Mathematics - Continuity - Differentiability Question with Solution | TestHub

$\begin{array}{rl}& f\left(0^{-}\right)={\mathrm{e}}^{{\lim }_{x\to 0}\frac{\mathrm{ax}}{\mathrm{x}}}={\mathrm{e}}^{\mathrm{a}}\\ & \mathrm{f}(0)=1+\mathrm{b}\\ & \mathrm{f}\left(0^{+}\right)=\frac{\frac{1}{2\sqrt{\mathrm{x}+4}}}{\frac{1}{3}(\mathrm{x}+\mathrm{c}{)}^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3}\cdot {\mathrm{c}}^{-\frac{2}{3}}}\\ & =\frac{3}{4}{\mathrm{c}}^{2/3}\end{array}$
Also at $\mathrm{x}=0$;
$c^{1/3}=2\Rightarrow c=8$
So $\mathrm{f}\left(0^{+}\right)=\frac{3}{4}(8{)}^{2/3}=3$
Now, ${\mathrm{e}}^{\mathrm{a}}=\mathrm{b}+1=3$
${\mathrm{e}}^{\mathrm{a}}.\mathrm{b}\cdot \mathrm{c}=3\cdot 2\cdot 8=48$