A2 June 2019 Paper 2 Q10
10
| Scheme | Marks | AO |
|---|---|---|
| \(f(0) = \ln\left(\frac{1}{2} + \cos 0\right) = \ln\left(\frac{3}{2}\right)\) | B1 | 1.1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\ln\left(\dfrac{1}{2} + \cos x\right)\right) = \dfrac{-\sin x}{\frac{1}{2} + \cos x} \Rightarrow f'(0) = 0\) | M1 | 3.1a |
| \(\dfrac{\mathrm{d}^2\ln\left(\frac{1}{2} + \cos x\right)}{\mathrm{d}x^2} = \dfrac{-\cos x\left(\frac{1}{2} + \cos x\right) + \sin x(-\sin x)}{\left(\frac{1}{2} + \cos x\right)^2}\) \(\ldots \Rightarrow f''(0) = -\frac{2}{3}\) | A1 | 1.1 |
| \(\ln\left(\dfrac{1}{2} + \cos x\right) = \ln\left(\dfrac{3}{2}\right) - \dfrac{x^2}{3} + \ldots\) | A1 | 1.1 |
| [4] |
Notes
B1: soi
M1: Differentiating using chain rule (or rule for \(\ln(f(x))\)) and evaluating when \(x = 0\). Allow sign error in numerator
A1: Differentiating again using quotient (or product/chain) rule. NB Simplifies to \(-\dfrac{\frac{1}{2}\cos x + 1}{\left(\frac{1}{2} + \cos x\right)^2}\)
A1: www
If zero scored then SC1 for correct expansion
| Scheme | Marks | AO |
|---|---|---|
| \(\ln\left(\dfrac{1}{2} + \cos x\right) = 0 \Rightarrow x = \dfrac{\pi}{3}\) (or \(-\dfrac{\pi}{3}\)) | B1 | 1.1 |
| \(\therefore \ln\left(\dfrac{3}{2}\right) - \dfrac{\left(\frac{\pi}{3}\right)^2}{3} \approx 0\) | M1 | 3.1a |
| \(\ln\left(\dfrac{3}{2}\right) - \dfrac{\pi^2}{27} \approx 0 \Rightarrow \pi \approx \sqrt{27\ln\left(\dfrac{3}{2}\right)} = 3\sqrt{3\ln\left(\dfrac{3}{2}\right)}\) | A1 | 3.2a |
| [3] |
Notes
B1: Finding either \(\pm\pi/3\) as a root. Allow \(60^\circ\) for B1. Ignore other roots
M1: Substituting their root, in radians, into their Maclaurin series and equating (approximately) to 0.
Or equating their expression (approximately) to 0 and rearranging for \(x\): \(\ln\left(\frac{3}{2}\right) - \frac{x^2}{3} \approx 0 \Rightarrow x \approx \sqrt{3\ln\left(\frac{3}{2}\right)}\)
\(\frac{\pi}{3} \approx \sqrt{3\ln\left(\frac{3}{2}\right)} \Rightarrow \pi \approx 3\sqrt{3\ln\left(\frac{3}{2}\right)}\)
A1: Could see \(\pm\) but must be removed by final conclusion. Must use approximately equals symbol (not just equals symbol)