Question

The function f :R → R , given by f (x) = ln | x + 1+ x2 | is neither even nor odd.

Answer 1

f (x ) = ln | x +1 +x^2 |

put x ===> -x

f ( -x ) = ln | -x + 1 + x^2 |

if we add f (x ) and f(-x)

f (x) +f (- x) =ln | x + 1 +x^2| +ln | -x + 1 +x^2|
=ln{(1 +x^2)^2-x^2} isn't equal zero
hence f (x ) isn't odd function

again subtract f (x ) -f ( -x)

f (x) -f ( -x) =ln (x + 1 + x^2)/( - x + 1 + x^2)
isn't equal to zero .

so, f ( x) isn't even function.

hence f (x ) is neither even nor odd