Q

# Find the intervals in which the function f given by f x = 4 sin x - 2x - x cos x/ 2 + cos x  is  decreasing

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

Views

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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