(iii) Let $ f:R \to R$ defined by $f(x)=1+x^2$
Let $x_1 x_2 \in R $ such that $f(x_1)=f(x_2)$
Since $f(x_1)=f(x_2)$ does not imply $x_1=x_2$
Hence f is not one-one
Consider an element -2 in codomain R.
We see that there does not exists any $ x \in R$
f defined by $ R \to R\; f(x)=1+x^2$ is not one one and not onto