A) \[\left( 0,\frac{\pi }{2} \right)\]
B) \[\left( \frac{\pi }{2},\pi \right)\]
C) \[\left( \pi ,\frac{3\pi }{2} \right)\]
D) \[\left( \frac{3\pi }{2},2\pi \right)\]
Correct Answer: C
Solution :
Let f(x) = tan x - x We know for \[0<x<\frac{\pi }{2}\Rightarrow \tan x>x\] \[\therefore \]\[f(x)=\tan x-x\]has no root in\[\left( 0,\frac{\pi }{2} \right)\]for \[\frac{\pi }{2}<x<\pi ,\tan x\]is negative. \[\therefore \]\[f(x)=\tan x-x<0\] So, \[f(x)=0\] has no root in \[\left( \frac{\pi }{2},\pi \right)\]. For\[\frac{3\pi }{2}<x<2\pi ,\tan x\]is negative \[\therefore \]\[f(x)=\tan x-x<0\] So, f(x) = 0 has no root in \[\left( \frac{3\pi }{2},2\pi \right).\] We have, \[f\pi =0-\pi <0\]and \[f\left( \frac{3\pi }{2} \right)=\tan \frac{3\pi }{2}-\frac{3\pi }{2}>0\] \[\therefore \]f(x) = 0 has at least one root between \[\pi \]and \[\frac{\pi 3}{2}.\]
You need to login to perform this action.
You will be redirected in
3 sec
