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Manipal Entrance Test 2014 Math Paper

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Question : 34 of 80

Marks: +1, -0

f(x)=\begin{cases} \frac{\sin^{3}(\sqrt{3}) \cdot \log(1+3x)}{(\tan^{-1}\sqrt{x})^{2} (e^{5\sqrt{x}}-1) x}, & x\neq 0 \\ a, & x=0 \end{cases}

is continuous in [0.1], if 'a' equals to

0
$\frac{3}{5}$
2
$\frac{5}{3}$

Solution:

We have,

f(x)=\begin{cases} \frac{\sin^{3}(\sqrt{3}) \cdot \log(1+3x)}{x \cdot (\tan^{-1}\sqrt{x})^{2} (e^{5\sqrt{x}}-1) x}, & x\neq 0 \\ a, & x=0 \end{cases}

For continuity in

[0,1], f(0)=f(0^{+})

, otherwise it is discontinuous.

∴

\lim\limits_{x\to 0^{+}} \frac{\sin^{3}(\sqrt{x}) \cdot \log(1+3x)}{x \cdot (\tan^{-1}\sqrt{x})^{2} \cdot (e^{5\sqrt{x}}-1)}

= \lim\limits_{x\to 0^{+}} \left[ \frac{3}{5} \cdot \frac{\sin^{3}\sqrt{x}}{(\sqrt{x})^{3}} \cdot \frac{(\sqrt{x})^{2}}{(\tan^{-1}\sqrt{x})^{2}} \right.

\times \frac{\log(1+3x)}{3x} \cdot \frac{5\sqrt{x}}{e^{5\sqrt{x}}-1}

= \frac{3}{5} \lim\limits_{x\to 0^{+}} \frac{\sin^{3}\sqrt{x}}{(\sqrt{x})^{3}} \cdot \frac{(\sqrt{x})^{2}}{\tan^{-1}\sqrt{x}}

\times \frac{\log(1+3x)}{3x} \cdot \frac{5\sqrt{x}}{e^{5\sqrt{x}}-1} = a

⇒

a = \frac{3}{5}

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