Question

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

Explanation for the correct answer:

Step 1: Solve for the first order derivative of $f\left(x\right)$ and apply condition for increasing / decreasing function

A function $f\left(x\right)$ is said to be increasing if $f\text{'}\left(x\right)>0$ in that interval and decreasing if $f\text{'}\left(x\right)<0$ in that interval

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

Differentiating with respect to $x$ we get

$\Rightarrow$$f\text{'}\left(x\right)=\frac{\left(1\right)\sin x-x\left({\displaystyle \frac{d}{dx}}\sin x\right)}{{\left(\sin x\right)}^2}$ $...\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

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

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

In the interval $(0,1]$ $\tan x>x$ and $\cos x>0$

Hence, $f\text{'}\left(x\right)=\frac{\left(\tan x-x\right)\cos x}{{\sin }^{2}x}>0$

Hence, $f\left(x\right)=\frac{x}{\sin x}$ is an increasing function in $(0,1]$

Step-2: Solve for the first order derivative of $g\left(x\right)$

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

Differentiating with respect to $x$ we get

$\Rightarrow$$g\text{'}\left(x\right)=\frac{\left(1\right)tanx-x\left({\displaystyle \frac{d}{dx}}\mathrm{ta}nx\right)}{{\left(\tan x\right)}^2}$ $...\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

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

Let $h\left(x\right)=tanx-xse{c}^{2}x$

Step- 3: Apply the condition for increasing/ decreasing function for $h\left(x\right)$ and further conclude about $g\left(x\right)$

Differentiating with respect to $x$ we get

$\Rightarrow$$h\text{'}\left(x\right)=se{c}^{2}x-se{c}^{2}x-2xse{c}^{2}x\tan x$

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

In the interval $(0,1]$ , $h\text{'}\left(x\right)=-2xse{c}^{2}x\tan x<0$

$h\left(x\right)=tanx-xse{c}^{2}x$ is a decreasing function in the interval $(0,1]$

$\Rightarrow$$h\left(x\right)<0$

$\Rightarrow$$g\text{'}\left(x\right)=\frac{tanx-xse{c}^{2}x}{{\tan }^{2}x}<0$

Hence, $g\left(x\right)=\frac{x}{tanx}$ is a decreasing function in the interval $(0,1]$. So, option (C) is the correct answer.