Question

Find the intervals in which the function f given by $f\left(x\right)=\frac{4sinx-2x-xcosx}{2+cosx}$ is
increasing

Find the intervals in which the function f given by $f\left(x\right)=\frac{4sinx-2x-xcosx}{2+cosx}$ is
decreasing

Answer 1

Given,$f\left(x\right)=\frac{4sinx-2x-xcosx}{2+cosx}=\frac{4sinx-x(2+cosx)}{2+cosx}=\frac{4sinx}{2+cosx}-x$
On differentiating w.r.t.x, we get
$f^{\prime }\left(x\right)=4\left\{\frac{(2+cosx)cosx-sinx(0-sinx)}{(2+cos{x}^2)}\right\}-1=4\left\{\frac{2cosx+co{s}^{2}x+si{n}^{2}x}{(2+cosx{)}^2}\right\}-1(\because co{s}^{2}x+si{n}^{2}x=1)=\frac{8cosx+4}{(2+cosx{)}^2}-1=\frac{8cosx+4-(2+cosx{)}^2}{(2+cosx{)}^2}=\frac{8cosx+4-4-co{s}^{2}x-4cosx}{(2+cosx{)}^2}=\frac{4cosx-co{s}^{2}x}{(2+cosx{)}^2}=\frac{8cosx+4-4-co{s}^{2}x-4cosx}{(2+cosx{)}^2}=\frac{4cosx-co{s}^{2}x}{(2+cosx{)}^2}=\frac{cosx(4-cosx)}{(2+cosx{)}^2}$
We know that - $1\le cosx\le 1$
$\Rightarrow 4-cosx>0and(2+cosx{)}^2>0$
(a) For increasing
f'(x) > 0 when cos x > 0 [$\because$ cos x is positive in 1st ~ and~ 4th quadrant]
$\therefore f\left(x\right)$ is increasing in the interval $(0,\frac{\pi }{2})and(\frac{3\pi }{2},2\pi ).$

Given,$f\left(x\right)=\frac{4sinx-2x-xcosx}{2+cosx}=\frac{4sinx-x(2+cosx)}{2+cosx}=\frac{4sinx}{2+cosx}-x$
On differentiating w.r.t.x, we get
$f^{\prime }\left(x\right)=4\left\{\frac{(2+cosx)cosx-sinx(0-sinx)}{(2+cos{x}^2)}\right\}-1=4\left\{\frac{2cosx+co{s}^{2}x+si{n}^{2}x}{(2+cosx{)}^2}\right\}-1(\because co{s}^{2}x+si{n}^{2}x=1)=\frac{8cosx+4}{(2+cosx{)}^2}-1=\frac{8cosx+4-(2+cosx{)}^2}{(2+cosx{)}^2}=\frac{8cosx+4-4-co{s}^{2}x-4cosx}{(2+cosx{)}^2}=\frac{4cosx-co{s}^{2}x}{(2+cosx{)}^2}=\frac{8cosx+4-4-co{s}^{2}x-4cosx}{(2+cosx{)}^2}=\frac{4cosx-co{s}^{2}x}{(2+cosx{)}^2}=\frac{cosx(4-cosx)}{(2+cosx{)}^2}$
We know that - $1\le cosx\le 1$
$\Rightarrow 4-cosx>0and(2+cosx{)}^2>0$
(b) For decreasing
f'(x) > 0 when cos x < 0
[$\because$ cos x is negative in 2nd~ and~ 3rd quadrant]
$\therefore$ f(x) is decreasing in the interval $(\frac{\pi }{2},\frac{3\pi }{2}).$