If Ι = ∫ t a n - 1 e x e x + e - x d x = t a n - 1 f x 2 2 + C (where C is the constant of integration), then the range of y = f x ∀ x ∈ R is

Question

If Ι = ∫ t a n - 1 e x e x + e - x d x = t a n - 1 f x 2 2 + C (where C is the constant of integration), then the range of y = f x ∀ x ∈ R is, - ∞ , ∞, 0 , ∞, 0 , ∞, - ∞ , 0

MARKS App · Accepted Answer

Given integral = ∫ t a n - 1 e x 1 + e 2 x e x d x Let, t a n - 1 e x = t ⇒ 1 1 + e 2 x ⋅ e x d x = d t ⇒ Ι = ∫ t d t = t 2 2 + C = t a n - 1 e x 2 2 + C Hence, f x = e x ∈ 0 , ∞

Search any question & find its solution

Looking for more such questions to practice?