JEE Main 2023 — Continuity and Differentiability Question with Solution

Since, $f\left(x\right)=\left\{\begin{array}{l}3x^2+k\sqrt{x+1}; 0<x<1\\ m{x}^2+k^2; x\ge 1\end{array}\right.$ is differentiable at $x=1$, so function must be continuous at $x=1$, hence

$\mathrm{LHL}=\mathrm{RHL}$

$3+k\sqrt{2}=m+k^2$

$\Rightarrow k^2-k\sqrt{2}+m-3=0 ....\left(1\right)$

And,

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}6x+\frac{k}{2\sqrt{x+1}}; 0<x<1\\ 2mx; x>1\end{array}\right.$

So,

$f^{\text{'}}\left(1^{-}\right)=f^{\text{'}}\left(1^{+}\right)$

$\Rightarrow 6+\frac{k}{2\sqrt{2}}=2m$

$\Rightarrow m=3+\frac{k}{4\sqrt{2}} ...\left(2\right)$

Putting in $\left(1\right)$, we get

$k^2-k\sqrt{2}+3+\frac{k}{4\sqrt{2}}-3=0$

$\Rightarrow k^2-\left(\sqrt{2}-\frac{1}{4\sqrt{2}}\right)k=0$

$\Rightarrow k^2-\left(\frac{7\sqrt{2}}{8}\right)k=0$

$\Rightarrow k\left(k-\frac{7\sqrt{2}}{8}\right)=0$

$\Rightarrow k=\frac{7\sqrt{2}}{8}$

So,

$m=1+\frac{k}{12\sqrt{2}}=1+\frac{7\sqrt{2}}{96\sqrt{2}}=\frac{103}{96}$

Hence,

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}6x+\frac{7\sqrt{2}}{16\sqrt{x+1}}; 0<x<1\\ \frac{103x}{16}; x>1\end{array}\right.$

So,

$\frac{f^{\text{'}}\left(8\right)}{f^{\text{'}}\left({\displaystyle \frac{1}{8}}\right)}=\frac{103\times 8}{16}\times \frac{1}{\left({\displaystyle \frac{6}{8}}+{\displaystyle \frac{7\sqrt{2}}{16\sqrt{{\displaystyle \frac{9}{8}}}}}\right)}$

$\Rightarrow \frac{f^{\text{'}}\left(8\right)}{f^{\text{'}}\left({\displaystyle \frac{1}{8}}\right)}=\frac{103}{2}\times \frac{3}{4}$

$\Rightarrow \frac{8f^{\text{'}}\left(8\right)}{f^{\text{'}}\left({\displaystyle \frac{1}{8}}\right)}=309$