FP2 June 2013 (R) Q5
5.
Given that \(y = 2\) at \(x = \dfrac{\pi}{3}\)
| Scheme | Marks |
|---|---|
| I.F. \(= \mathrm{e}^{\int 2\tan x\,\mathrm{d}x} = \mathrm{e}^{2\ln\sec x} = \sec^2 x\) | M1A1 |
| \(y\sec^2 x = \displaystyle\int\sec^2 x\sin 2x\,\mathrm{d}x\) | M1 |
| \(y\sec^2 x = \displaystyle\int\dfrac{2\sin x\cos x}{\cos^2 x}\,\mathrm{d}x = 2\int\tan x\,\mathrm{d}x\) | |
| \(y\sec^2 x = 2\ln\sec x\ (+c)\) | M1depA1 |
| \(y = \dfrac{2\ln\sec x + c}{\sec^2 x}\) | A1ft |
| (6) |
Notes
M1 for the \(\mathrm{e}^{\int 2\tan x\,\mathrm{d}x}\) or \(\mathrm{e}^{\int\tan x\,\mathrm{d}x}\) and attempting the integration – \(\mathrm{e}^{(2)\ln\sec x}\) should be seen if final result is not \(\sec^2 x\)
A1 for IF \(= \sec^2 x\)
M1 for multiplying the equation by their IF and attempting to integrate the lhs
M1dep for attempting the integration of the rhs \(\sin 2x = 2\sin x\cos x\) and \(\sec x = \dfrac{1}{\cos x}\) needed. Dependent on the second M mark
A1cao for all integration correct ie \(y\sec^2 x = 2\ln\sec x\ (+c)\) constant not needed
A1ft for re-writing their answer in the form \(y = \ldots\) Accept any equivalent form but the constant must be present. eg \(y = \dfrac{\ln\left(A\sec^2 x\right)}{\sec^2 x}\), \(\quad y = \cos^2 x\left[\ln\left(\sec^2 x\right) + c\right]\)
| Scheme | Marks |
|---|---|
| \(y = 2,\ x = \dfrac{\pi}{3}\) | |
| \(2 = \dfrac{2\ln\sec\left(\frac{\pi}{3}\right) + c}{\sec^2\left(\frac{\pi}{3}\right)}\) \(2 = \dfrac{2\ln(2) + c}{4}\) | |
| \(c = 8 - 2\ln 2\) | M1A1 |
| \(x = \dfrac{\pi}{6}\quad y = \dfrac{2\ln\sec\left(\frac{\pi}{6}\right) + 8 - 2\ln 2}{\sec^2\left(\frac{\pi}{6}\right)}\) | |
| \(y = \dfrac{2\ln\frac{2}{\sqrt{3}} + 8 - 2\ln 2}{\frac{4}{3}}\) | M1 |
| \(y = \dfrac{3}{4}\left(8 + 2\ln\dfrac{1}{\sqrt{3}}\right) = 6 + \dfrac{3}{2}\ln\dfrac{1}{\sqrt{3}} = 6 - \dfrac{3}{4}\ln 3\) | A1 |
| (4) | |
| (10 marks) |
Notes
Alternative: \(c\) may not appear explicitly
| Scheme | Marks |
|---|---|
| \(y\sec^2\tfrac{\pi}{6} - 2\sec^2\tfrac{\pi}{3} = 2\ln\left(\dfrac{\sec\tfrac{\pi}{6}}{\sec\tfrac{\pi}{3}}\right)\) | M1A1 |
| \(\tfrac{4}{3}y - 8 = 2\ln\tfrac{1}{\sqrt{3}}\) | |
| \(y = \dfrac{3}{4}\left(8 + 2\ln\dfrac{1}{\sqrt{3}}\right) = 6 + \dfrac{3}{2}\ln\dfrac{1}{\sqrt{3}} = 6 - \dfrac{3}{4}\ln 3\) | M1A1 |
M1 for using the given values \(y = 2,\ x = \dfrac{\pi}{3}\) in their general solution to obtain a value for the constant of integration
A1 for eg \(c = 8 - 2\ln 2\) or \(A = \dfrac{1}{4}\mathrm{e}^8\) (Check the constant is correct for their correct answer for (a)).
Answers to 3 significant figures acceptable here and can include \(\cos\dfrac{\pi}{3}\) or \(\sec\dfrac{\pi}{3}\)
M1 for using their constant and \(x = \dfrac{\pi}{6}\) in their general solution and attempting the simplification to the required form.
A1cao for \(y = 6 - \dfrac{3}{4}\ln 3\) \(\left(\dfrac{3}{4} \text{ or } 0.75\right)\)
Alternative to 5b
M1 for finding the difference between \(y\sec^2\tfrac{\pi}{6}\) and \(2\sec^2\tfrac{\pi}{3}\) (or equivalent with their general solution)
A1 for \(y\sec^2\tfrac{\pi}{6} - 2\sec^2\tfrac{\pi}{3} = 2\ln\left(\dfrac{\sec\tfrac{\pi}{6}}{\sec\tfrac{\pi}{3}}\right)\)
M1 for re-arranging to \(y = \ldots\) and attempting the simplification to the required form
A1cao for \(y = 6 - \dfrac{3}{4}\ln 3\) \(\left(\dfrac{3}{4} \text{ or } 0.75\right)\)