Question: 30. Let Q2 = f(Q1) be the absolute maximum and Q4 = f(Q3) the absolute minimum of f(x) = 513 + (22.5)x2 + (30)x +

30. Let Q2 = f(Q1) be the absolute maximum and Q4 = f(Q3) the absolute minimum of f(x) = 513 + (22.5)x2 + (30)x + (-8) for x in [-5, 0]. Let Q = In(3 + 1Q1| + 2|Q2/ + 3 Q3| + 41Q41). Then T = 5 sin (100Q) satisfies:- (A) 0

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