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f(x) = cos λx. If f(1/2) = -1, then determine λ and hence evaluate f(k) for natural number k. Also, ∑(k=1 to 20) 1/[sin k sin(k+1)] and ∑(k=1 to 20) [sin(k+1) - 1]/[sin k·sin(k+1)]

1. λ = 2π and f(k) = cos 2πk
2. λ = π and f(k) = cos πk
3. λ = 4π and f(k) = cos 4πk
4. λ = 0 and f(k) = 1

Correct answer: λ = 2π and f(k) = cos 2πk

Solution

The correct option is right because if f(1/2) = cos(λ/2) = -1, then λ must be 2π, as cos(π) = -1. Consequently, f(k) = cos(2πk) evaluates to 1 for any integer k, confirming the periodic nature of the cosine function.

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