C3 June 2005 Q2
2.
(a) Differentiate with respect to \(x\)
(i) \(3\sin^2 x + \sec 2x\), (3)
(ii) \(\{x + \ln(2x)\}^3\). (3)
Given that \(y = \dfrac{5x^2 - 10x + 9}{(x - 1)^2}\), \(x \neq 1\),
(b) show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{8}{(x - 1)^3}\). (6)
| Scheme | Marks |
|---|---|
| (i) \(6\sin x\cos x + 2\sec 2x\tan 2x\) or \(3\sin 2x + 2\sec 2x\tan 2x\) [M1 for \(6\sin x\)] | M1 A1 A1 (3) |
| (ii) \(3(x + \ln 2x)^2\left(1 + \dfrac{1}{x}\right)\) [B1 for \(3(x + \ln 2x)^2\)] | B1 M1 A1 (3) |
| Scheme | Marks |
|---|---|
| Differentiating numerator to obtain \(10x - 10\) | B1 |
| Differentiating denominator to obtain \(2(x - 1)\) | B1 |
| Using quotient rule formula correctly: | M1 |
| To obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(x - 1)^2(10x - 10) - (5x^2 - 10x + 9)2(x - 1)}{(x - 1)^4}\) | A1 |
| Simplifying to form \(\dfrac{2(x - 1)[5(x - 1)^2 - (5x^2 - 10x + 9)]}{(x - 1)^4}\) | M1 |
| \(= -\dfrac{8}{(x - 1)^3}\) * (c.s.o.) | A1 |
| (6) | |
| (12 marks) |
Alternatives for (b)
| Either Using product rule formula correctly: | M1 |
| Obtaining \(10x - 10\) | B1 |
| Obtaining \(-2(x - 1)^{-3}\) | B1 |
| To obtain \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (5x^2 - 10x + 9)\{-2(x - 1)^{-3}\} + (10x - 10)(x - 1)^{-2}\) | A1 cao |
| Simplifying to form \(\dfrac{10(x - 1)^2 - 2(5x^2 - 10x + 9)}{(x - 1)^3}\) | M1 |
| \(= -\dfrac{8}{(x - 1)^3}\) * (c.s.o.) | A1 (6) |
| Or Splitting fraction to give \(5 + \dfrac{4}{(x - 1)^2}\) | M1 B1 B1 |
| Then differentiating to give answer | M1 A1 A1 (6) |