Question

If $f\left(x\right)=\frac{x}{\sin x}$ and $g\left(x\right)=\frac{x}{\tan x}$, where $0<x\le 1$, then in the interval

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

Answer 1

The correct option is D

$f\left(x\right)$ is increasing function

Explanation for the correct option.

Step 1. Find the nature of the function $f\left(x\right)$.

For the function $f\left(x\right)=\frac{x}{\sin x}$, the derivative is given as:

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

In the interval $(0,1]$ it is known that $\tan x>x$ and so

$$\begin{gathered} \frac{\sin x}{\cos x}>x \\ \Rightarrow \sin x>x\cos x \\ \Rightarrow \sin x-x\cos x>0 \\ \Rightarrow \frac{\sin x-x\cos x}{{\sin }^{2}x}>0 \\ \Rightarrow f\text{'}\left(x\right)>0 \end{gathered}$$

So, in the interval $(0,1]$ it is found that $f\text{'}\left(x\right)>0$ and so the function $f\left(x\right)=\frac{x}{\sin x}$ is increasing in that interval.

Step 2. Find the nature of the function $g\left(x\right)$.

For the function $g\left(x\right)=\frac{x}{\tan x}$, the derivative is given as:

$g\text{'}\left(x\right)=\frac{\tan x-xse{c}^{2}x}{{\tan }^{2}x}$

In the interval $(0,1]$ it is known that $x>\sin x\cos x$. So

$$\begin{gathered} x>\frac{{\displaystyle \frac{\sin x\cos x}{{\cos }^{2}x}}}{{\displaystyle \frac{{\displaystyle 1}}{{\displaystyle {\cos }^{2}x}}}} \\ \Rightarrow x>\frac{\tan x}{se{c}^{2}x} \\ \Rightarrow xse{c}^{2}x>\tan x \\ \Rightarrow xse{c}^{2}x-\tan x>0 \\ \Rightarrow \tan x-xse{c}^{2}x<0 \end{gathered}$$

So in the interval $(0,1]$, $\tan x-xse{c}^{2}x<0$ and so $g\text{'}\left(x\right)<0$ and thus the function $g\left(x\right)=\frac{x}{\tan x}$ is decreasing in that interval.

So in the interval $(0,1]$ , function $f\left(x\right)$ is increasing while the function $g\left(x\right)$ is decreasing.

Hence, the correct option is D.