IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kislay

determine the nature of function
f(x)= log(x +

(1 +x²)).

ans is odd.pls show the working.

krishna.gopal

For odd function f(-x)=-f(x)
f(-x) =log(sqrt(1+x^2)-x)
We can write sqrt(1+x^2)-x =
(sqrt(1+x^2)-x)*(sqrt(1+x^2)+x)/(sqrt(1+x^2)+x)
=1/(sqrt(1+x^2)+x)
So f(-x)=log(1/(sqrt(1+x^2)+x))=-log(sqrt(1+x^2)+x)=-f(x)

kishan12

a bit simpler method kislay
f(x)+f(-x)=0
this is for odd function...right na.
put values and get ans..always use when log function comes..

joyfrancis

for odd function f(x)=-f(-x) i.e f(x)+f(-x)=0

log(

(1 +x²)+x).+ log(

(1 +x²)-x)

=log (1+x²-x²)

=log1=0

Hence the function is odd.

spideyunlimited

correct.. but according to question u should determine the nature and not prove that nature is odd.

spideyunlimited

find f(x), f(-x) and then see what nature is true :)

kislay

yeah spidy is corrrect.........
gud approach kishan

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