2019 H2 Mathematics Paper 2 Question 4

Maclaurin Series

Answers

{f'(x)=2 \sec 2x \tan 2x}

f''(x)\allowbreak{=4\sec 2x \tan^2 2x} \allowbreak {+ 4 \sec^3 2x}

{0.02001}

{0.02001}

Since

{0 \leq x \leq 0.02, }

the value of

{x}

is small so the terms

{x^3}

and above omitted in the Maclaurin series in (i) and (ii) are negligible.

Hence the answers to part (ii) and (iii) are close to each other and the approximation is accurate.

{g(x) = \cosec 2x = \frac{1}{\sin 2x}}

{g(0)}

is undefined. Hence a Maclaurin series cannot be found.

\begin{align*} f(x) &= \sec 2x \\ f'(x) &= 2 \sec 2x \tan 2x \; \blacksquare \\ f''(x) &= 4 \sec 2x \tan^2 2x + 4 \sec^3 2x \; \blacksquare \end{align*}

\begin{align*} f(0) &= \sec 2(0) \\ &= 1 \\ f'(0) &= 2 \sec 2(0) \tan 2(0) \\ &= 0 \\ f''(0) &= 0 + 4 \sec^3 2(0) \\ &= 4\end{align*}

Maclaurin series for

{f(x):}

\begin{align*} f(x) &= 1 + 0x + \frac{4}{2!}x^2 + \ldots \\ &= 1 + 2x^2 + \ldots \; \blacksquare \end{align*}

\begin{align*} & \int_0^{0.02} \sec 2x \; \mathrm{d}x \\ & \approx \int_0^{0.02} 1 + 2x^2 \; \mathrm{d}x \\ & = \left[ x + \frac{2x^3}{3} \right]_0^{0.02} \\ & = 0.02001 \textrm{ (5 dp)} \; \blacksquare \end{align*}

Using a GC,

\begin{align*} & \int_0^{0.02} \sec 2x \; \mathrm{d}x \\ & \approx 0.02001 \textrm{ (5 dp)} \; \blacksquare \end{align*}