Let f : - π 2 , π 2 → R be given by f x = log ⁡ sec ⁡ x + tan ⁡ x 3 . Then

Step-by-step solution

f x = log ⁡ sec ⁡ x + tan ⁡ x 3 ∀ x ∈ - π 2 , π 2 f - x = - f x , hence f x is odd function Let g x = sec ⁡ x + tan ⁡ x ∀ x ∈ - π 2 , π 2 ⇒ g ′ x = sec ⁡ x sec ⁡ x + tan ⁡ x > 0 ∀ x ∈ - π 2 , π 2 ⇒ g x is one - one function Hence log e ⁡ g x 3 is one - one function. And g x ∈ 0 , ∞ ∀ x ∈ - π 2 , π 2 ⇒ log ⁡ g x ∈ R Hence f x is an onto function.