2019 H2 Maths Paper 2, Question 4

Let $f(x) = \sec 2x$. Differentiating using the chain rule, with $\frac{d}{du}\left( \sec u \right) = \sec u\tan u$ and $u = 2x$,

Differentiating again, using the product rule on $2\sec 2x\tan 2x$,

At $x = 0$: $\sec 0 = 1$ and $\tan 0 = 0$, so

Using the Maclaurin expansion $f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2}$,

By GC,

The estimate from (ii) and the calculator value from (iii) round to the same $0.02001$ to 5 decimal places; the unrounded values agree to about 8 significant figures, differing only by roughly $2 \times 10^{- 9}$.

This is because the Maclaurin series was truncated after the $x^{2}$ term, so the error in the approximation comes from the next term of the series, $\frac{10}{3}x^{4}$; integrating this over $\lbrack 0,\ 0.02\rbrack$ contributes only $\frac{2}{3}(0.02)^{5} \approx 2.13 \times 10^{- 9}$ to the integral, which is negligible. Since $x = 0.02$ is very close to $0$, i.e. close to the point about which the series was expanded, the truncated series gives an excellent approximation to the true value.

$g(x) = cosec2x = \frac{1}{\sin 2x}$ is not defined at $x = 0$, since $\sin 0 = 0$ makes the denominator $0$; in fact $g(x) \rightarrow \pm \infty$ as $x \rightarrow 0$. A Maclaurin series requires $g(0)$ (and $g'(0)$, $g''(0)$, …) to exist, so no Maclaurin series for $g$ can be found.