A2 October 2020 Paper 1 Q8
8
(a) Using exponentials, show that \(\cosh 2u \equiv 2\sinh^2 u + 1\). [2]
(b) By differentiating both sides of the identity in part (a) with respect to \(u\), show that
\(\sinh 2u \equiv 2\sinh u\cosh u\). [1]
\(\sinh 2u \equiv 2\sinh u\cosh u\). [1]
(c) Use the substitution \(x = \sinh^2 u\) to find \(\displaystyle\int \sqrt{\frac{x}{x + 1}}\,\mathrm{d}x\). Give your answer in the form \(a\sinh^{-1} b\sqrt{x} + \mathrm{f}(x)\) where \(a\) and \(b\) are integers and \(\mathrm{f}(x)\) is a function to be determined. [5]
(d) Hence determine the exact area of the region between the curve \(y = \sqrt{\dfrac{x}{x + 1}}\), the \(x\)-axis, the line \(x = 1\) and the line \(x = 2\). Give your answer in the form \(p + q\ln r\) where \(p\), \(q\) and \(r\) are numbers to be determined. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(2\sinh^2 u + 1 \equiv 2\left(\dfrac{\mathrm{e}^u - \mathrm{e}^{-u}}{2}\right)^2 + 1\) | M1 | 2.1 |
| \(= \dfrac{\mathrm{e}^{2u} - 2 + \mathrm{e}^{-2u}}{2} + 1 \equiv \dfrac{\mathrm{e}^{2u} + \mathrm{e}^{-2u}}{2} \equiv \cosh 2u\) AG | A1 | 2.1 |
| [2] |
Notes
M1: Use of exponential form for \(\sinh u\)
| Scheme | Marks | AO |
|---|---|---|
| \(\cosh 2u \equiv 2\sinh^2 u + 1\) \(\Rightarrow 2\sinh 2u \equiv 4\sinh u\cosh u\) \(\Rightarrow \sinh 2u \equiv 2\sinh u\cosh u\) AG | B1 | 2.1 |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \(x = \sinh^2 u \Rightarrow \mathrm{d}x = 2\sinh u\cosh u\ \mathrm{d}u\) \(\Rightarrow \displaystyle\int \sqrt{\frac{x}{x + 1}}\,\mathrm{d}x = \int \sqrt{\frac{\sinh^2 u}{\sinh^2 u + 1}}\,2\sinh u\cosh u\ \mathrm{d}u\) | M1 | 3.1a |
| \(= 2\displaystyle\int \sinh^2 u\,\mathrm{d}u\) | A1 | 1.1 |
| \(= \displaystyle\int (\cosh 2u - 1)\,\mathrm{d}u\) | M1 | 1.1a |
| \(= \dfrac{1}{2}\sinh 2u - u + c = \sinh u\cosh u - u + c\) | A1 | 1.1 |
| \(= \sqrt{x(1 + x)} - \sinh^{-1}\sqrt{x} + c\) So \(\mathrm{f}(x) = \sqrt{x(1 + x)} + c,\ a = -1,\ b = 1\) | A1 | 1.1 |
| [5] |
Notes
M1: Attempt to find \(\dfrac{\mathrm{d}x}{\mathrm{d}u}\)
M1: Use double angle formulae and attempt to integrate.
A1: Ignore \(c\).
A1: \(c\) must be included here as part of \(\mathrm{f}(x)\) – allow \(a\) and \(b\) not being stated explicitly but \(\mathrm{f}(x)\) must be [the guidance ends here in the printed mark scheme]
Alternative method
| Scheme | Marks | AO |
|---|---|---|
| \(= 2\displaystyle\int \left(\frac{\mathrm{e}^u - \mathrm{e}^{-u}}{2}\right)^2\mathrm{d}u = \tfrac{1}{2}\int \mathrm{e}^{2u} - 2 + \mathrm{e}^{-2u}\,\mathrm{d}u\) | M1 | 1.1a |
| \(= \tfrac{1}{4}\mathrm{e}^{2u} - \tfrac{1}{4}\mathrm{e}^{-2u} - u + c = \tfrac{1}{2}\sinh 2u - u + c\) | A1 | 1.1 |
| \(= \sqrt{x(1 + x)} - \sinh^{-1}\sqrt{x} + c\) So \(\mathrm{f}(x) = \sqrt{x(1 + x)} + c,\ a = -1,\ b = 1\) | A1 | 1.1 |
| [5] |
M1: Use exponentials and attempt to integrate.
A1: Ignore \(c\).
A1: \(c\) must be included here as part of \(\mathrm{f}(x)\) – allow \(a\) and \(b\) not being stated explicitly but \(\mathrm{f}(x)\) must be [the guidance ends here in the printed mark scheme]
| Scheme | Marks | AO |
|---|---|---|
| Area \(= \left[\sqrt{x(1 + x)} - \sinh^{-1}\sqrt{x}\right]_1^2\) \(= \left(\sqrt{6} - \ln(\sqrt{2} + \sqrt{3})\right) - \left(\sqrt{2} - \ln(1 + \sqrt{2})\right)\) \(= \sqrt{6} - \sqrt{2} + \ln\left(\dfrac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}}\right)\) | M1 | 1.1 |
| So \(p = \sqrt{6} - \sqrt{2},\ q = 1,\ r = \dfrac{1 + \sqrt{2}}{\sqrt{2} + \sqrt{3}}\) | A1 | 1.1 |
| [2] |
Notes
M1: Correct limits substituted and subtracted into their answer to (c) soi
A1: \(p\), \(q\), \(r\) must be stated