Question

Find out whether the given function is even, odd or neither even nor odd, where
$f\left(x\right)=\{\begin{array}{cc}x|x|,& x\le -1\\ [1+x]+[1-x],& -1<x<1\\ -x|x|,& x\ge 1\end{array}$
where $||$ and $\left[\right]$ represent the modulus and greatest integral functions.

• A. $f\left(x\right)$ is even
• B. $f\left(x\right)$ is odd
• C. $f\left(x\right)$ is neither even nor odd
• D. $f\left(x\right)$ is both even and odd

Answer 1

The correct option is A $f\left(x\right)$ is even

The given function can be written as
$f\left(x\right)=\{\begin{array}{ccc}-x^2& ,& x\le -1\\ 2+\left[x\right]+[-x]& ,& -1<x<1\\ -x^2& ,& x\ge 1\end{array}$
(Using definition of modulus and the greatest integral functions)
Now,
$f(-x)=\{\begin{array}{ccc}-x^2& ,& x\le -1\\ 2+[-x]+\left[x\right]& ,& -1<x<1\\ -x^2& ,& x\ge 1\end{array}$
$\Rightarrow f(-x)=f\left(x\right)$
Hence f is an even function