IIT JEE , AIEEE Forum - Mathematics , Algebra

hary

pls solve dis f(x)=log(1+x) - log(1-x) and g(x) =log(3x+x^3) - log(1+3x^2)

then f(g(x)) =

-f(x)

3f(x)

[f(x)]^3

-3f(x)

atish22

Is the answer zero???

hary

plsssss help!!!!!!

hary

arey koi nahi hai kya???

studyid

3f(x)

hary

yes da ans is 3f(x) but how pls explain in detail plss

studyid

$f(x) = log (1+x) - log (1-x):\\hence f(x) = log(\frac{1+x}{1-x})$

$similarly\:g(x) = log(\frac{{x}^{3}+3x}{1+3{x}^{2}})$

hence

$f(g(x)) = log(\frac{1+\frac{{x}^{3}+3x}{1+3{x}^{2}}}{1-\frac{{x}^{3}+3x}{1+3{x}^{2}}})$

$so f(g(x)) = log (\frac{{x}^{3}+3{x}^{2}+3x+1}{-{x}^{3}+3{x}^{2}-3x+1})$

$so\:f(g(x)) = log (\frac{(1+x)^{3}}{(1-x)^{3}})\\ = 3log(\frac{1+x}{1-x})\\=3f(x)$

hary

yaa got it thanx