Question

If $f\left(x\right)=\frac{x}{sinx}$ & $g\left(x\right)=\frac{x}{tanx}$, where $0<x\le 1$, then in this interval

• A. both $f\left(x\right)$ & $g\left(x\right)$ are increasing functions
• B. both $f\left(x\right)$ & $g\left(x\right)$ are decreasing functions
• C. $f\left(x\right)$ is an increasing function
• D. $g\left(x\right)$ is an increasing function

Answer 1

The correct option is C $f\left(x\right)$ is an increasing function

$f\left(x\right)=\frac{x}{\sin x}$

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

$=\frac{\cos x(\tan x-x)}{{\sin }^{2}x}>0$

$(\begin{array}{c}\cos x>0,{\sin }^{2}x>0,x\in (0,1)\\ \text{and}\tan x>x\end{array}$

$f\left(x\right)\text{is an increasing function.}$

$g\left(x\right)=\frac{x}{\tan x}$

$g^{\prime }\left(x\right)=\frac{\tan x-x{\sec }^{2}x}{{\tan }^{2}x}$

$=\frac{\sin x\cos x-x}{{\sin }^{2}x}$

$=\frac{\sin 2x-2x}{2{\sin }^{2}x}<0$

$(\begin{array}{c}{\sin }^{2}x>0,\\ x\in (0,1)\end{array}\sin 2x<2x$, $)$

$g\left(x\right)$ is a decreasing function.

Hence,

$C$ is correct option.