Question

Write whether f : R → R, given by $f\left(x\right)=x+\sqrt{x^2}$, is one-one, many-one, onto or into.

Answer 1

$$\begin{gathered} f\left(x\right)=x+\sqrt{x^2}=x\pm x=0or2x \\ \text{So, each element}x\text{in the domain may contain 2 images.} \\ \text{For example,} \\ f\left(0\right)=0+\sqrt{0^2}=0 \\ f\left(-1\right)=-1+\sqrt{{\left(-1\right)}^2}=-1+\sqrt{1}=-1+1=0 \\ \text{Here, the image of 0 and -1 is 0.} \end{gathered}$$
Hence, f is may-one.