2008 H2 Mathematics Paper 2 Question 1

Maclaurin Series

Answers

(i)

(ii)

{x + x^2 + \frac{1}{3} x^3 + \ldots}

(iii)

(iv)

{\{ x \in \mathbb{R}:}\allowbreak {-1.96 < x < 1.56 \}}

Full solutions

(i)

(ii)

\begin{align*} & \mathrm{e}^x \sin x \\ & = \left( 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 +\ldots \right) \left( x - \frac{1}{6} x^3 +\ldots \right) \\ & = x + x^2 + \left( \frac{1}{2}-\frac{1}{6} \right)x^3 + \ldots \\ & = x + x^2 + \frac{1}{3} x^3 + \ldots \; \blacksquare \end{align*}

(iii)

(iv)

For

{g(x)}

to be within

{\pm 0.5}

of

{f(x)}

-0.5 < f(x)-g(x) < 0.5

Considering the graphs of

{y=f(x)-g(x), \;}

{y=-0.5 }

and

{y=0.5, }

Hence the set of values of

{x}

required:

\{ x \in \mathbb{R}: -1.96 < x < 1.56 \} \; \blacksquare

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