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Mathematics Question on Applications of Derivatives

Find the intervals in which the function f given by f(x)=4sinx−2x−x2+cosxf(x)=\frac{4sinx-2x-x}{2+cosx} is (i)increasing (ii)decreasing

Answer

f(x)=4sinx−2x−xcosx2+cosxf(x)=\frac{4sinx-2x-xcosx}{2+cosx}
∴ f′(x)=(2+cosx)(2cosx−2−cosx+xsinx)−(4sinx−2x−xcosx)(−sinx)(2+cosx)2f'(x)=\frac{(2+cosx)(2cosx-2-cosx+xsinx)-(4sinx-2x-xcosx)(-sinx)}{(2+cosx)^2}
=(2+cosx)(3cosx−2+xsinx)+sinx(4sinx−2x−xcosx)(2+cosx)2\frac{(2+cosx)(3cosx-2+xsinx)+sinx(4sinx-2x-xcosx)}{(2+cosx)^2}
=6cosx−4+2xsinx+3cos2x−2cosx+xsinxcosx+4sin2x−2xsinx−xsinxcosx(2+cosx)2=\frac{6cosx-4+2xsinx+3cos^2x-2cosx+xsinxcosx+4sin^2x-2xsinx-xsinxcosx}{(2+cosx)^2}
=4cosx−4+3cos2x+4sin2x(2+cosx)2=\frac{4cosx-4+3cos^2x+4sin^2x}{(2+cosx)^2}
=4cosx−4+3cos2x+4−4cos2x(2+cosx)2=\frac{4cosx-4+3cos^2x+4-4cos^2x}{(2+cosx)^2}
=4cosx−cos2x(2+cosx)2=\frac{4cosx-cos^2x}{(2+cosx)^2}
=cosx(4−cosx)(2+cosx)2=\frac{cosx(4-cosx)}{(2+cosx)2}
Now f′(x)=0f'(x)=0
⇒cosx=0 or cosx=4
But, cosx≠4
∴cosx=0
⇒x=π2,3π2⇒x=\frac{\pi}{2},\frac{3\pi}{2}
Now,x=π2x=\frac{\pi}{2} and x=3π2x=\frac{3\pi}{2} divides (0, 2π) into three disjoint intervals i.e.,
(0,π2)(0,\frac{π}{2}),(π2,3π2)(\frac{π}{2},\frac{3π}{2}),and (3π2,2π)(\frac{3π}{2},2π)
In intervals (0,π2)(0,\frac{π}{2})and (3π2,2π)(\frac{3π}{2},2π),f′(x)>0f'(x)>0
Thus, f(x) is increasing for 0<x<x20<x<\frac{x}{2} and 3π2\frac{3π}{2}<x<2π
In the interval π2,3π2\frac{π}{2},\frac{3π}{2},f′(x)<0.f'(x)<0.
Thus, f(x) is decreasing for π2\frac{π}{2}<x<3π2\frac{3π}{2}