Exams › JEE Main › General
Let f : R → R be a continuous function satisfying f(x) + f(x + k) = n, for all x ∈ R where k > 0 and n is a positive integer. If dx ) x ( f I nk 4 0 1 ∫ = and dx ) x ( f I k 3 k 2 ∫ − = then
- I 1 + 2I 2 = 4nk
- I 1 + 2I 2 = 2nk
- I 1 + nI 2 = 4n 2 k
- I 1 + nI 2 = 6n 2 k
Correct answer: I 1 + nI 2 = 4n 2 k
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