2018 H2 Mathematics Paper 2 Question 4

Maclaurin Series

Answers

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

The expansion is not valid for

{x=\frac{\pi}{4}}

{-1.0644}

{-1.0670}

\begin{align*} & \ln (\cos 2x) \\ & = \ln \left( 1 - 2 x^2 + \frac{2}{3} x^4 - \frac{4}{45} x^6 + \ldots \right) \\ & = \left( - 2 x^2 + \frac{2}{3} x^4 - \frac{4}{45} x^6 \right) \\ & \phantom{= \Big(-} - \frac{\left( - 2 x^2 + \frac{2}{3} x^4 - \frac{4}{45} x^6 \right)^2}{2} \\ & \phantom{= \Big(- -} + \frac{\left( - 2 x^2 + \frac{2}{3} x^4 - \frac{4}{45} x^6 \right)^3}{3} + \ldots \\ & = - 2 x^2 + \frac{2}{3} x^4 - \frac{4}{45} x^6 - \frac{4x^4 -\frac{8}{3}x^6}{2} - \frac{8x^6}{3} + \ldots \\ &= - 2 x^2 - \frac{4}{3} x^4 - \frac{64}{45} x^6 + \ldots \; \blacksquare \end{align*}

Since

{\ln X}

is undefined for

{X=0}

, the expansion is not valid if

\begin{align*} \cos 2x &= 0 \\ 2x &= \frac{\pi}{2} \\ x &= \frac{\pi}{4} \; \blacksquare \end{align*}

\begin{align*} & \int_0^{0.5} \frac{\ln(\cos 2x)}{x^2} \; \mathrm{d}x \\ &\approx \int_0^{0.5} \frac{- 2 x^2 - \frac{4}{3} x^4 - \frac{64}{45} x^6}{x^2} \; \mathrm{d}x \\ &= \int_0^{0.5} - 2 - \frac{4}{3} x^2 - \frac{64}{45} x^4 \; \mathrm{d}x \\ &= \left[ - 2 x - \frac{4}{9} x^3 - \frac{64}{225} x^5 \right]_0^{0.5} \\ &\approx -1.0644 \textrm{ (4 dp)} \; \blacksquare \end{align*}

Using a GC,

\begin{gather*} \int_0^{0.5} \frac{\ln(\cos 2x)}{x^2} \; \mathrm{d}x \\ \approx -1.0670 \textrm{ (4 dp)} \; \blacksquare \end{gather*}