Question

$f\left(x\right)=\{\begin{array}{cc}\frac{{\sin }^3\left(\sqrt{3}\right)\cdot \log (1+3x)}{\left({\tan }^{-1}\sqrt{x}{)}^2\right(e^{5\sqrt{x}}-1)x},& x\ne 0\\ a,& x=0\end{array}$ continuous in $[0,1]$, if ${}^{\prime }{a}^{\prime }$ equals to

• A. $0$
• B. $\frac{3}{5}$
• C. $2$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

$\frac{3}{5}$

We have, $f\left(x\right)=\{\begin{array}{cc}\frac{{\sin }^3\left(\sqrt{3}\right)\cdot \log (1+3x)}{\left({\tan }^{-1}\sqrt{x}{)}^2\right(e^{5\sqrt{x}}-1)x},& x\ne 0\\ a,& x=0\end{array}$
For continuity in $[0,1],f\left(0\right)=f\left(0^{+}\right)$, otherwise it is discontinuous.
Therefore, $\underset{x\to 0^{+}}{lim}\frac{{\sin }^3\left(\sqrt{x}\right)\cdot \log (1+3x)}{x\cdot ({\tan }^{-1}\sqrt{x}{)}^2\cdot (e^{5\sqrt{x}}-1)}$
$=\underset{x\to 0^{+}}{lim}[\frac{3}{5}\cdot \frac{{\sin }^3\sqrt{x}}{(\sqrt{x}{)}^3}\cdot \frac{(\sqrt{x}{)}^2}{({\tan }^{-1}\sqrt{x}{)}^2}\times \frac{\log (1+3x)}{3x}\cdot \frac{5\sqrt{x}}{e^{5\sqrt{x}}-1}]$
$=\frac{3}{5}\underset{x\to 0^{+}}{lim}\frac{{\sin }^3\sqrt{x}}{(\sqrt{x}{)}^3}\cdot \frac{(\sqrt{x}{)}^2}{{\tan }^{-1}\sqrt{x}}\times \frac{\log (1+3x)}{3x}\cdot \frac{5\sqrt{x}}{e^5\sqrt{x}-1}=a$
$\Rightarrow a=\frac{3}{5}$.