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OABC is a square where O is the origin and AB = 1. Consider the set of points $$s = {(x_{i},y_{i})}$$ in the square such that $$x_{i}+y_{i}$$≤1. Let $$P (x_{1}, y_{1})$$ and $$Q (x_{2}, y_{2})$$ be two such points. Two operations addition (+) and multiplication (.) on S are defined as
$$P + Q = (x_{1}+x_{2} - x_{1}x_{2},y_{1}y_{2})$$
$$P.Q = (x_{1}x_{2},y_{1}+y_{2} - y_{1}y_{2})$$
Let $$P=\left(x_1,\ y_1\right)$$ and $$Q=\left(x_2,\ y_2\right)$$
$$2P=P+P=(x_1+x_1-x_1^2, y_1^2)=(x_1(2-x_1), y_1^2)$$
Since $$x_1$$ and $$y_1$$ are less than 1, $$x_1(2-x_1)>x_1$$ and $$y_1^2<y_1$$.
As the value of n increases the value of x-coordinate tends towards 1 and the value of y-coordinate tends towards 0.
Thus, $$nP=(1, 0)$$
Similarly, $$nQ=(1, 0)$$
Thus, $$nP+nQ=(1+1-1\times 1, 0\times 0)=(1, 0)$$
Hence, the answer is option B.
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