Complex Numbers And Quadratic Equations question 629
Question: If $ f(x_1)-f(x_2)=f( \frac{x_1-x_2}{1-x_1x_2} ) $ for then what is $ f(x) $ equal to?
Options:
A) $ \ln ( \frac{1-x}{1+x} ) $
B) $ \ln ( \frac{2+x}{1-x} ) $
C) $ {{\tan }^{-1}}( \frac{1-x}{1+x} ) $
D) $ {{\tan }^{-1}}( \frac{1+x}{1-x} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x_1)-f(x_2)=f( \frac{x_1-x_2}{1-x_1x_2} ) $ $ x_1,x_2\in (-1,1) $ then $ f(x)=\log \frac{(1-x)}{(1+x)} $ $ f(x_1)=\log \frac{1-x_1}{1+x_1} $ $ f(x_2)=\log \frac{1-x_2}{1+x_2} $ $ f(x_1)-f(x_2)=\log \frac{1-x_1}{1+x_1}-\log \frac{1-x_2}{1+x_2} $ $ =\log \frac{(1-x_1)}{(1+x_1)}\times \frac{(1+x_2)}{(1-x_2)} $ $ =\log \frac{(1-x_1+x_2-x_1x_2)}{(1+x_1-x_2-x_1x_2)} $ $ =\log \frac{(1-x_1x_2)-(x_1-x_2)}{(1-x_1x_2)+(x_1-x_2)} $ $ =\log \frac{1-( \frac{x_1-x_2}{1-x_1x_2} )}{1+( \frac{x_1-x_2}{1-x_1x_2} )} $ $ f(x_1)-f(x_2)=f( \frac{x_1-x_2}{1-x_1x_2} ) $
BETA
