Question

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

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

Answer 1

The correct option is B $\frac{3}{5}$

We have,
$f\left(x\right)=\{\begin{array}{cc}\frac{{\sin }^3\left(\sqrt{x}\right)\cdot \log (1+3x)}{{\left({\tan }^{-1}\sqrt{x}\right)}^2(e^{5\sqrt{x}}-1)x},& x\ne 0\\ a,& x=0\end{array}$

For continuity in $[0,1],f\left(0\right)=f\left(0^{+}\right)$, otherwise it is discontinuous.

$\therefore f\left(0\right)=\underset{x\to 0^{+}}{\text{lim}}\frac{{\sin }^3\left(\sqrt{x}\right)\cdot \log (1+3x)}{x\cdot {\left({\tan }^{-1}\sqrt{x}\right)}^2\cdot (e^{5\sqrt{x}}-1)}$

$=\underset{x\to 0^{+}}{\text{lim}}[\frac{3}{5}\cdot \frac{{\sin }^3\sqrt{x}}{{\left(\sqrt{x}\right)}^3}\cdot \frac{{\left(\sqrt{x}\right)}^2}{{\left({\tan }^{-1}\sqrt{x}\right)}^2}\cdot \frac{\log (1+3x)}{3x}\cdot \frac{5\sqrt{x}}{e^{5\sqrt{x}}-1}]$
$=\frac{3}{5}\cdot 1\cdot 1\cdot 1\cdot 1$
$\Rightarrow a=\frac{3}{5}$.