Question

Let A = {x ∈ R : x ≠ 0, −4 ≤ x ≤ 4} and f : A ∈ R be defined by $f\left(x\right)=\frac{\left|x\right|}{x}$for x ∈ A.
Then th (is

(a) [1, −1]
(b) [x : 0 ≤ x ≤ 4]
(c) {1}
(d) {x : −4 ≤ x ≤ 0}

Answer 1

$$\begin{gathered} \mathrm{As},\left|x\right|=\left\{\begin{array}{l}x,x\ge 0\\ -x<0\end{array}\right. \\ \mathrm{So},f\left(x\right)=\frac{x}{\left|x\right|} \\ \mathrm{When}x<0\mathrm{i}.\mathrm{e}.x\in [-4,0) \\ f\left(x\right)=\frac{x}{-x}=-1 \\ \mathrm{and}\mathrm{when}x>0\mathrm{i}.\mathrm{e}.x\in (0,4] \\ f\left(x\right)=\frac{x}{x}=1 \\ \mathrm{So},\mathrm{range}\left(f\right)=\left\{-1,1\right\} \end{gathered}$$
Disclaimer: The question in the book has some error. The solution is created according to the question given in the book.