The extremum values of the function f(x) = fr | vedclass

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(C) Given $f(x) = \frac{1}{\sin x + 4} - \frac{1}{\cos x - 4} = \frac{\cos x - 4 - \sin x - 4}{(\sin x + 4)(\cos x - 4)} = \frac{\cos x - \sin x - 8}{

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$\frac{2\sqrt{2}}{4\sqrt{2} + 1}$

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(C) Given $f(x) = \frac{1}{\sin x + 4} - \frac{1}{\cos x - 4} = \frac{\cos x - 4 - \sin x - 4}{(\sin x + 4)(\cos x - 4)} = \frac{\cos x - \sin x - 8}{\sin x \cos x - 4\sin x + 4\cos x - 16}$.To find the extremum,we differentiate $f(x)$ with respect to $x$ and set $f'(x) = 0$.Let $u = \sin x + \cos x$. Then $u \in [-\sqrt{2}, \sqrt{2}]$.The expression simplifies to $f(u) = \frac{1}{4 + \sin x} + \frac{1}{4 - \cos x}$.By analyzing the critical points where $\sin x = \cos x$,we find the extrema occur at $x = n\pi + \frac{\pi}{4}$.For $x = 2n\pi + \frac{3\pi}{4}$,$\sin x = \frac{1}{\sqrt{2}}$ and $\cos x = -\frac{1}{\sqrt{2}}$.Substituting these,we get $f(x) = \frac{1}{4 + 1/\sqrt{2}} - \frac{1}{-1/\sqrt{2} - 4} = \frac{\sqrt{2}}{4\sqrt{2} + 1} + \frac{\sqrt{2}}{4\sqrt{2} + 1} = \frac{2\sqrt{2}}{4\sqrt{2} + 1}$.Rationalizing $\frac{2\sqrt{2}}{4\sqrt{2} + 1} 	imes \frac{4\sqrt{2} - 1}{4\sqrt{2} - 1} = \frac{16 - 2\sqrt{2}}{32 - 1} = \frac{16 - 2\sqrt{2}}{31}$.Upon re-evaluating the options provided,the value $\frac{2\sqrt{2}}{4\sqrt{2} + 1}$ is correct.

The extremum values of the function $f(x) = \frac{1}{\sin x + 4} - \frac{1}{\cos x - 4}$ for $x \in R$ are:

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