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MHT CET 2021 Math Solved Paper

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Question : 19 of 100

Marks: +1, -0

If

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

is continuous in

[0,1]

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

Solution:

We have,

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

For continuity in

[0,1], f(0)=\lim\limits_{x\to 0}f(x)

otherwise it is discontinuous.

\therefore \;\; a \;\; =\lim\limits_{x\to 0}\frac{\sin^{3}(\sqrt{x}) \cdot \log(1+3x)}{x(\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})^{3}}{(\tan^{-1}\sqrt{x})^{3}}\right]

\left[\times \;\frac{\log(1+3x)}{3x} \cdot \;\frac{5\sqrt{x}}{3^{5\sqrt{x}}-1}\right]

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

\therefore \;\; a \;\; =\;\frac{3}{5}

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