A2 October 2020 Paper 2 Q5
5.
| Scheme | Marks | AO |
|---|---|---|
| \(y = \tan^{-1}x \Rightarrow \tan y = x \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}y} = \sec^2 y\) \(y = \tan^{-1}x \Rightarrow \tan y = x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x}\sec^2 y = 1\) | M1 | 3.1a |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 1 + \tan^2 y\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\left(1 + \tan^2 y\right) = 1\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{1 + \tan^2 y} = \dfrac{1}{1 + x^2}\,*\) | A1* | 2.1 |
| (3) |
Notes
(a)
M1: Makes progress in establishing the derivative by taking the tan of both sides and differentiating with respect to \(y\) or implicitly with respect to \(x\)
M1: Use of the correct identity
A1*: Fully correct proof
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}\left(\tan^{-1}4x\right)}{\mathrm{d}x} = \dfrac{4}{1 + 16x^2}\) | B1 | 1.1b |
| \(\displaystyle\int x\tan^{-1}4x\,\mathrm{d}x = \alpha x^2\tan^{-1}4x - \int \alpha x^2 \times \text{‘}\frac{4}{1 + 16x^2}\text{’}\,\mathrm{d}x\) | M1 | 2.1 |
| \(\displaystyle\int x\tan^{-1}4x\,\mathrm{d}x = \frac{x^2}{2}\tan^{-1}4x - \int \frac{x^2}{2} \times \frac{4}{1 + 16x^2}\,\mathrm{d}x\) | A1 | 1.1b |
| \(\displaystyle = \ldots - \frac{1}{8}\int \frac{16x^2 + 1 - 1}{1 + 16x^2}\,\mathrm{d}x = \ldots - \frac{1}{8}\int \left(1 - \frac{1}{1 + 16x^2}\right)\mathrm{d}x\) or let \(4x = \tan u \Rightarrow \displaystyle\frac{1}{8}\int \frac{\tan^2 u}{1 + \tan^2 u} \times \frac{1}{4}\sec^2 u\,\mathrm{d}u\) \(\displaystyle\Rightarrow \frac{1}{32}\int \tan^2 u\,\mathrm{d}u = \frac{1}{32}\int \left(\sec^2 u - 1\right)\mathrm{d}u\) | M1 | 3.1a |
| \(= \dfrac{x^2}{2}\tan^{-1}4x - \dfrac{1}{8}x + \dfrac{1}{32}\tan^{-1}4x + k\) | A1 | 2.1 |
| (5) |
Notes
(b)
B1: Correct derivative
M1: Uses integration by parts in the correct direction
A1: Correct expression
M1: Adopts a correct strategy for the integration by splitting into two fractions or using a substitution of \(4x = \tan u\) to get to an integrable form
A1: Correct answer
(corrected from the printed mark scheme: \(\displaystyle\frac{1}{32}\int \left(\sec^2 u - 1\right)\mathrm{d}u\) is printed as \(\displaystyle\frac{1}{32}\int \sec^2 u - u\,\mathrm{d}u\))
| Scheme | Marks | AO |
|---|---|---|
| Mean value \(= \left(\dfrac{1}{\frac{\sqrt{3}}{4} - 0}\right)\left[\dfrac{x^2}{2}\tan^{-1}4x - \dfrac{1}{8}x + \dfrac{1}{32}\tan^{-1}4x\right]_0^{\frac{\sqrt{3}}{4}}\) \(= \dfrac{4}{\sqrt{3}}\left(\left(\dfrac{3}{32} \times \dfrac{\pi}{3} - \dfrac{1}{8} \times \dfrac{\sqrt{3}}{4} + \dfrac{1}{32} \times \dfrac{\pi}{3}\right) - 0\right)\) | M1 | 2.1 |
| \(= \dfrac{\sqrt{3}}{72}\left(4\pi - 3\sqrt{3}\right)\) or \(\dfrac{\sqrt{3}}{18}\pi - \dfrac{1}{8}\) oe | A1 | 1.1b |
| (2) | ||
| (10 marks) |
Notes
(c)
M1: Correctly applies the method for the mean value for their integration. The limit of zero can be implied if it comes to 0.
A1: Correct exact answer. Allow exact equivalents e.g. \(\dfrac{4\pi\sqrt{3} - 9}{72},\ \dfrac{\pi\sqrt{3}}{18} - \dfrac{1}{8}\)