Question

Let $f\left(x\right)=x+|x-100|-|x+100|$ and $g\left(x\right)=|f\left(x\right)|-1,$ then

• A. $f\left(x\right)$ is neither even nor odd
• B. $g\left(x\right)=0$ has $6$ solutions
• C. $g\left(x\right)$ is an even function
• D. $f\left(x\right)$ is an odd function

Answer 1

Answer: (B), (C), (D)

$f\left(x\right)$ is an odd function

$f\left(x\right)=x+|x-100|-|x+100|\Rightarrow f\left(x\right)=\{\begin{array}{c}200+x;x<-100\\ -x;-100\le x\le 100\\ x-200;x>100\end{array}$
Graph of $f\left(x\right)$


Clearly, $f\left(x\right)$ is an odd function.
Now, $g\left(x\right)=|f\left(x\right)|-1$
Graph of $|f\left(x\right)|$ is


Graph of $g\left(x\right)$ is


It can be observed that $g\left(x\right)$ is symmetrical with respect to $y-$ axis. So $g\left(x\right)$ is even function.
Therefore, $g\left(x\right)=0$ has $6$ solutions