Applications Of Derivatives Question 245
Question: If $ f(x)=\frac{x}{\sin x} $ and $ g(x)=\frac{x}{\tan x} $ , where $ 0<x\le 1 $ , then in this interval
[IIT 1997 Re-Exam]
Options:
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:
Correct Answer: C
Solution:
$ f’(x)=\frac{\sin x-x\cos x}{{{\sin }^{2}}x}=\frac{\cos x(\tan x-x)}{{{\sin }^{2}}x} $
$ 0<x\le 1\Rightarrow x\in Q_1\Rightarrow \tan x>x,\cos x>0 $
$ \therefore f’(x)>0 $ for $ 0<x\le 1 $
$ \therefore $ $ f(x) $ is an increasing function. $ g’(x)=\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} $
$ (\sin 2x-2x)’=2\cos 2x-2=2[\cos 2x-1]<0 $
therefore $ \sin 2x-2x $ is decreasing
therefore $ \sin 2x-2x<0 $
$ \therefore $ $ g’(x)<0\Rightarrow g(x) $ is decreasing.
BETA
