If a^2+b^2-a b-a-b+1 ≤ 0, a, b ≤ R^+, then a+b is equal to:

Given that $a^2+b^2-a b-a-b+1 \leq 0$

$\Rightarrow 2 a^2+2 b^2-2 a b-2 a-2 b+2 \leq 0$

$(a-b)^2+(a-1)^2+(b-1)^2 \leq 0$

⇒ a = 1 and b = 1

This is because of the fact that the square of a real number cannot be negative

Hence (2) is the correct answer.