JEE Main 2025 — Continuity and Differentiability Question with Solution

$\begin{aligned}
& f\left(0^{-}\right)=\mathrm{e}^{\lim _{x \rightarrow 0} \frac{\mathrm{ax}}{\mathrm{x}}}=\mathrm{e}^{\mathrm{a}} \\ & \mathrm{f}(0)=1+\mathrm{b} \\ & \mathrm{f}\left(0^{+}\right)=\frac{\frac{1}{2 \sqrt{\mathrm{x}+4}}}{\frac{1}{3}(\mathrm{x}+\mathrm{c})^{-\frac{2}{3}}}=\frac{\frac{1}{2(2)}}{\frac{1}{3} \cdot \mathrm{c}^{-\frac{2}{3}}} \\ & =\frac{3}{4} \mathrm{c}^{2 / 3}
\end{aligned}$<br/>Alsoat$\mathrm{x}=0$;<br/>$c^{1 / 3}=2 \Rightarrow c=8$<br/>So$\mathrm{f}\left(0^{+}\right)=\frac{3}{4}(8)^{2 / 3}=3$<br/>Now,$\mathrm{e}^{\mathrm{a}}=\mathrm{b}+1=3$<br/>$\mathrm{e}^{\mathrm{a}} . \mathrm{b} \cdot \mathrm{c}=3 \cdot 2 \cdot 8=48$