Question

Which of the following real valued function is/are not even functions?

• A. $f\left(x\right)=x^2\cos x$
• B. $f\left(x\right)=x^3\sin x$
• C. $f\left(x\right)=e^x{x}^3\sin x$
• D. $f\left(x\right)=x-\left[x\right]$, where $\left[x\right]$ denotes the greatest integer less than or equal to $x$

Answer 1

Answer: (C), (D)

We know that, if $f(-x)=f\left(x\right)$ then function is even and if $f(-x)=-f\left(x\right)$ then function is odd
(a) $f\left(x\right)=x^3\sin x$
$f(-x)={(-x)}^3\sin -x$
$-x^3(-\sin x)$
$=x^3\sin x=f\left(x\right)$
So, $f\left(x\right)$ is even

(b) $f\left(x\right)=x^2\cos x$
$f(-x)={(-x)}^2\cos -x=x^2\cos x=f\left(x\right)$
So, $f\left(x\right)$ is even

(c) $f\left(x\right)=e^x{x}^3\sin x$
$f(-x)=e^{-x}{(-x)}^3\sin -x$
$=e^{-x}{x}^3\sin x\ne f\left(x\right)$
$\therefore$ $f\left(x\right)$ is not even

(d)$f\left(x\right)=x-\left[x\right]$
$f(-x)=(-x)-[-x]\ne f\left(x\right)$
$\therefore$ $f\left(x\right)$ is not even.