FP2 June 2009 Q5
5. \[y = \sec^2 x\]
(a) Show that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\sec^4 x - 4\sec^2 x\). (4)
(b) Find a Taylor series expansion of \(\sec^2 x\) in ascending powers of \(\left(x - \dfrac{\pi}{4}\right)\), up to and including the term in \(\left(x - \dfrac{\pi}{4}\right)^3\). (6)
| Scheme | Marks |
|---|---|
| \(y = \sec^2 x = (\sec x)^2\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2(\sec x)^1(\sec x\tan x) = 2\sec^2 x\tan x\) Either \(2(\sec x)^1(\sec x\tan x)\) or \(2\sec^2 x\tan x\) | B1 aef |
| Apply product rule: \(\left\{\begin{array}{ll} u = 2\sec^2 x & v = \tan x \\ \dfrac{\mathrm{d}u}{\mathrm{d}x} = 4\sec^2 x\tan x & \dfrac{\mathrm{d}v}{\mathrm{d}x} = \sec^2 x \end{array}\right\}\) | |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4\sec^2 x\tan^2 x + 2\sec^4 x\) Two terms added with one of either \(A\sec^2 x\tan^2 x\) or \(B\sec^4 x\) in the correct form. Correct differentiation | M1 A1 |
| \(= 4\sec^2 x(\sec^2 x - 1) + 2\sec^4 x\) | |
| Hence, \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6\sec^4 x - 4\sec^2 x\) Applies \(\tan^2 x = \sec^2 x - 1\) leading to the correct result. | A1 AG |
| (4) |
| Scheme | Marks |
|---|---|
| \(y_{\frac{\pi}{4}} = (\sqrt{2})^2 = \underline{2}\), \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{4}} = 2(\sqrt{2})^2(1) = \underline{4}\) Both \(y_{\frac{\pi}{4}} = \underline{2}\) and \(\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_{\frac{\pi}{4}} = \underline{4}\) | B1 |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{\frac{\pi}{4}} = 6(\sqrt{2})^4 - 4(\sqrt{2})^2 = 24 - 8 = 16\) Attempts to substitute \(x = \frac{\pi}{4}\) into both terms in the expression for \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\). | M1 |
| \(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3} = 24\sec^3 x(\sec x\tan x) - 8\sec x(\sec x\tan x)\) \(= 24\sec^4 x\tan x - 8\sec^2 x\tan x\) Two terms differentiated with either \(24\sec^4 x\tan x\) or \(-8\sec^2 x\tan x\) being correct | M1 |
| \(\left(\dfrac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{4}} = 24(\sqrt{2})^4(1) - 8(\sqrt{2})^2(1) = 96 - 16 = 80\) \(\left(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{4}} = \underline{80}\) | B1 |
| \(\sec^2 x \approx 2 + 4\left(x - \frac{\pi}{4}\right) + \frac{16}{2}\left(x - \frac{\pi}{4}\right)^2 + \frac{80}{6}\left(x - \frac{\pi}{4}\right)^3 + \ldots\) Applies a Taylor expansion with at least 3 out of 4 terms ft correctly. Correct Taylor series expansion. | M1 A1 |
| \(\left\{\sec^2 x \approx 2 + 4\left(x - \frac{\pi}{4}\right) + 8\left(x - \frac{\pi}{4}\right)^2 + \frac{40}{3}\left(x - \frac{\pi}{4}\right)^3 + \ldots\right\}\) | |
| (6) | |
| (10 marks) |
Notes
(corrected from the printed mark scheme: the line giving 80 is printed with \(\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{\frac{\pi}{4}}\) instead of \(\left(\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right)_{\frac{\pi}{4}}\), and the two expansion lines are printed as \(\sec x \approx \ldots\) instead of \(\sec^2 x \approx \ldots\))