NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

If ∫ d x e x - 1 = 2 t a n - 1 f x + C , (where x > 0 and C is the constant of integration) then the range of f x is

Options

A0 , ∞
B0 , ∞
C1 , ∞
D1 , ∞

Correct answer

A. 0 , ∞

Step-by-step solution

Le,t e x - 1 = t 2 ⇒ e x ⋅ d x = 2 t d t ∴ Ι = ∫ 2 t d t t e x = ∫ 2 d t t 2 + 1 = 2 t a n - 1 t + C = 2 t a n - 1 e x - 1 + C ∴ f x = e x - 1 For x ∈ 0 , ∞ ; e x ∈ 1 , ∞ ⇒ e x - 1 ∈ 0 , ∞

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