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

Question

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, 0 , ∞, 0 , ∞, 1 , ∞, 1 , ∞

MARKS App · Accepted Answer

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 , ∞

Search any question & find its solution

Looking for more such questions to practice?