the function f:R gives R, given by f(x)=ln|x+√1+x^2| is even or odd function.?
Answers
Answered by
0
f(x)=ln|x+√(1+x^2)|
put x= -x
f(-x)=ln|-x+√(1+x^2)|
now if we add this
f(x)+f(-x)=ln|x+√(1+x^2)|+ln|-x+√(1+x^2)|
=ln{x+√(1+x^2)}{-x+√(1+x^2)}
use here
(a-b)(a+b)=a^2-b^2
=ln{(√(1+x^2))^2-x^2}
=ln{1+x^2-x^2}
=ln(1)=0
here we see f(x)+f(-x)=0
so f(x) is an odd function
put x= -x
f(-x)=ln|-x+√(1+x^2)|
now if we add this
f(x)+f(-x)=ln|x+√(1+x^2)|+ln|-x+√(1+x^2)|
=ln{x+√(1+x^2)}{-x+√(1+x^2)}
use here
(a-b)(a+b)=a^2-b^2
=ln{(√(1+x^2))^2-x^2}
=ln{1+x^2-x^2}
=ln(1)=0
here we see f(x)+f(-x)=0
so f(x) is an odd function
Similar questions