A2 October 2021 Paper 2 Q3
3.
\[\mathrm{f}(x) = \arcsin x \qquad -1 \leqslant x \leqslant 1\]Give your answer in the form \(\dfrac{p}{q}\) where \(p\) and \(q\) are integers to be determined. (2)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}^\prime(x) = A\left(1 - x^2\right)^{-\frac{1}{2}}\), \(\mathrm{f}^{\prime\prime}(x) = Bx\left(1 - x^2\right)^{-\frac{3}{2}}\) and \(\mathrm{f}^{\prime\prime\prime}(x) = C\left(1 - x^2\right)^{-\frac{3}{2}} + Dx^2\left(1 - x^2\right)^{-\frac{5}{2}}\) or \(\dfrac{C\left(1 - x^2\right)^{\frac{3}{2}} + Dx^2\left(1 - x^2\right)^{\frac{1}{2}}}{\left(1 - x^2\right)^3}\) | M1 | 2.1 |
\(\mathrm{f}^\prime(x) = \left(1 - x^2\right)^{-\frac{1}{2}}\) or \(\dfrac{1}{\sqrt{1 - x^2}}\) \(\mathrm{f}^{\prime\prime}(x) = x\left(1 - x^2\right)^{-\frac{3}{2}}\) or \(\dfrac{x}{\left(1 - x^2\right)^{\frac{3}{2}}}\) and \(\mathrm{f}^{\prime\prime\prime}(x) = \left(1 - x^2\right)^{-\frac{3}{2}} + 3x^2\left(1 - x^2\right)^{-\frac{5}{2}}\) or \(\dfrac{1}{\left(1 - x^2\right)^{\frac{3}{2}}} + \dfrac{3x^2}{\left(1 - x^2\right)^{\frac{5}{2}}}\) from quotient rule \(\dfrac{\left(1 - x^2\right)^{\frac{3}{2}} + 3x^2\left(1 - x^2\right)^{\frac{1}{2}}}{\left(1 - x^2\right)^3}\) | A1 | 1.1b |
| Finds \(\mathrm{f}(0)\), \(\mathrm{f}^\prime(0)\), \(\mathrm{f}^{\prime\prime}(0)\) and \(\mathrm{f}^{\prime\prime\prime}(0)\) and applies the formula \(\mathrm{f}(x) = \mathrm{f}(0) + \mathrm{f}^\prime(0)x + \dfrac{\mathrm{f}^{\prime\prime}(0)}{2}x^2 + \dfrac{\mathrm{f}^{\prime\prime\prime}(0)}{6}x^3\) \(\left\{\mathrm{f}(0) = 0,\ \mathrm{f}^\prime(0) = 1,\ \mathrm{f}^{\prime\prime}(0) = 0,\ \mathrm{f}^{\prime\prime\prime}(0) = 1\right\}\) | M1 | 1.1b |
| \(\mathrm{f}(x) = x + \dfrac{x^3}{6}\) cso | A1 | 1.1b |
| (4) |
Notes
(a)
M1: Finds the correct form of the first three derivatives, may be unsimplified – the third may come later.
A1: Correct first three derivatives, may be unsimplified – the third may come later.
M1: Finds \(\mathrm{f}(0)\), \(\mathrm{f}^\prime(0)\), \(\mathrm{f}^{\prime\prime}(0)\) and \(\mathrm{f}^{\prime\prime\prime}(0)\) and applies to the correct formula, needs to go up to \(x^3\)
A1: \(x + \dfrac{x^3}{6}\) cso ignore any higher terms whether correct or not.
Special case: If they think that their \(\mathrm{f}^{\prime\prime}(0) \neq 0\) then maximum score M1 A0 M1 A0. M1 for correct form of the first two derivatives. M1 Correctly uses their \(\mathrm{f}(0)\), \(\mathrm{f}^\prime(0)\), \(\mathrm{f}^{\prime\prime}(0)\) and applies to the correct formula.
Note: If candidates do not find the first three derivatives but use \(\mathrm{f}(0) = 0\), \(\mathrm{f}^\prime(0) = 1\), \(\mathrm{f}^{\prime\prime}(0) = 0\), \(\mathrm{f}^{\prime\prime\prime}(0) = 1\) and use these correctly in the formula this can score M0 A0 M1 A0
| Scheme | Marks | AO |
|---|---|---|
| \(\arcsin\left(\dfrac{1}{2}\right) = \dfrac{1}{2} + \dfrac{\left(\frac{1}{2}\right)^3}{6} = \dfrac{\pi}{6} \Rightarrow \pi = \ldots\) | M1 | 1.1b |
| \(\pi = \dfrac{25}{8}\) o.e. | A1ft | 2.2b |
| (2) | ||
| (6 marks) |
Notes
(b)
M1: Substitutes \(x = \dfrac{1}{2}\) into both sides and rearranges to find \(\pi = \ldots\)
A1ft: Infers that \(\pi = \dfrac{25}{8}\) o.e. Follow through their \(6\mathrm{f}\left(\dfrac{1}{2}\right)\)