C3 June 2009 Q4
4.
| Scheme | Marks |
|---|---|
| \(y = x^2\cos 3x\) | |
| Apply product rule: \(\left\{\begin{aligned} u &= x^2 & v &= \cos 3x \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= 2x & \frac{\mathrm{d}v}{\mathrm{d}x} &= -3\sin 3x \end{aligned}\right\}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\cos 3x - 3x^2\sin 3x\) | M1 A1 A1 |
| (3) |
Notes
M1: Applies \(vu' + uv'\) correctly for their \(u, u', v, v'\) AND gives an expression of the form \(\alpha x\cos 3x \pm \beta x^2\sin 3x\)
A1: Any one term correct
A1: Both terms correct and no further simplification to terms in \(\cos\alpha x^2\) or \(\sin\beta x^3\).
| Scheme | Marks |
|---|---|
| \(y = \dfrac{\ln(x^2 + 1)}{x^2 + 1}\) | |
| \(u = \ln(x^2 + 1) \Rightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{2x}{x^2 + 1}\) | M1 A1 |
| Apply quotient rule: \(\left\{\begin{aligned} u &= \ln(x^2 + 1) & v &= x^2 + 1 \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{2x}{x^2 + 1} & \frac{\mathrm{d}v}{\mathrm{d}x} &= 2x \end{aligned}\right\}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\left(\frac{2x}{x^2 + 1}\right)\left(x^2 + 1\right) - 2x\ln(x^2 + 1)}{\left(x^2 + 1\right)^2}\) | M1 A1 |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x - 2x\ln(x^2 + 1)}{\left(x^2 + 1\right)^2}\right\}\) | |
| (4) |
Notes
M1: \(\ln(x^2 + 1) \to \dfrac{\text{something}}{x^2 + 1}\)
A1: \(\ln(x^2 + 1) \to \dfrac{2x}{x^2 + 1}\)
M1: Applying \(\dfrac{vu' - uv'}{v^2}\)
A1: Correct differentiation with correct bracketing but allow recovery.
{Ignore subsequent working.}
| Scheme | Marks |
|---|---|
| \(y = \sqrt{4x + 1},\ x \gt -\tfrac{1}{4}\) | |
| At \(P\), \(y = \sqrt{4(2) + 1} = \underline{\sqrt{9}} = \underline{3}\) | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \underline{\tfrac{1}{2}(4x + 1)^{-\frac{1}{2}}(4)}\) | M1* A1 aef |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{(4x + 1)^{\frac{1}{2}}}\) | |
| At \(P\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{\left(4(2) + 1\right)^{\frac{1}{2}}}\) | M1 |
| Hence \(\mathrm{m}(\mathbf{T}) = \dfrac{2}{3}\) | |
| Either \(\mathbf{T}\): \(y - 3 = \tfrac{2}{3}(x - 2)\); or \(y = \tfrac{2}{3}x + c\) and \(3 = \tfrac{2}{3}(2) + c \Rightarrow c = 3 - \tfrac{4}{3} = \tfrac{5}{3}\); | dM1*; |
| Either \(\mathbf{T}\): \(3y - 9 = 2(x - 2)\); \(\mathbf{T}\): \(3y - 9 = 2x - 4\) \(\mathbf{T}\): \(\underline{2x - 3y + 5 = 0}\) | A1 |
| or \(\mathbf{T}\): \(y = \tfrac{2}{3}x + \tfrac{5}{3}\) \(\mathbf{T}\): \(3y = 2x + 5\) \(\mathbf{T}\): \(\underline{2x - 3y + 5 = 0}\) | |
| (6) | |
| (13 marks) |
Notes
B1: At \(P\), \(y = \underline{\sqrt{9}}\) or 3
M1*: \(\pm k(4x + 1)^{-\frac{1}{2}}\)
A1 aef: \(2(4x + 1)^{-\frac{1}{2}}\)
M1: Substituting \(x = 2\) into an equation involving \(\frac{\mathrm{d}y}{\mathrm{d}x}\);
dM1*: \(y - y_1 = m(x - 2)\) or \(y - y_1 = m(x - \text{their stated } x)\) with ‘their TANGENT gradient’ and their \(y_1\); or uses \(y = mx + c\) with ‘their TANGENT gradient’, their \(x\) and their \(y_1\).
A1: \(\underline{2x - 3y + 5 = 0}\). Tangent must be stated in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.