NTA Abhyas JEE Main2020MathematicsDefinite IntegrationPractice
Consider Ι 1 = ∫ π 4 π 2 e sin ⁡ x + 1 e cos ⁡ x + 1 d x and Ι 2 = ∫ π 4 π 2 e cos ⁡ x + 1 e sin ⁡ x + 1 d x , then
Options
- AΙ 1 > Ι 2
- BΙ 1 < Ι 2
- CΙ 1 = Ι 2
- DΙ 1 + Ι 2 = 0
Correct answer
A. Ι 1 > Ι 2
Step-by-step solution
As sin ⁡ x > cos ⁡ x ,   ∀ x ∈ π 4 , π 2 ⇒ e sin ⁡ x > e cos ⁡ x ∴ e sin ⁡ x + 1 e cos ⁡ x + 1 > e cos ⁡ x + 1 e sin ⁡ x + 1 ⇒ ∫ π 4 π 2 e sin ⁡ x + 1 e cos ⁡ x + 1 d x > ∫ π 4 π 2 e cos ⁡ x + 1 e sin ⁡ x + 1 ⇒ Ι 1 > Ι 2 > 0