Question

Which of the following is an odd function?

Answer 1

Given:

$f(x)=x^{3}+5 x^{2}+x$

$f(x)=\sqrt{x}$

$f(x)=x^{2}+x$

$f(x)=-x$

To find:

The odd function.

Solution:

If f(-x) = -f(x), then the function is odd.

If f(-x) = f(x), then the function is even.

Option A: $f(x)=x^{3}+5 x^{2}+x$

Substitute x = -x

$f(-x)=(-x)^{3}+5 (-x)^{2}+(-x)$

          $=-x^3+5x^2-x$

          $=-(x^3-5x^2+x)$

          ≠ - f(x)

It is not odd function.

Option B: $f(x)=\sqrt{x}$

Substitute x = -x

$f(-x)=\sqrt{-x}$

          $=\sqrt{x} i$   (Since $\sqrt{-1} =i$)

          ≠ - f(x)

It is not odd function.

Option C: $f(x)=x^{2}+x$

Substitute x = -x

$f(-x)=(-x)^{2}+(-x)$

          $=x^2-x$

          $=-(-x^2+x)$

          ≠ - f(x)

It is not odd function.

Option D: $f(x)=-x$

Substitute x = -x

$f(-x)=-(-x)$

          $=-f(x)$

It is odd function.

Therefore, f(x) = -x is an odd function.