JEE Main202228 Jun 2022Evening ShiftMathematicsDefinite IntegrationActual
Let f : R → R be continuous function satisfying f x + f x + k = n , for all x ∈ R where k > 0 and n is a positive integer. If I 1 = ∫ 0 4 n k f x d x and I 2 = ∫ - k 3 k f x d x , then
Options
- AI 1 + 2 I 2 = 4 n k
- BI 1 + 2 I 2 = 2 n k
- CI 1 + n I 2 = 4 n 2   K
- DI 1 + n I 2 = 6 n 2 k
Correct answer
C. I 1 + n I 2 = 4 n 2   K
Step-by-step solution
Given, f x + f x + k = n ....(i) Replacing x → x + k in above equation we get, f x + k + f x + 2 k = n .....(ii) Subtraction equation (i) from equation (ii) we get, ⇒ f x = f x + 2 k f x is periodic with period 2 k Now, I 1 = ∫ 0 4 n k f x d x = 2 n ∫ 0 2 k f x d x I 2 = ∫ - k 3 k f x d x = 2 ∫ 0 2 k f x d x Now integrating both side of f x + f x + k = n with limit ∫ 0 k we get, ⇒ ∫ 0 k f x d x + ∫ 0 k f x + k d x = n k ⇒ ∫ 0 k f x d x + ∫