A2 October 2021 Paper 1 Q16
16
(a) Show using exponentials that \(\cosh 2u = 1 + 2\sinh^2 u\). [4]
(b) Show that \(\displaystyle\int_0^2 \frac{x^2}{\sqrt{4 + x^2}}\,\mathrm{d}x = 2\sqrt{2} - 2\ln\left(1 + \sqrt{2}\right)\). [10]
| Scheme | Marks | AO |
|---|---|---|
| \(\text{LHS} = \dfrac{\mathrm{e}^{2u} + \mathrm{e}^{-2u}}{2}\) | B1 | 2.1 |
| \(\text{RHS} = 1 + 2\left(\dfrac{\mathrm{e}^u - \mathrm{e}^{-u}}{2}\right)^2\) | B1 | 2.1 |
| \(= 1 + 2\left(\dfrac{\mathrm{e}^{2u} - 2 + \mathrm{e}^{-2u}}{4}\right)\) | B1 | 2.1 |
| \(= 1 + \left(\dfrac{\mathrm{e}^{2u} - 2 + \mathrm{e}^{-2u}}{2}\right) = \dfrac{\mathrm{e}^{2u} + \mathrm{e}^{-2u}}{2} = \text{LHS}\) | B1 | 2.2a |
| [4] |
| Scheme | Marks | AO |
|---|---|---|
| Let \(x = 2\sinh u \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}u} = 2\cosh u\) | M1 | 3.1a |
| \(\displaystyle\int_0^2 \frac{x^2}{\sqrt{4 + x^2}}\,\mathrm{d}x = \displaystyle\int_0^{\operatorname{arsinh}1} \frac{4\sinh^2 u}{\sqrt{4 + 4\sinh^2 u}}\,2\cosh u\,\mathrm{d}u\) | A1 | 1.1 |
| \(= \displaystyle\int_0^{\operatorname{arsinh}1} \frac{4\sinh^2 u}{2\cosh u}\,2\cosh u\,\mathrm{d}u\) | M1 | 2.1 |
| \(= \displaystyle\int_0^{\operatorname{arsinh}1} 4\sinh^2 u\,\mathrm{d}u\) | A1 | 2.1 |
| \(= \displaystyle\int_0^{\operatorname{arsinh}1} (2\cosh 2u - 2)\,\mathrm{d}u\) | M1 | 2.1 |
| \(= \left[\sinh 2u - 2u\right]_0^{\operatorname{arsinh}1}\) | A1 | 2.1 |
| \(\operatorname{arsinh}1 = \ln\left(1 + \sqrt{2}\right)\) | B1 | 2.1 |
| \(\Rightarrow \frac{1}{2}\left(\mathrm{e}^{2\ln(1 + \sqrt{2})} - \mathrm{e}^{-2\ln(1 + \sqrt{2})}\right) - 2\ln\left(1 + \sqrt{2}\right)\) \(= \frac{1}{2}\left[\left(\mathrm{e}^{\ln(1 + \sqrt{2})}\right)^2 - \left(\mathrm{e}^{\ln(1 + \sqrt{2})}\right)^{-2}\right] - 2\ln\left(1 + \sqrt{2}\right)\) | M1 | 2.1 |
| \(= \frac{1}{2}\left[\left(1 + \sqrt{2}\right)^2 - \left(1 + \sqrt{2}\right)^{-2}\right] - 2\ln\left(1 + \sqrt{2}\right)\) \(\left(1 + \sqrt{2}\right)^{-1} = \dfrac{1}{1 + \sqrt{2}} = \dfrac{1 - \sqrt{2}}{\left(1 + \sqrt{2}\right)\left(1 - \sqrt{2}\right)} = \sqrt{2} - 1\) | M1 | 2.1 |
| giving \(= \frac{1}{2}\left[\left(3 + 2\sqrt{2}\right) - \left(3 - 2\sqrt{2}\right)\right] - 2\ln\left(1 + \sqrt{2}\right)\) \(= 2\sqrt{2} - 2\ln\left(1 + \sqrt{2}\right)\) | A1 | 2.2a |
| [10] |
Notes
M1: \(1 + \sinh^2 u = \cosh^2 u\) used (third mark)
M1: \(\cosh 2u = 1 + 2\sinh^2 u\) used (fifth mark)
B1: soi
For the last two M1 marks: or \(\sinh 2u = 2\sinh u\cosh u\) M1
\(= 2\sinh u\sqrt{1 + \sinh^2 u}\) M1
\(= 2\sqrt{2}\)
A1: AG
Alternative for last 6 marks
| Scheme | Marks |
|---|---|
| \(= \displaystyle\int_0^{\operatorname{arsinh}1} \left(\mathrm{e}^u - \mathrm{e}^{-u}\right)^2\,\mathrm{d}u\) \(= \displaystyle\int_0^{\operatorname{arsinh}1} \left(\mathrm{e}^{2u} - 2 + \mathrm{e}^{-2u}\right)\mathrm{d}u\) | M1 |
| \(= \left[\frac{1}{2}\mathrm{e}^{2u} - 2u - \frac{1}{2}\mathrm{e}^{-2u}\right]_0^{\operatorname{arsinh}1}\) | A1 |
| \(\operatorname{arsinh}1 = \ln\left(1 + \sqrt{2}\right)\) | B1 |
| \(\mathrm{e}^{2\ln(1 + \sqrt{2})} = \left(1 + \sqrt{2}\right)^2\) | M1 |
| \(\mathrm{e}^{-2\ln(1 + \sqrt{2})} = \left(1 + \sqrt{2}\right)^{-2} = \left(\sqrt{2} - 1\right)^2\) | M1 |
| giving \(\dfrac{3 + 2\sqrt{2}}{2} - \dfrac{3 - 2\sqrt{2}}{2} - 2\ln\left(1 + \sqrt{2}\right)\) \(= 2\sqrt{2} - 2\ln\left(1 + \sqrt{2}\right)\) | A1 |
| [10] |
M1: \(\sinh u = \frac{1}{2}\left(\mathrm{e}^u - \mathrm{e}^{-u}\right)\) used
A1: AG