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
$\frac{dy}{d\theta }=\frac{d}{d\theta }\left[\frac{4\sin \theta }{2+\cos \theta }-\theta \right]$
$=\frac{(2+\cos \theta )\frac{d}{d\theta }(4\sin \theta )-4\sin \theta \frac{d}{d\theta }(2+\cos \theta )}{(2+\cos \theta {)}^2}-1$
$=\frac{4\cos \theta (2+\cos \theta )-4\sin \theta (-\sin \theta )}{(2+\cos \theta {)}^2}-1$
$=\frac{8\cos \theta +4{\cos }^2\theta +4{\sin }^2\theta }{{\left(2+{\cos }^2\theta \right)}^2}-1$
$=\frac{8\cos \theta +4\left({\cos }^2\theta +{\sin }^2\theta \right)}{(2+\cos \theta {)}^2}-1$
$=\frac{8\cos \theta +4}{(2+\cos 0{)}^2}-1=\frac{8\cos \theta +4-(2+\cos \theta {)}^2}{(2+\cos 0{)}^2}$
$=\frac{8\cos \theta +4-4-{\cos }^2\theta -4\cos \theta }{(2+\cos \theta {)}^2}$
$=\frac{4\cos \theta -{\cos }^2\theta }{(2+\cos \theta {)}^2}$
$\frac{dy}{d\theta }=\frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}$
$\because -1\le -\cos \theta \le 1$
$3\le 4-\cos \theta \le 5$
$\Rightarrow 4-\cos \theta >0$
Hence, $\frac{4-\cos \theta }{(2+\cos \theta {)}^2}>0$
$\because y$ is increasing.
$\therefore \frac{dy}{d\theta }\ge 0$
$\Rightarrow \frac{\cos \theta (4-\cos \theta )}{(2+\cos \theta {)}^2}>0\Rightarrow \cos \theta >0\left[\because \frac{4-\cos \theta }{(2+\cos \theta {)}^2}>0\right]$
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, $f^{\prime }(x)\ge 0$. (ii) for strictly increasing function, $f^{\prime }(x)>0$.