C3 January 2011 Q7
7. The curve \(C\) has equation
\[y = \frac{3 + \sin 2x}{2 + \cos 2x}\](a) Show that\[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6\sin 2x + 4\cos 2x + 2}{(2 + \cos 2x)^2}\] (4)
(b) Find an equation of the tangent to \(C\) at the point on \(C\) where \(x = \dfrac{\pi}{2}\). Write your answer in the form \(y = ax + b\), where \(a\) and \(b\) are exact constants. (4)
| Scheme | Marks |
|---|---|
| \(y = \dfrac{3 + \sin 2x}{2 + \cos 2x}\) | |
| Apply quotient rule: \(\left\{\begin{aligned} u &= 3 + \sin 2x & v &= 2 + \cos 2x \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= 2\cos 2x & \frac{\mathrm{d}v}{\mathrm{d}x} &= -2\sin 2x \end{aligned}\right\}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\cos 2x(2 + \cos 2x) - -2\sin 2x(3 + \sin 2x)}{(2 + \cos 2x)^2}\) | M1 A1 A1 |
| \(= \dfrac{4\cos 2x + 2\cos^2 2x + 6\sin 2x + 2\sin^2 2x}{(2 + \cos 2x)^2}\) | |
| \(= \dfrac{4\cos 2x + 6\sin 2x + 2(\cos^2 2x + \sin^2 2x)}{(2 + \cos 2x)^2}\) | |
| \(= \dfrac{4\cos 2x + 6\sin 2x + 2}{(2 + \cos 2x)^2}\) (as required) | A1* |
| (4) |
Notes
M1: Applying \(\dfrac{vu' - uv'}{v^2}\)
A1: Any one term correct on the numerator
A1: Fully correct (unsimplified).
A1*: For correct proof with an understanding that \(\cos^2 2x + \sin^2 2x = 1\). No errors seen in working.
| Scheme | Marks |
|---|---|
| When \(x = \frac{\pi}{2}\), \(y = \dfrac{3 + \sin\pi}{2 + \cos\pi} = \dfrac{3}{1} = 3\) | B1 |
| At \(\left(\frac{\pi}{2}, 3\right)\), \(\mathrm{m}(\mathbf{T}) = \dfrac{6\sin\pi + 4\cos\pi + 2}{(2 + \cos\pi)^2} = \dfrac{-4 + 2}{1^2} = -2\) | B1 |
| Either \(\mathbf{T}\): \(y - 3 = -2\left(x - \frac{\pi}{2}\right)\) or \(y = -2x + c\) and \(3 = -2\left(\frac{\pi}{2}\right) + c \Rightarrow c = 3 + \pi\); | M1 |
| \(\mathbf{T}\): \(y = -2x + (\pi + 3)\) | A1 |
| (4) | |
| (8 marks) |
Notes
B1: \(y = 3\)
B1: \(\mathrm{m}(\mathbf{T}) = -2\)
M1: \(y - y_1 = m\left(x - \frac{\pi}{2}\right)\) with ‘their TANGENT gradient’ and their \(y_1\); or uses \(y = mx + c\) with ‘their TANGENT gradient’;
A1: \(y = -2x + \pi + 3\)