Let $f:R\to R$  be a function. Define $g:R\to R$  by $g\left(x\right)=\left|f\left(x\right)\right|$  for all $x$ . Then, $g$  is:

$g\left(x\right)=\left|f\left(x\right)\right|\ge 0$ . So, $g\left(x\right)$  cannot be onto. If $f\left(x\right)$  is one-one and $f\left(x_1\right)=-f\left(x_2\right)$  then, $g\left(x_1\right)=g\left(x_2\right)$ . So, $f\left(x\right)$ is non-one’ does not ensure that $g\left(x\right)$  is one-one.

If $f\left(x\right)$  is continuous for $x\in R$ ,$\left|f\left(x\right)\right|$ is also continuous for $x\in R$ . This is obvious from the following graphical consideration.

So the answer (c) is correct. The fourth answer (d) is not correct from the above graphs $y=f\left(x\right)$  is differentiable at $P$  while $y=\left|f\left(x\right)\right|$  has two tangents at $P$ , i.e., not differentiable at $P$ .