Prove that the function f: NN, defined by f(x) = x^{2} + x+1 is one one but not onto. Share with your friends Share 6 Manish Bharti answered this 27 View Full Answer Bhumika Jain answered this f(x) = x^{2}+x+1 Let x_{1}, y_{1} N such that f(x_{1}) = f(y_{1}) x_{1}^{2}+ x_{1}+1 =y_{1}^{2} + y_{1 }+ 1=(x_{1 }- y_{1}) (x_{1}+y_{1}+1) = 0 [As x_{1}+y_{1}+1 for any N] x_{1}=y_{1 }f is one-one function Clearly f(x) = x_{1}^{2} + x+1 > 3 for x E N But f(x) does not assume values 1 and 2 f:N N is not onto function f:N N is not onto functionf(x) = x_{2}+x+1 6