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Answer
f(1) = $$f(1)^2$$ => f(1) = 0 or 1. But, if f(1) = 0, then f(x*1) = f(1) * f(x) = 0, so the function becomes f(x) = 0, which is not the case. So, f(1) = 1
f(2*1/2) = f(2)*f(1/2) => 1 = 1/4 * f(1/2) => f(1/2) = 4
