JEE Main202225 Jun 2022Evening ShiftMathematicsDefinite IntegrationActual
If b n = ∫ 0 π 2 cos 2 n x sin x d x , n ∈ ℕ , then
Options
- Ab 3 - b 2 , b 4 - b 3 , b 5 - b 4 are in an A.P. with common difference - 2
- B1 b 3 - b 2 , 1 b 4 - b 3 , 1 b 5 - b 4 are in an A.P. with common difference 2
- Cb 3 - b 2 , b 4 - b 3 , b 5 - b 4 are in a G.P.
- D1 b 3 - b 2 , 1 b 4 - b 3 , 1 b 5 - b 4 are in an A.P. with common difference - 2
Correct answer
D. 1 b 3 - b 2 , 1 b 4 - b 3 , 1 b 5 - b 4 are in an A.P. with common difference - 2
Step-by-step solution
Given, b n = ∫ 0 π 2 1 + cos 2 n x sin x d x b n + 1 - b n = ∫ 0 π 2 cos 2 n + 1 x - cos 2 n x sin x d x = ∫ 0 π 2 - sin 2 n + 1 x sin x sin x d x = cos ( 2 n + 1 ) x 2 n + 1 0 π 2 = - 1 2 n + 1 So, 1 b 3 - b 2 = - 5 1 b 4 - b 3 = - 7 1 b 5 - b 4 = - 9 So, 1 b 3 - b 2 , 1 b 4 - b 3 , 1 b 5 - b 4 are in A.P. with c . d = - 2