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Given that tan(2*pi*|sin(theta)|) = cot(2*pi*|cos(theta)|) for theta in R, define f(x) = (|sin(theta)| + |cos(theta)|)^x. Find the value of lim(x->infinity) floor(2/f(x)), where floor(.) denotes the greatest integer function.

1. 0
2. 1
3. 2
4. 3

Correct answer: 1

Solution

From the condition, 2*pi*(|sin(theta)| + |cos(theta)|) = pi/2 + n*pi, giving |sin(theta)| + |cos(theta)| = (2n+1)/4. The value that lies in [1, sqrt(2)] is (2n+1)/4 for appropriate n; specifically n=2 gives 5/4. Then f(x) = (5/4)^x -> infinity, so 2/f(x) -> 0 and floor(2/f(x)) = 0... Recheck: tan(A)=cot(B) => tan(A)=tan(pi/2 - B) => A = pi/2 - B + n*pi => A+B = pi/2 + n*pi. So 2*pi*|sin(theta)| + 2*pi*|cos(theta)| = pi/2 + n*pi => |sin(theta)|+|cos(theta)| = (2n+1)/4. For n=1: 3/4 (less than 1, not achievable). For n=2: 5/4 in [1,sqrt(2)]. So |sin(theta)|+|cos(theta)| = 5/4. f(x) = (5/4)^x -> infinity as x->inf. 2/f(x) -> 0⁺. floor(0⁺) = 0. Answer is 0.

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