This function is not an one-one function as for x=a,b and c, the functional value is 0. As f(a) = f(b), but a≠b, it is not an one-one function.
It is a unto function as its range is R. For x=-∞, f(x) can be -∞ and for x=∞, f(x) can be ∞. And since it is a continuous function, all the real numbers are its range.
So the function is not one-one but onto. answer is (a).