To determine the set A for which the function f: R A defined as f(x) = x2 + 1 is onto, we need to find the range of the function f(x).

The function is given by $f(x) = x^2 + 1$.

For any real number $x$, we know that $x^2 \ge 0$.
Adding 1 to both sides of the inequality, we get:

This means that the minimum value of $f(x)$ is 1, and $f(x)$ can take any value greater than or equal to 1.
Therefore, the range of $f(x)$ is $[1, \infty)$.

For a function $f: \mathbb{R} \to A$ to be onto (or surjective), its codomain $A$ must be equal to its range.
Since the range of $f(x) = x^2 + 1$ is $[1, \infty)$, the set $A$ must be $[1, \infty)$ for the function to be onto.

The options are:

a) $(-\infty, \infty)$

b) $(-\infty, 1]$

c) $[1, \infty)$

d) $[1, \infty)$

The correct option is c) or d).

The final answer is $\boxed{\text{c) } [1, \infty)}$.