A2 October 2020 Paper 2 Q10
10 Let \(\mathrm{f}(x) = \sin^{-1}(x)\).
(a)
(i) Determine \(\mathrm{f}''(x)\). [2]
(ii) Determine the first two non-zero terms of the Maclaurin expansion for \(\mathrm{f}(x)\). [3]
(iii) By considering the first two non-zero terms of the Maclaurin expansion for \(\mathrm{f}(x)\), find an approximation to \(\displaystyle\int_0^{\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x\). Give your answer correct to 6 decimal places. [2]
(b) By writing \(\mathrm{f}(x)\) as \(\sin^{-1}(x) \times 1\), determine the value of \(\displaystyle\int_0^{\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x\). Give your answer in exact form. [3]
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{f}'(x) = \dfrac{1}{\left(1 - x^2\right)^{\frac{1}{2}}}\) from the formula book | M1 | 1.1 |
| so \(\mathrm{f}''(x) = -\frac{1}{2}.\dfrac{1}{\left(1 - x^2\right)^{\frac{3}{2}}}.(-2x)\) | ||
| \(= \dfrac{x}{\left(1 - x^2\right)^{\frac{3}{2}}}\) | A1 | 1.1 |
| [2] | ||
| (ii) \(\mathrm{f}(0) = 0,\ \mathrm{f}'(0) = 1\) and \(\mathrm{f}''(0) = 0\) | B1 | 1.1 |
| \(\mathrm{f}'''(x) = \dfrac{(1 - x^2)^{\frac{3}{2}} - x.\frac{3}{2}(1 - x^2)^{\frac{1}{2}}.(-2x)}{(1 - x^2)^3}\) | M1 | 3.1a |
| so \(\mathrm{f}'''(0) = 1\) and \(\mathrm{f}(x) = x + \frac{1}{6}x^3 + \ldots\) | A1 | 2.1 |
| [3] | ||
| (iii) \(\displaystyle\int_0^{\frac{1}{2}} \mathrm{f}(x)\,\mathrm{d}x \approx \int_0^{\frac{1}{2}} x + \tfrac{1}{6}x^3\,\mathrm{d}x\) | M1 | 1.1 |
| \(= 0.127604167\ldots\) \(= 0.127604\) to 6 dp | A1 | 1.1 |
| [2] |
Notes
(a)(i)
M1: Formula from the Formula Booklet and attempt differentiation. To within sign error
(a)(ii)
B1: or \(a_0 = 0\), \(a_1 = 1\) and \(a_2 = 0\). Ignore sign error in \(\mathrm{f}''(x)\)
M1: Differentiate and simplify far enough to be able to justify value 1. Either full derivative or “zero term” denoted as such
A1: Condone 3! In place of 6. Not BC. If M0 then SC1 for correct expansion
(a)(iii)
M1: Integral of their 2 term cubic with limits
A1: Could be BC
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int 1 \times \sin^{-1}x\,\mathrm{d}x = x\sin^{-1}x - \int \dfrac{x}{\sqrt{1 - x^2}}\,\mathrm{d}x\) | M1 | 3.1a |
| \(= x\sin^{-1}x + \left(1 - x^2\right)^{\frac{1}{2}}\) (+c) | A1 | 1.1 |
| \(\displaystyle\int_0^{\frac{1}{2}} \mathrm{f}(x) = \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1\) | A1 | 1.1 |
| [3] |
Notes
M1: Attempt integration by parts. ignore limits. Formula for parts must be correct