solution:
f : R → R, given by f(x) = | x |, characterized as
f contains values like (- 1, 1),(1, 1),(- 2, 2)(2,2)
f(- 1) = f(1), yet – 1
f isn’t one-one.
R contains some adverse numbers which are not pictures of any genuine number since f(x) = |x| is consistently non-negative. So f isn’t onto.
Henceforth, Modulus Function is neither one-one nor onto.