A2 June 2019 Paper 1 Q6
6 You are given that \(y = \tan^{-1}\sqrt{2x}\).
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). [2]
(b) Show that \(\displaystyle\int_{\frac{1}{6}}^{\frac{1}{2}} \frac{\sqrt{x}}{(x + 2x^2)}\,\mathrm{d}x = k\pi\) where \(k\) is a number to be determined in exact form. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(= \dfrac{1}{1 + \left(\sqrt{2x}\right)^2} \times \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{2x}\right)\) oe | M1 | 1.1a |
| \(= \dfrac{1}{1 + 2x} \times \dfrac{\sqrt{2}}{2\sqrt{x}} = \dfrac{1}{1 + 2x} \times \dfrac{1}{\sqrt{2x}}\) | A1 | 1.1 |
| [2] |
Notes
M1: Attempt to differentiate using chain rule, i.e. product of 2 terms
Alternative method
| Scheme | Marks |
|---|---|
| \(\tan y = \sqrt{2x}\) \(\Rightarrow \sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{2}}{2\sqrt{x}} \Rightarrow (1 + \tan^2 y)\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{2}}{2\sqrt{x}}\) | M1 |
| \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + 2x} \times \dfrac{\sqrt{2}}{2\sqrt{x}}\) | A1 |
| [2] |
M1: Make a substitution
(corrected from the printed mark scheme: the printed alternative has \((1 + \tan^2 x)\), which should be \((1 + \tan^2 y)\), and shows a total of [4] for this 2-mark part.)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int_{\frac{1}{6}}^{\frac{1}{2}} \frac{1}{(1 + 2x)\sqrt{x}}\,\mathrm{d}x = \sqrt{2}\int_{\frac{1}{6}}^{\frac{1}{2}} \frac{1}{(1 + 2x)\sqrt{2x}}\,\mathrm{d}x\) | M1 A1 | 3.1a 1.1 |
| \(= \sqrt{2}\Big[\tan^{-1}\sqrt{2x}\Big]_{\frac{1}{6}}^{\frac{1}{2}} = \sqrt{2}\left(\tan^{-1}1 - \tan^{-1}\dfrac{1}{\sqrt{3}}\right)\) | M1 | 1.1 |
| \(= \sqrt{2}\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right) = \dfrac{\sqrt{2}}{12}\pi\) So \(k = \dfrac{\sqrt{2}}{12}\) | A1 | 1.1 |
| [4] |
Notes
M1: Get into form of (a). Ignore limits
A1: Correct form
M1: Use (a) and correct limits in correct order.
A1: oe
Alternative method
| Scheme | Marks |
|---|---|
| Let \(u = \sqrt{x}\) \(\mathrm{d}u = \dfrac{1}{2\sqrt{x}}\mathrm{d}x \Rightarrow \mathrm{d}x = 2\sqrt{x}\,\mathrm{d}u = 2u\,\mathrm{d}u\) | M1 |
| \(\displaystyle\int_{\frac{1}{6}}^{\frac{1}{2}} \frac{\sqrt{x}}{(x + 2x^2)}\,\mathrm{d}x = \int_{x=\frac{1}{6}}^{x=\frac{1}{2}} \frac{u}{(u^2 + 2u^4)}2u\,\mathrm{d}u = 2\int_{x=\frac{1}{6}}^{x=\frac{1}{2}} \frac{1}{(1 + 2u^2)}\,\mathrm{d}u\) | A1 |
| \(= \sqrt{2}\Big[\tan^{-1}u\sqrt{2}\Big]_{x=\frac{1}{6}}^{x=\frac{1}{2}}\) | M1 |
| \(= \sqrt{2}\Big[\tan^{-1}\sqrt{2x}\Big]_{x=\frac{1}{6}}^{x=\frac{1}{2}} = \sqrt{2}\left(\tan^{-1}1 - \tan^{-1}\dfrac{1}{\sqrt{3}}\right) = \sqrt{2}\left(\dfrac{\pi}{4} - \dfrac{\pi}{6}\right)\) \(= \dfrac{\pi\sqrt{2}}{12}\) | A1 |
| [4] |
M1: Make a substitution
A1: Get into correct form
M1: Use standard result with correct limits in correct order