Let k and m be positive real numbers such tha | vedclass

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(A) Since $f(x)$ is differentiable for all $x > 0$,it must be continuous at $x = 1$.Thus,$\lim_{x 	o 1^-} f(x) = f(1)$.$3(1)^2 + k\sqrt{1+1} = m(1)^2

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$309$

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(A) Since $f(x)$ is differentiable for all $x > 0$,it must be continuous at $x = 1$.Thus,$\lim_{x 	o 1^-} f(x) = f(1)$.$3(1)^2 + k\sqrt{1+1} = m(1)^2 + k^2 \implies 3 + k\sqrt{2} = m + k^2 \quad \dots(1)$Also,$f'(x)$ must be continuous at $x = 1$,so $f'_-(1) = f'_+(1)$.For $0 For $x > 1$,$f'(x) = 2mx$.At $x = 1$,$6(1) + \frac{k}{2\sqrt{2}} = 2m(1) \implies 2m = 6 + \frac{k}{2\sqrt{2}} \implies m = 3 + \frac{k}{4\sqrt{2}} \quad \dots(2)$Substitute $(2)$ into $(1)$:$3 + k\sqrt{2} = (3 + \frac{k}{4\sqrt{2}}) + k^2$$k^2 + k(\frac{1}{4\sqrt{2}} - \sqrt{2}) = 0$$k^2 + k(\frac{1-8}{4\sqrt{2}}) = 0 \implies k^2 - \frac{7k}{4\sqrt{2}} = 0$.Since $k > 0$,$k = \frac{7}{4\sqrt{2}}$.Then $m = 3 + \frac{7/4\sqrt{2}}{4\sqrt{2}} = 3 + \frac{7}{32} = \frac{103}{32}$.Now,$f'(8) = 2m(8) = 16m = 16(\frac{103}{32}) = \frac{103}{2}$.And $f'(\frac{1}{8}) = 6(\frac{1}{8}) + \frac{k}{2\sqrt{1/8+1}} = \frac{3}{4} + \frac{7/4\sqrt{2}}{2\sqrt{9/8}} = \frac{3}{4} + \frac{7/4\sqrt{2}}{2(3/2\sqrt{2})} = \frac{3}{4} + \frac{7}{12} = \frac{9+7}{12} = \frac{16}{12} = \frac{4}{3}$.Therefore,$\frac{8f'(8)}{f'(1/8)} = \frac{8(103/2)}{4/3} = \frac{412}{4/3} = 103 	imes 3 = 309$.

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

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