Given, function \\\\(f: R\\rightarrow R \\\\) such that \\\\(f(x) = 1+ x^2\\\\), Let A and B be two sets of real numbers.Let \\\\(x_1, x_2 \\in A\\\\) such that \\\\(f(x_1)= f(x_2)\\\\).\\\\(\\Rightarrow 1+ x_1^2= 1+ x_2^2\\Rightarrow x_1^2- x_2^2=0\\Rightarrow (x_1-x_2)(x_1+x_2)=0\\\\)\\\\(\\Rightarrow x_1 =\\pm x_2.\\\\) Thus \\\\(f(x_1)= f(x_2)\\\\) does not imply that \\\\(x_1= x_2\\\\).For instance, \\\\(f(1) = f(-1) =2\\\\), i.e. , two elements (1, -1) of A have the same image in B. So, \\\\(f\\\\) is many-one function.Now, \\\\(y= 1+x^2\\Rightarrow x= \\sqrt{y-1} \\Rightarrow \\\\)elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, \\\\(f\\\\) is not onto.Hence, \\\\(f\\\\) is neither one-one onto. So, it is not bijective.