Question

lf $f\left(x\right)=\{\begin{array}{cc}|x|\text{if}-3<x\le -1& \\ -|x|\text{if}-1<x<1& \\ 1-|-x|\text{if}1\le x<3& \end{array}$ Then $\frac{x}{f\left(x\right)}\ge 0$ is true for

• A. $[1,3]$
• B. $[-1,3]$
• C. $(-1,3]$
• D. $[-3,0)$

Answer 1

Answer: (A)

$[1,3]$

Given,

$f\left(x\right)=\{\begin{array}{cc}|x|& -3<x\le -1\\ -|x|& -1<x<1\\ 1-|-x|& 1\le x<3\end{array}$

$\frac{x}{f\left(x\right)}$ for $-3<x\le -1$

$=\frac{x}{|x|}\to x\in 1$

$\frac{x}{f\left(x\right)}$ for $-1<x<1$

$=\frac{x}{-|x|}\to x\in -1$

$\frac{x}{f\left(x\right)}$ for $1\le x<3$

$=\frac{x}{1-|-x|}\to x\in 3$

Since,

$\frac{x}{f\left(x\right)}\ge 0$

$\therefore \frac{x}{f\left(x\right)}\in [1,3]$