Functions Question 545
Question: Let $ f:(-1,1)\to B $ , be a function defined by $ f(x)={{\tan }^{-1}}\frac{2x}{1-x^{2}}, $ then f is both one- one and onto when B is the interval
[AIEEE 2005]
Options:
A) $ [ -\frac{\pi }{2},\frac{\pi }{2} ] $
B) $ ( -\frac{\pi }{2},\frac{\pi }{2} ) $
C) $ ( 0,\frac{\pi }{2} ) $
D) $ [ 0,\frac{\pi }{2} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
For ? 1< x< 1, $ {{\tan }^{-1}}\frac{2x}{1-x^{2}}=2{{\tan }^{-1}}x $ Range of $ f(x)=( -\frac{\pi }{2},\frac{\pi }{2} ) $ . \Co-domain of function = B $ =( -\frac{\pi }{2},\frac{\pi }{2} ) $ .
BETA
