For real numbers x and y, such that xy ≠ 0, the following functions are defined. f(x, y) = root(x + y) if x+y > 0 and f(x, y) = (x + y)^2 otherwise. g(x, y) = (x + y)^2 if x + y ≮ 0 and g(x, y) = –( x + y) otherwise. Find the number of ordered pairs of the form (x, y), such that x, y belongs to Z; – 4 ≤ x ≤ 4; –3 ≤ y ≤ 3 and f(x, y) = g(x, y).

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f(x,y) = $$\sqrt{x+y}$$ if x+y > 0
= $$(x+y)^2$$ if x+y <= 0
g(x,y) = $$(x+y)^2$$ if x+y >= 0
= -(x+y) if x+y < 0

Since – 4 ≤ x ≤ 4; –3 ≤ y ≤ 3, the range of x+y is all integers from -7 to +7
When x+y <= -1: f(x,y) = $$(x+y)^2$$; g(x,y) = -(x+y) => Possible when x+y = -1. Number of solutions = 8 ( (-4,3), (-3,2), and so on)
When x+y = 0, f = 0 and g = 0 => x+y = 0. Number of possible solutions = 9
When x+y > 0, f = $$\sqrt{x+y}$$; g = $$(x+y)^2$$; f = g => x+y = 1. Number of solutions = 8.
So, total number of solutions = 8 + 9 + 8 = 25

Options are : 13, 12, 16, NOT