Application of Derivatives — Class 12 Maths Solution

We have, $f(x) = \cfrac{{4\sin x - 2x - x\cos x}}{{2 + \cos x}}$
$= \cfrac{{4\sin x - x(2 + \cos x)}}{{2 + \cos x}}$
$\Rightarrow \cfrac{{4\sin x}}{{2 + \cos x}} - \cfrac{{x(2 + \cos x)}}{{2 + \cos x}} = \cfrac{{4\sin x}}{{2 + \cos x}} - x$ …(i)

Differentiating (i) w.r.t. $x$, we get
$f'(x) = 4\left\{ {\cfrac{{(2 + \cos x)\cos x - \sin x(0 - \sin x)}}{{{{(2 + \cos x)}^2}}}} \right\} - 1$

$= 4{\rm{ }}\left\{ {\cfrac{{2\cos x + {{\cos }^2}x + {{\sin }^2}x}}{{{{(2 + \cos x)}^2}}}} \right\} - 1$
$= \cfrac{{8\cos x + 4 - {{(2 + \cos x)}^2}}}{{{{(2 + \cos x)}^2}}} = \cfrac{{\cos x(4 - \cos x)}}{{{{(2 + \cos x)}^2}}}$

Now, ${(2 + \cos x)^2}$ being a perfect square is always non-negative.

Also, ${(2 + \cos x)^2} \ne 0$ as $(2 + \cos x) \ne 0$

Therefore in the interval : ${(2 + \cos x)^2}$ is always $+ ve$ and $4 - \cos x$ is also always $+ ve$, as $\cos x$ is always numerically $\le 1$

(i) For $f(x)$ to be increasing function of $x,f'(x) > 0.$
$\Rightarrow \cos x > 0 \Rightarrow x \in \left( {0,\cfrac{\pi }{2}} \right) \cup \left( {\cfrac{{3\pi }}{2},2\pi } \right)$

Therefore, $f(x)$ is increasing in $\left( {0,\cfrac{\pi }{2}} \right) \cup \left( {\cfrac{{3\pi }}{2},2\pi } \right)$.

(ii) For $f(x)$ to be decreasing, $f'(x) < 0$

$\Rightarrow \cos x < 0 \Rightarrow x \in \left( {\cfrac{\pi }{2},\cfrac{{3\pi }}{2}} \right)$

Therefore, $f(x)$ is decreasing in $\left( {\cfrac{\pi }{2},\cfrac{{3\pi }}{2}} \right)$ .