Functions Question 605
Question: If $ f(x)=\log
[ \frac{1+x}{1-x} ] $ , then $ f[ \frac{2x}{1+x^{2}} ] $ is equal to [MP PET 1999; RPET 1999; UPSEAT 2003]
Options:
A) $ {{[f(x)]}^{2}} $
B) $ {{[f(x)]}^{3}} $
C) $ 2f(x) $
D) $ 3f(x) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=\log (x+\sqrt{x^{2}+1}) $
$ \therefore ,f( \frac{2x}{1+x^{2}} )=\log ,[ \frac{1+\frac{2x}{1+x^{2}}}{1-\frac{2x}{1+x^{2}}} ]=\log ,[ \frac{x^{2}+1+2x}{x^{2}+1-2x} ] $ $ =\log ,{{[ \frac{1+x}{1-x} ]}^{2}}=2,\log ,[ \frac{1+x}{1-x} ]=2,f(x) $ .
BETA
