Question

Find the intervals in which the function $f\left(x\right)=\frac{4\sin x-2x-x\cos x}{2+\cos x}$ is (i) increasing
(ii) decreasing

Answer 1

$f\left(x\right)=\frac{4\sin x-2x-x\cos x}{2+\cos x}=\frac{4\sin x}{2+\cos x}-\frac{2x+x\cos x}{2+\cos x}=\frac{4\sin x}{2+\cos x}-\frac{x(2+\cos x)}{(2+\cos x)}=\frac{4\sin x}{2+\cos x}-x$

Now let's find out $f^{{}^{\prime }}\left(x\right)$

$f^{{}^{\prime }}\left(x\right)=\frac{4\cos x(2+\cos x)+\sin x\left(4\sin x\right)}{{(2+\cos x)}^2}-1=\frac{8\cos x+4(\cos {}^{2}x+\sin {}^{2}x)}{{(2+\cos x)}^2}-1=\frac{8\cos x+4-{(2+\cos x)}^2}{{(2+\cos x)}^2}=\frac{8\cos x+4-4-4\cos x-\cos {}^{2}x}{{(2+\cos x)}^2}=\frac{\cos x(4-\cos x)}{{(2+\cos x)}^2}$

Now,${(2+\cos x)}^2>0\forall xϵR$

Similarly,$4-\cos x>0\forall xϵR$

as $\cos xϵ[-1,1]\Rightarrow 4-\cos xϵ[3,5]>0$

$\therefore$ the function $f\left(x\right)$ will be increasing if $f^{{}^{\prime }}\left(x\right)>0\Rightarrow \cos x>0$ and decreasing if $f^{{}^{\prime }}\left(x\right)<0\Rightarrow \cos x<0$

(i) Increasing

$f^{{}^{\prime }}\left(x\right)>0\Rightarrow \cos x>0\therefore xϵ(0,\frac{\pi }{4})\cup (\frac{3\pi }{4},2\pi )$

(ii) Decreasing

$f^{{}^{\prime }}\left(x\right)<0\Rightarrow \cos x<0\therefore xϵ(\frac{\pi }{4},\pi )\cup (\pi ,\frac{3\pi }{2})$