x + 2\sec^4 x$ | M1 | Two terms added with one of either $A\sec^2 x\tan^2 x$ or $B\sec^4 x$ in correct form |
| | A1 | Correct differentiation |
| $= 4\sec^2 x(\sec^2 x-1)+2\sec^4 x = 6\sec^4 x - 4\sec^2 x$ | A1 **AG** | Applies $\tan^2 x = \sec^2 x-1$ leading to correct result (4 marks) |
## Part (b)
| $y_{\frac{\pi}{4}} = (\sqrt{2})^2 = 2$, $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_{\!\frac{\pi}{4}} = 2(\sqrt{2})^2(1) = 4$ | B1 | Both $y_{\frac{\pi}{4}}=2$ and $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_{\!\frac{\pi}{4}}=4$ |
|---|---|---|
| $\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right)_{\!\frac{\pi}{4}} = 6(\sqrt{2})^4 - 4(\sqrt{2})^2 = 24-8 = 16$ | M1 | Attempts to substitute $x=\frac{\pi}{4}$ into both terms of $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ |
| $\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 24\sec^3 x(\sec x\tan x) - 8\sec x(\sec x\tan x) = 24\sec^4 x\tan x - 8\sec^2 x\tan x$ | M1 | Two terms differentiated with either $24\sec^4 x\tan x$ or $-8\sec^2 x\tan x$ being correct |
| $\left(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\right)_{\!\frac{\pi}{4}} = 24(\sqrt{2})^4(1)-8(\sqrt{2})^2(1) = 96-16 = 80$ | B1 | $\left(\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}\right)_{\!\frac{\pi}{4}}=80$ |
| $\sec^2 x \approx 2+4\!\left(x-\frac{\pi}{4}\right)+\frac{16}{2}\!\left(x-\frac{\pi}{4}\right)^2+\frac{80}{6}\!\left(x-\frac{\pi}{4}\right)^3+\ldots$ | M1 | Applies Taylor expansion with at least 3 of 4 terms correctly |
| $\sec^2 x \approx 2+4\!\left(x-\frac{\pi}{4}\right)+8\!\left(x-\frac{\pi}{4}\right)^2+\frac{40}{3}\!\left(x-\frac{\pi}{4}\right)^3+\ldots$ | A1 | Correct Taylor series expansion (6 marks) |5.
$$y = \sec ^ { 2 } x$$
\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 Find a Taylor series expansion of $\sec ^ { 2 } x$ in ascending powers of $\left( x - \frac { \pi } { 4 } \right)$, up to and including the term in $\left( x - \frac { \pi } { 4 } \right) ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2009 Q5 [10]}}