C3 June 2008 Q6
6.
(a) Differentiate with respect to \(x\),
(i) \(\mathrm{e}^{3x}(\sin x + 2\cos x)\), (3)
(ii) \(x^3\ln(5x + 2)\). (3)
Given that \(y = \dfrac{3x^2 + 6x - 7}{(x + 1)^2}\), \(x \neq -1\),
(b) show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{20}{(x + 1)^3}\). (5)
(c) Hence find \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and the real values of \(x\) for which \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{15}{4}\). (3)
| Scheme | Marks |
|---|---|
| (i) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{3x}(\sin x + 2\cos x)\right) = 3\mathrm{e}^{3x}(\sin x + 2\cos x) + \mathrm{e}^{3x}(\cos x - 2\sin x)\) \(\left(= \mathrm{e}^{3x}(\sin x + 7\cos x)\right)\) | M1 A1 A1 (3) |
| (ii) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(x^3\ln(5x + 2)\right) = 3x^2\ln(5x + 2) + \dfrac{5x^3}{5x + 2}\) | M1 A1 A1 (3) |
| (6) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(x + 1)^2(6x + 6) - 2(x + 1)(3x^2 + 6x - 7)}{(x + 1)^4}\) | M1 A1 A1 |
| \(= \dfrac{(x + 1)(6x^2 + 12x + 6 - 6x^2 - 12x + 14)}{(x + 1)^4}\) | M1 |
| \(= \dfrac{20}{(x + 1)^3}\ \ \ast\) cso | A1 |
| (5) |
Notes
Note: The simplification in part (b) can be carried out as follows\(\dfrac{(x + 1)^2(6x + 6) - 2(x + 1)(3x^2 + 6x - 7)}{(x + 1)^4}\)
\(= \dfrac{(6x^3 + 18x^2 + 18x + 6) - (6x^3 + 18x^2 - 2x - 14)}{(x + 1)^4}\)
\(= \dfrac{20x + 20}{(x + 1)^4} = \dfrac{20(x + 1)}{(x + 1)^4} = \dfrac{20}{(x + 1)^3}\) M1 A1
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{60}{(x + 1)^4} = -\dfrac{15}{4}\) | M1 |
| \((x + 1)^4 = 16\) | M1 |
| \(x = 1, -3\) both | A1 |
| (3) | |
| (14 marks) |