SATHEE: Application of Derivatives 2 Question 14

Application of Derivatives 2 Question 14

14. 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

(d) $g (x)$ is an increasing function

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Solution:

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

$\begin{aligned} 0 < x < 1 \\ g^{'} (x) = \frac{\tan x - x \sec^{2} x}{\tan^{2} x} \end{aligned}$

Again, $H (x) = \tan x - x \sec^{2} x, 0 \leq x \leq 1$

Now, $H (x)$ is continuous in $[0, 1]$ and differentiable in ]0, $1 [$ .

$\begin{aligned} For 0 < x < 1, H (x) = \tan x - x \sec^{2} x, 0 \leq x \leq 1 \\ \Rightarrow H^{'} (x) = \sec^{2} x - \sec^{2} x - 2 x \sec^{2} x \tan x \\ = - 2 x \sec^{2} x \tan x < 0 \end{aligned}$

Hence, $H (x)$ is decreasing function in $[0, 1]$ .

Thus, $H (x) < H (0)$ for $0 < x < 1$

$\begin{array}{llll} \Rightarrow & H (x) < 0 & for & 0 < x < 1 \\ \Rightarrow & g^{'} (x) < 0 & for & 0 < x < 1 \end{array}$

$\Rightarrow g (x)$ is decreasing function in $(0, 1]$ .

Therefore, $g (x) = \frac{x}{\tan x}$ is a decreasing function in $0 < x \leq 1$ .

$\begin{aligned} Also, g (x) < g (0) for 0 < x \leq 1 \\ \Rightarrow \frac{x}{\tan x} < 1 for 0 < x \leq 1 \\ \Rightarrow x < \tan x for 0 < x \leq 1 \\ Now, let f (x) = \begin{array}{ccc} x / \sin x & for & 0 < x \leq 1 \\ 1 & for & x = 0 \end{array} \end{aligned}$