Application Of Derivatives Ques 31
31. If $f(x)=\frac{x}{\sin x}$ and $g(x)=\frac{x}{\tan x}$, where $0<x \leq 1$, then in this interval
(a) both $f(x)$ and $g(x)$ are increasing functions
(b) both $f(x)$ and $g(x)$ are decreasing functions
(c) $f(x)$ is an increasing function
(d) $g(x)$ is an increasing function
Show Answer
Answer:
(b)
Solution:
Formula:
Increasing and decreasing of a function:
- Given, $g(x)=\frac{x}{\tan x}$, where $0<x \leq 1$
Now, $g(x)$ is continuous in $[0,1]$ and differentiable in ]0, $1[$.
For
$ \begin{aligned} & 0<x<1 \\ & g^{\prime}(x)=\frac{\tan x-x \sec ^{2} x}{\tan ^{2} x} \end{aligned} $
Again, $\quad H(x)=\tan x-x \sec ^{2} x, 0 \leq x \leq 1$
Now, $H(x)$ is continuous in $[0,1]$ and differentiable in ]0, $1[$.
$ \begin{alignedat} & \text { For } \quad 0<x<1, H(x)=\tan x-x \sec ^{2} x, 0 \leq x \leq 1 \\ & \Rightarrow \quad H^{\prime}(x)=\sec ^{2} x-\sec ^{2} x-2 x \sec x \tan x \\ & =-2 x \sec ^{2} x \tan x<0 \end{aligned} $
Hence, $H(x)$ is a decreasing function in $[0,1]$.
Thus, $H(x)<H(0)$ for $0<x<1$
$ \begin{array}{llll} \Rightarrow & H(x)<0 & \text { for } & 0<x<1 \\ \Rightarrow & g^{\prime}(x)<0 & \text { for } & 0<x<1 \end{array} $
$\Rightarrow g(x)$ is decreasing function in $(0,1]$.
Therefore, $g(x)=\frac{x}{\tan x}$ is not necessarily a decreasing function in $0<x \leq 1$.
$ \begin{alignedat} & \text { Also, } \quad g(x)<g(0) \text { for } 0<x \leq 1 \\ & \Rightarrow \quad \frac{x}{\tan x}<1 \quad \text { for } 0<x \leq 1 \\ & \Rightarrow \quad x<\tan x \quad \text { for } \quad 0<x \leq 1 \\ & \text { Now, let } \quad f(x)=\begin{array}{ccc} x / \sin x & \text{for} & 0 < x \leq 1 \\ 1 & \text{for} & x=0 \end{array} \end{aligned} $
Now, $f$ is continuous in $[0,1]$ and differentiable in $] 0,1[$. For $0<x<1$,
$f^{\prime}(x)=\frac{\sin x-x \cos x}{\sin ^{2} x}=\frac{(\tan x-x) \cos x}{\sin ^{2} x}>0$ for $0<x<\frac{\pi}{2}$
$\Rightarrow f(x)$ increases in $[0,1]$.
Thus, $f(x)=\frac{x}{\sin x}$ increases in $0<x<\pi$.
Therefore, option (c) is the answer.
BETA
