f:R→R is given by,f(x)=∣x∣={x, if x≥0−x, if x<0​It is seen that f(−1)=∣−1∣=1,f(1)=∣1∣=1∴f(−1)=f(1), but −1=1∴f is not one-one.Now, consider −1∈R.It is known that f(x)=∣x∣ is always non-negative. Thus, there does not exist any element x in domain R such that f(x)=∣x∣=−1∴f is not onto.Hence, the modulus function is neither one-one nor onto.