Functions Question 664
Question: If $ f(x_1)-f(x_2)=f( \frac{x_1-x_2}{1-x_1x_2} ) $ for $ x_1,x_2\in
[-1,,1] $ , then $ f(x) $ is constant [Roorkee 1998]
Options:
A) $ \log \frac{(1-x)}{(1+x)} $
B) $ {{\tan }^{-1}}\frac{(1-x)}{(1+x)} $
C) $ \log \frac{(1+x)}{(1-x)} $
D) $ {{\tan }^{-1}}\frac{(1+x)}{(1-x)} $
Show Answer
Answer:
Correct Answer: B
Solution:
When $ x_1=-1 $ and $ x_2=1 $ , then $ f(-1)-f(1)=f[ \frac{-1-1}{1+1} ]=f(-1)\Rightarrow f(1)=0 $ Which is satisfied when $ f(x)={{\tan }^{-1}}( \frac{1-x}{1+x} ) $ When $ x_1=x_2=0 $ , then $ f(0)-f(0)=f[ \frac{0-0}{1-0} ]=f(0)\Rightarrow f(0)=0 $ When $ x_1=-1 $ and $ x_2=0 $ then $ f(-1)-f(0)=f( \frac{-1-0}{1-0} )=f(-1)\Rightarrow f(0)=0 $ Which is satisfied when $ f(x)=\log ( \frac{1-x}{1+x} ) $ and $ f(x)=\log ( \frac{1+x}{1-x} ) $ .
BETA
