2018 H2 Maths Paper 2 Q4 — Worked Solution

Part (i)

9 marks

Approach

Rather than differentiating $\ln\left( \cos 2x \right)$ repeatedly, it is quicker to substitute the known Maclaurin series for $\cos x$ (from MF26) into $\cos 2x$ , then substitute that into the known series for $\ln(1 + v)$ , keeping terms only up to $x^{6}$ .

From MF26, $\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \ldots$ , so

\cos 2x

= 1 - \frac{(2x)^{2}}{2!} + \frac{(2x)^{4}}{4!} - \frac{(2x)^{6}}{6!} + \ldots

= 1 - 2x^{2} + \frac{2}{3}x^{4} - \frac{4}{45}x^{6} + \ldots

Let $v = - 2x^{2} + \frac{2}{3}x^{4} - \frac{4}{45}x^{6}$ , so that $\cos 2x = 1 + v$ . From MF26, $\ln(1 + v) = v - \frac{v^{2}}{2} + \frac{v^{3}}{3} - \ldots$ . Keeping terms up to $x^{6}$ ,

v^{2} = \left( - 2x^{2} + \frac{2}{3}x^{4} \right)^{2} + O\left( x^{8} \right) = 4x^{4} - \frac{8}{3}x^{6} + O\left( x^{8} \right)

v^{3} = \left( - 2x^{2} \right)^{3} + O\left( x^{8} \right) = - 8x^{6} + O\left( x^{8} \right)

So

\ln\left( \cos 2x \right)

= v - \frac{v^{2}}{2} + \frac{v^{3}}{3}

= \left( - 2x^{2} + \frac{2}{3}x^{4} - \frac{4}{45}x^{6} \right) - \frac{1}{2}\left( 4x^{4} - \frac{8}{3}x^{6} \right) + \frac{1}{3}\left( - 8x^{6} \right)

= - 2x^{2} + \left( \frac{2}{3} - 2 \right)x^{4} + \left( - \frac{4}{45} + \frac{4}{3} - \frac{8}{3} \right)x^{6}

= - 2x^{2} - \frac{4}{3}x^{4} - \frac{64}{45}x^{6}

$\therefore$ $\ln\left( \cos 2x \right) \approx - 2x^{2} - \frac{4}{3}x^{4} - \frac{64}{45}x^{6}$ .

Since $\ln\left( \cos 2x \right)$ is only defined where $\cos 2x > 0$ , and on $0 \leq x \leq \frac{1}{4}\pi$ we have $2x \in \left\lbrack 0,\frac{\pi}{2} \right\rbrack$ with $\cos 2x = 0$ only at $x = \frac{1}{4}\pi$ ,

$\therefore$ the expansion is not valid at $x = \frac{\pi}{4}$ .

Part (ii)

Dividing the expansion by $x^{2}$ ,

\frac{\ln\left( \cos 2x \right)}{x^{2}} \approx - 2 - \frac{4}{3}x^{2} - \frac{64}{45}x^{4}

Integrating term by term,

\int\frac{\ln\left( \cos 2x \right)}{x^{2}}\, dx

\approx \int\left( - 2 - \frac{4}{3}x^{2} - \frac{64}{45}x^{4} \right)dx

= - 2x - \frac{4}{9}x^{3} - \frac{64}{225}x^{5} + C

Using this approximate antiderivative between $x = 0$ and $x = 0.5$ (the series value at $x = 0$ is $0$ , matching the removable behaviour of $\frac{\ln\left( \cos 2x \right)}{x^{2}}$ there),

\int_{0}^{0.5}\frac{\ln\left( \cos 2x \right)}{x^{2}}\, dx

\approx \left\lbrack - 2x - \frac{4}{9}x^{3} - \frac{64}{225}x^{5} \right\rbrack_{0}^{0.5}

= - 1 - \frac{1}{18} - \frac{2}{225}

= - \frac{479}{450}

$\therefore$ $\int_{0}^{0.5}\frac{\ln\left( \cos 2x \right)}{x^{2}}\, dx \approx - \frac{479}{450} = - 1.0644$ (4 d.p.).

Part (iii)

1 mark

By GC, $\int_{0}^{0.5}\frac{\ln\left( \cos 2x \right)}{x^{2}}\, dx = - 1.0670$ (4 d.p.).

Note: this GC value is obtained directly from the original function (the GC evaluates the improper-looking integral fine, since $\frac{\ln\left( \cos 2x \right)}{x^{2}} \rightarrow - 2$ as $x \rightarrow 0$ ), and is close to but not identical to the series-based approximation in (ii), since (ii) drops terms in $x^{8}$ and higher.

2018 H2 Maths Paper 2, Question 4

Part (i)

Part (ii)

Part (iii)

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