f: R → R given by $f (x) = x + \sqrt{x^2}$, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → none of these
We have,
$f(x) = x + \sqrt{x^2} = x+|x|=\left\{\begin{matrix}2x,&x≥0\\x-x=0,&x<0\end{matrix}\right.$
Clearly, f is many-one into function.