Let f:R→R  be a function. Define g:R→R  by g(x)=|f(x)|  for all x . Then, g  is:

g(x)=|f(x)|≥0 . So, g(x)  cannot be onto. If f(x)  is one-one and f(x1)=−f(x2)  then, g(x1)=g(x2) . So, f(x) is non-one’ does not ensure that g(x)  is one-one.                 If f(x)  is continuous for x∈R ,|f(x)| is also continuous for x∈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(x)  is differentiable at P  while y=|f(x)|  has two tangents at P , i.e., not differentiable at P .