Question

Which of the following functions is an odd function

• A. $y=x^4-2x^2$
• B. $y=x-x^2$
• C. $y=\cos x$
• D. $y=x-\frac{x^3}{6}+\frac{x^5}{40}$

Answer 1

Answer: (D)

If $f(x)=x^4-2x^2$ then $f(-x)=(-x{)}^4-2(-x{)}^2=$ $x^4-2x^2=f(x)$. Hence $f$ is an even function.
If $u(x)=x-x^2$ then $u(-x)=x-x^2$ so $u$ is neither even nor an odd function.
$p(x)=\cos x,p(-x)=\cos (-x)=\cos x=p(x)$ so $p$ is even function.
If $s(x)=x-\frac{x^3}{6}+\frac{x^5}{40}$ then $s(-x)=-x-\frac{x^3}{6}-\frac{x^5}{40}=$ $-\left(x-\frac{x^3}{6}+\frac{x^5}{40}\right)=-s(x)$, so $s$ is an odd function.