It is given that f:R.⟶R. is defined by f(x)=x1​.One-one:​f(x)=f(y)⇒x1​=y1​⇒x=y​af is one-one.Onto:x=y1​∈R.( Exists as y=0)such thatf(x)=(y1​)1​=yaf is onto.Thus, the given function (t) is one-one and onto.Now, consider function g:N→ R.defined byg(x)=x1​We have,g(x1​)=g(x2​)⇒x1​1​=x2​1​⇒x1​=x2​sg is one-one.Further, it is clear that g is not onto as for 1.2∈R. there does not exit any x in N such that g(x)=1.21​Hence, function g is one-one but not onto.