Let k and m be positive real numbers such that the function f(x) =...

Since, f\left(x\right) = \left\{$\begin{array}{ll}3{x}^{2} + k\sqrt{x + 1}; 0 < x < 1 \\ m{x}^{2} + {k}^{2}; x \ge 1\end{array}$\right) is differentiable atx = 1$, so function must be continuous at$x = 1$, hence</p><p>$LHL = RHL$</p><p>$3 + k\sqrt{2} = m + {k}^{2}$</p><p>$⇒ {k}^{2} - k\sqrt{2} + m - 3 = 0 . . . . \left(1\right)$</p><p>And,</p><p>${f}^{'}\left(x\right) = \left\{$$\begin{array}{ll}6x + \dfrac{k}{2\sqrt{x + 1}}; 0 1\end{array}$$\right)

So,

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

$⇒ 6 + \dfrac{k}{2\sqrt{2}} = 2m$

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

Putting in \left(1\right)$, we get</p><p>${k}^{2} - k\sqrt{2} + 3 + \dfrac{k}{4\sqrt{2}} - 3 = 0$</p><p>$⇒ {k}^{2} - \left(\sqrt{2} - \dfrac{1}{4\sqrt{2}}\right)k = 0$</p><p>$⇒ {k}^{2} - \left(\dfrac{7\sqrt{2}}{8}\right)k = 0$</p><p>$⇒ k\left(k - \dfrac{7\sqrt{2}}{8}\right) = 0$</p><p>$⇒ k = \dfrac{7\sqrt{2}}{8}$</p><p>So,</p><p>$m = 1 + \dfrac{k}{12\sqrt{2}} = 1 + \dfrac{7\sqrt{2}}{96\sqrt{2}} = \dfrac{103}{96}$</p><p>Hence,</p><p>${f}^{'}\left(x\right) = \left\{$$\begin{array}{ll}6x + \dfrac{7\sqrt{2}}{16\sqrt{x + 1}}; 0 1\end{array}$$\right)

So,

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

$⇒ \dfrac{{f}^{'}\left(8\right)}{{f}^{'}\left(\dfrac{1}{8}\right)} = \dfrac{103}{2} \times \dfrac{3}{4}$

⇒ \dfrac{8{f}^{'}\left(8\right)}{{f}^{'}\left(\dfrac{1}{8}\right)} = 309