6)  Find the intervals in which the function f given by f x is equal to 

 f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x } is 

(ii) decreasing

Answers (1)

Given function is
f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }
f^{'}(x) = \frac{(4\cos x - 2-\cos x+x\sin x)(2+\cos x)-(4\sin x - 2x - x\cos x )(-\sin x)}{(2+\cos x)^2}
             =\frac{4\cos x - \cos^2x}{2+\cos x}
f^{'}(x)=0\\ \frac{4\cos x - \cos^2x}{2+\cos x} =0\\ \cos x(4-\cos x) = 0\\ \cos x = 0 \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ \ \cos x =4
But    \cos x \neq 4
So,
\cos x = 0 \\ x = \frac{\pi}{2} \ and \ \frac{3\pi}{2}
Now three ranges are there \left ( 0,\frac{\pi}{2} \right ),\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )
In interval \left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )  , f^{'}(x) > 0

Hence, given  function  f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }  is increasing in interval  \left ( 0,\frac{\pi}{2} \right ) \ and \ \left ( \frac{3\pi}{2},2\pi \right )
in interval ,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right ), f^{'}(x) < 0
Hence, given  function  f (x) = \frac{4 \sin x - 2x - x \cos x }{2 + \cos x }   is decreasing in interval  ,\left ( \frac{\pi}{2},\frac{3\pi}{2} \right )

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