C3 January 2009 Q1
1.
(a) Find the value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at the point where \(x = 2\) on the curve with equation\[y = x^2\sqrt{(5x - 1)}.\] (6)
(b) Differentiate \(\dfrac{\sin 2x}{x^2}\) with respect to \(x\). (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{(5x - 1)}\right) = \dfrac{\mathrm{d}}{\mathrm{d}x}\left((5x - 1)^{\frac{1}{2}}\right)\) \(= 5\times\dfrac{1}{2}(5x - 1)^{-\frac{1}{2}}\) | M1 A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\sqrt{(5x - 1)} + \dfrac{5}{2}x^2(5x - 1)^{-\frac{1}{2}}\) | M1 A1ft |
| At \(x = 2\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\sqrt{9} + \dfrac{10}{\sqrt{9}} = 12 + \dfrac{10}{3}\) | M1 |
| \(= \dfrac{46}{3}\) Accept awrt 15.3 | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\sin 2x}{x^2}\right) = \dfrac{2x^2\cos 2x - 2x\sin 2x}{x^4}\) | M1 A1+A1 A1 |
| (4) | |
| (10 marks) |
Alternative to (b)
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sin 2x\times x^{-2}\right) = 2\cos 2x\times x^{-2} + \sin 2x\times(-2)x^{-3}\) | M1 A1 + A1 |
| \(= 2x^{-2}\cos 2x - 2x^{-3}\sin 2x \quad \left(= \dfrac{2\cos 2x}{x^2} - \dfrac{2\sin 2x}{x^3}\right)\) | A1 (4) |