Let $\mathrm{k}$ and $\mathrm{m}$ be positive real numbers such that the function $$f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$$ is differentiable for all $x > 0$. Then $\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}$ is equal to ____________.

Here, $$f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$$ <br/><br/>$\because f(x)$ is differentiable at $x>0$ <br/><br/>So, $f(x)$ is differentiable at $x=1$ <br/><br/>$$ \begin{gathered} f\left(1^{-}\right)=f(1)=f\left(1^{+}\right) \\\\ 3+k \sqrt{2}=m+k^2 ......(i) \end{gathered} $$ <br/><br/>$$ \begin{aligned} & f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\\\ & 6(1)+\frac{k}{2 \sqrt{1+1}}=2 m(1) \\\\ & \Rightarrow 6+\frac{k}{2 \sqrt{2}}=2 m ......(ii) \end{aligned} $$ <br/><br/>Using (i) and (ii), <br/><br/>$$ \begin{aligned} & 3+k \sqrt{2}=3+\frac{k}{4 \sqrt{2}}+k^2 \\\\ & \Rightarrow k^2+k\left[\frac{1}{4 \sqrt{2}}-\sqrt{2}\right]=0 \end{aligned} $$ <br/><br/>$$ \Rightarrow k\left[k+\frac{1-8}{4 \sqrt{2}}\right]=0 \Rightarrow k=0, \frac{7}{4 \sqrt{2}} $$ <br/><br/>As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = $\frac{7}{4 \sqrt{2}}$ <br/><br/>$$ \begin{aligned} & \text { for } k=\frac{7}{4 \sqrt{2}}, m=3+\frac{\frac{7}{4 \sqrt{2}}}{4 \sqrt{2}} \\\\ & =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32} \end{aligned} $$ <br/><br/>$$ \text { So, } \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=\frac{8 \times\left[2 \times \frac{103}{32} \times 8\right]}{6 \times \frac{1}{8}+\frac{7}{4 \sqrt{2}} \times 2 \sqrt{918}} $$ <br/><br/>$$ =\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412 \times 12}{16}=309 $$ <br/><br/><b>Concept :</b> <br/><br/>$f(x)$ is differentiable at $x=a$, if $f^{\prime}\left(a^{-}\right)=f'\left(a^{+}\right)$