If the extreme value of the function f(x) = f | vedclass

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(C) Let $u = \sin x$. Since $x \in (0, \frac{\pi}{2})$,$u \in (0, 1)$.Define $g(u) = \frac{4}{u} + \frac{1}{1-u}$.To find the extreme value,differenti

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$\frac{\sqrt{5}}{\sqrt{m}}$

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(C) Let $u = \sin x$. Since $x \in (0, \frac{\pi}{2})$,$u \in (0, 1)$.Define $g(u) = \frac{4}{u} + \frac{1}{1-u}$.To find the extreme value,differentiate $g(u)$ with respect to $u$:$g'(u) = -\frac{4}{u^2} + \frac{1}{(1-u)^2}$.Set $g'(u) = 0$ to find critical points:$\frac{1}{(1-u)^2} = \frac{4}{u^2} \implies u^2 = 4(1-u)^2 \implies u^2 = 4(1 - 2u + u^2)$.$u^2 = 4 - 8u + 4u^2 \implies 3u^2 - 8u + 4 = 0$.Solving the quadratic equation: $(3u - 2)(u - 2) = 0$.Since $u \in (0, 1)$,we have $u = \frac{2}{3}$.Thus,$\sin k = \frac{2}{3}$.Then $\cos^2 k = 1 - \sin^2 k = 1 - \frac{4}{9} = \frac{5}{9}$,so $\cos k = \frac{\sqrt{5}}{3}$.The value $m = f(k) = g(\frac{2}{3}) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9$.We need to find $\cos k$ in terms of $m=9$.$\cos k = \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{\sqrt{9}} = \frac{\sqrt{5}}{\sqrt{m}}$.Thus,the correct option is $C$.

If the extreme value of the function $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$ in the interval $[0, \frac{\pi}{2}]$ is $m$ and it exists at $x = k$,then $\cos k =$

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