Question

Find the intervals in which the function f given by \(f(x)=\frac{4sinx-2x-x}{2+cosx}\) is (i)increasing (ii)decreasing

Answer 1

\(f(x)=\frac{4sinx-2x-xcosx}{2+cosx}\)
∴ \(f'(x)=\frac{(2+cosx)(2cosx-2-cosx+xsinx)-(4sinx-2x-xcosx)(-sinx)}{(2+cosx)^2}\)
=\(\frac{(2+cosx)(3cosx-2+xsinx)+sinx(4sinx-2x-xcosx)}{(2+cosx)^2}\)
\(=\frac{6cosx-4+2xsinx+3cos^2x-2cosx+xsinxcosx+4sin^2x-2xsinx-xsinxcosx}{(2+cosx)^2}\)
\(=\frac{4cosx-4+3cos^2x+4sin^2x}{(2+cosx)^2}\)
\(=\frac{4cosx-4+3cos^2x+4-4cos^2x}{(2+cosx)^2}\)
\(=\frac{4cosx-cos^2x}{(2+cosx)^2}\)
\(=\frac{cosx(4-cosx)}{(2+cosx)2}\)
Now \(f'(x)=0\)
⇒cosx=0 or cosx=4 
But, cosx≠4 
∴cosx=0
\(⇒x=\frac{\pi}{2},\frac{3\pi}{2}\)
Now,\(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\) divides (0, 2π) into three disjoint intervals i.e.,
\((0,\frac{π}{2})\),\((\frac{π}{2},\frac{3π}{2})\),and \((\frac{3π}{2},2π)\)
In intervals \((0,\frac{π}{2})\)and \((\frac{3π}{2},2π)\),\(f'(x)>0\)
Thus, f(x) is increasing for \(0<x<\frac{x}{2}\) and \(\frac{3π}{2}\)<x<2π
In the interval \(\frac{π}{2},\frac{3π}{2}\),\(f'(x)<0.\)
Thus, f(x) is decreasing for \(\frac{π}{2}\)<x<\(\frac{3π}{2}\)