A2 June 2020 Paper 2 Q11
11
(a) Starting from the series given in the formulae booklet, show that the general term of the Maclaurin series for\[\frac{\sin x}{x} - \cos x\]
is
\[(-1)^{r+1}\frac{2r}{(2r + 1)!}x^{2r}\][4 marks]
(b) Show that\[\lim_{x \to 0}\left[\frac{\dfrac{\sin x}{x} - \cos x}{1 - \cos x}\right] = \frac{2}{3}\]
[4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Finds the second or third simplified term of the series for \(\dfrac{\sin x}{x}\) or finds the general term | M1 | 1.1a |
| Finds general term of series for \(\dfrac{\sin x}{x}\) with \(x^{2r}\) | M1 | 1.1a |
| Subtracts general terms of \(\dfrac{\sin x}{x}\) and \(\cos x\) | M1 | 1.1a |
| Completes a rigorous argument to show the required result, including \(\dfrac{1}{(2r + 1)!} - \dfrac{1}{(2r)!} = \dfrac{1 - (2r + 1)}{(2r + 1)!} = \dfrac{-2r}{(2r + 1)!}\) AG | R1 | 2.1 |
Typical solution
\[\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots + \frac{(-1)^r x^{2r}}{(2r + 1)!} + \cdots\]\[\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + \frac{(-1)^r x^{2r}}{(2r)!}\]\[\frac{1}{(2r + 1)!} - \frac{1}{(2r)!} = \frac{1 - (2r + 1)}{(2r + 1)!} = \frac{-2r}{(2r + 1)!}\]\[\therefore \frac{(-1)^r x^{2r}}{(2r + 1)!} - \frac{(-1)^r x^{2r}}{(2r)!} = \frac{(-1)^r x^{2r}(-2r)}{(2r + 1)!} = (-1)^{r+1}\frac{2r}{(2r + 1)!}x^{2r}\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to determine the value of the limit by finding the first non-zero term of one series. | M1 | 3.1a |
| Explains or shows that each series has terms in higher powers of \(x\) | E1 | 2.4 |
| Deduces that the required limit can be determined by dividing numerator and denominator by the lowest power of \(x\) or by using l’Hôpital’s rule. | M1 | 2.2a |
| Completes a rigorous argument to show the required result \(\displaystyle \lim_{x \to 0}\left[\frac{\dfrac{\sin x}{x} - \cos x}{1 - \cos x}\right] = \frac{2}{3}\) | R1 | 2.1 |
| (8 marks) |
Typical solution
First non-zero terms of series expansion of \(\dfrac{\sin x}{x} - \cos x\) are \(\dfrac{x^2}{3}\) and \(-\dfrac{x^4}{30}\)
First non-zero terms of series expansion of \(1 - \cos x\) are \(\dfrac{x^2}{2}\) and \(-\dfrac{x^4}{24}\)
\[\lim_{x \to 0}\left[\frac{\dfrac{x^2}{3} - \dfrac{x^4}{30} + \cdots}{\dfrac{x^2}{2} - \dfrac{x^4}{24} + \cdots}\right] = \lim_{x \to 0}\left[\frac{\dfrac{1}{3} - \dfrac{x^2}{30} + \cdots}{\dfrac{1}{2} - \dfrac{x^2}{24} + \cdots}\right] = \frac{1/3}{1/2} = \frac{2}{3}\]