Question

$f\left(x\right)=\{\begin{array}{c}x|x|x\le -1\\ [1+x]+[1-x]-1<x<1\\ -x|x|x\ge 1\end{array}$, then $f\left(x\right)$ is

• A. both even as well as odd function
• B. an even function
• C. an odd function
• D. neither even nor odd function

Answer 1

The correct option is B an even function

$f\left(x\right)=\{\begin{array}{c}-x^{2}x\le -1\\ 0+1-1<x<0\\ 2x=0\\ 1+00<x<1\\ -x^2,x\ge 1\end{array}\Rightarrow f\left(x\right)=\{\begin{array}{c}-x^{2}x\le -1\\ 1-1<x<0\\ 2x=0\\ 10<x<1\\ -x^{2}x\ge 1\end{array}$
This is clearly of the form $f(-x)=f\left(x\right)$
Hence $f\left(x\right)$ is even function.