Explanation

(c)
Since, $f(n) = 1 + n^{2}$
For one-to-one, $1 + n_{1}^{2} = 1 + n_{2}^{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{1}^{2} - n_{2}^{2} = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n_{1} = n_{2}\ \ \ \ \ \ \ \ \ \ (\because n_{1} + n_{2} \neq 0)$
$\therefore f(n)$ is one-to-one.
But $f(n)$is not onto as every element of codomain is not the image of any element of domains.
Hence, $f(n)$ is one-to-one but not onto.