Question

判断函数f(x)=ln[x+(根号X^2+1)]的奇偶性

Answer 1

因为f(x)=ln[x+(x^2+1)^(1/2)]

所以f(-x)=ln[-x+(x^2+1)^(1/2)]

所以f(x)+f(-x)=ln[x+(x^2+1)^(1/2)]+ln[-x+(x^2+1)^(1/2)]

=ln{[x+(x^2+1)^(1/2)]*[-x+(x^2+1)^(1/2)]}

=ln[(x^2+1)-x^2]

=ln1=0

所以-f(x)=f(-x)

所以函数f(x)=ln[x+(x^2+1)^(1/2)]为奇函数