Question

The function $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ is increasing on $\theta$ in the interval

• A. $[0, \pi]$
• B. $\left[0, \frac{\pi}{2}\right]$
• C. $\left(0, \frac{\pi}{2}\right)$
• D. None of these

Answer 1

Answer: (B)

We have, $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$
On differentiating w.r.t. $\theta$, we get
$\frac{d y}{d \theta} =\frac{d}{d \theta}\left[\frac{4 \sin \theta}{2+\cos \theta}-\theta\right] $
$ =\frac{(2+\cos \theta) \frac{d}{d \theta}(4 \sin \theta)-4 \sin \theta \frac{d}{d \theta}(2+\cos \theta)}{(2+\cos \theta)^2}-1$
$ =\frac{4 \cos \theta(2+\cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^2}-1 $
$ =\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta}{\left(2+\cos ^2 \theta\right)^2}-1$
$=\frac{8 \cos \theta+4\left(\cos ^2 \theta+\sin ^2 \theta\right)}{(2+\cos \theta)^2}-1$
$=\frac{8 \cos \theta+4}{(2+\cos 0)^2}-1=\frac{8 \cos \theta+4-(2+\cos \theta)^2}{(2+\cos 0)^2}$
$=\frac{8 \cos \theta+4-4-\cos ^2 \theta-4 \cos \theta}{(2+\cos \theta)^2}$
$=\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}$
$\frac{d y}{d \theta}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}$
$\because -1 \leq-\cos \theta \leq 1 $
$3 \leq 4-\cos \theta \leq 5 $
$\Rightarrow 4-\cos \theta >0$
Hence, $ \frac{4-\cos \theta}{(2+\cos \theta)^2} >0$
$\because y$ is increasing.
$\therefore \frac{d y}{d \theta} \geq 0 $
$\Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}>0 \Rightarrow \cos \theta>0\left[\because \frac{4-\cos \theta}{(2+\cos \theta)^2}>0\right]$
We know that $\cos \theta$ is positive when it lies on first quadrant.
$\therefore \theta \in\left[0, \frac{\pi}{2}\right]$
Hence, given function is increasing in interval $\left[0, \frac{\pi}{2}\right]$.
Note Since, it is a continuous function, for increasing function, $f^{\prime}(x) \geq 0$. (ii) for strictly increasing function, $f^{\prime}(x)>0$.