FP3 June 2011 Q2
2.
(a) Given that \(y = x\arcsin x,\ 0 \leqslant x \leqslant 1\), find
(i) an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\),
(ii) the exact value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) when \(x = \dfrac{1}{2}\). (3)
(b) Given that \(y = \arctan(3e^{2x})\), show that \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{5\cosh 2x + 4\sinh 2x}\] (5)
| Scheme | Marks |
|---|---|
| (i) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x}{\sqrt{(1 - x^2)}} + \arcsin x\) | M1 A1 |
| (2) | |
| (ii) At given value derivative \(= \dfrac{1}{\sqrt{3}} + \dfrac{\pi}{6} = \dfrac{2\sqrt{3} + \pi}{6}\) | B1 |
| (1) |
Notes
M1 Differentiating getting an \(\arcsin x\) term and a \(\dfrac{1}{\sqrt{1 \pm x^2}}\) term
A1 CAO
B1 CAO any correct form
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6e^{2x}}{1 + 9e^{4x}}\) | 1M1 A1 |
| \(= \dfrac{6}{e^{-2x} + 9e^{2x}}\) | 2M1 |
| \(= \dfrac{3}{\frac{5}{2}(e^{2x} + e^{-2x}) + \frac{4}{2}(e^{2x} - e^{-2x})}\) | 3M1 |
| \(\therefore \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{5\cosh 2x + 4\sinh 2x}\) * | A1 cso |
| (5) | |
| (8 marks) |
Notes
1M1 Of correct form \(\dfrac{ae^{2x}}{1 \pm be^{4x}}\)
1A1 CAO
2M1 Getting from expression in \(e^{4x}\) to \(e^{2x}\) and \(e^{-2x}\) only
3M1 Using \(\sinh 2x\) and \(\cosh 2x\) in terms of \((e^{2x} + e^{-2x})\) and \((e^{2x} - e^{-2x})\)
2A1 CSO – answer given