June Revision Exercise 4 Q7

By KS Teng

(i)

f(x) = \frac{4}{(1-3)\sqrt{x^2+4}}

= 4(1-3x)^{-1} (4)^{-\frac{1}{2}} (1 + \frac{x^2}{4})^{-\frac{1}{2}}

= 2[1 + 3x + (3x)^2 + (3x)^3 + \ldots][1 - \frac{1}{2} \frac{x^2}{4} + \ldots]

= 2[1 + 3x + (-\frac{1}{8}+9)x^2 + (-\frac{3}{8}+27)x^3 + \ldots]

= 2 + 6x + \frac{71}{4}x^2 + \frac{213}{4}x^2 + \ldots

|3x| \textless 1 \text{ and } |\frac{x^2}{4}| \textless 1

\Rightarrow \text{Range of values} = (-\frac{1}{3}, \frac{1}{3})

2 \mathrm{tan} y = f(x)

2 \mathrm{sec}^2 y \frac{dy}{dx} = f'(x)

2[2\mathrm{sec}^2 y \mathrm{tan}y (\frac{dy}{dx})^2 + \mathrm{sec}^2 y \frac{d^2y}{dx^2}] = f''(x)

When x = 0, f(0) = 2, f'(0)=6, f''(0)=\frac{71}{2}

2 \mathrm{tan}y=2 \Rightarrow y= \frac{\pi}{4}

2(2)\frac{dy}{dx}=6 \Rightarrow \frac{dy}{dx}=\frac{3}{2}

2[2(2)(1)(\frac{3}{2})^2 + 2 \frac{d^2y}{dx^2}]= \frac{71}{2} \Rightarrow = \frac{35}{8}

Hence y = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{8}\frac{x^2}{2!}+ \ldots = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{16}x^2 + \ldots

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