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

Let f :-π2,π2→R be given by fx=log⁡sec⁡x+tan⁡x3. Then
  1. fx is an odd function
  2. fx is a one - one function
  3. fx is an onto function
  4. fx is an even function

Solution

fx=log⁡sec⁡x+tan⁡x3 ∀ x ∈-π2,π2
f-x= -fx, hence fx is odd function
Let gx=sec⁡x+tan⁡x ∀ x ∈-π2,π2
⇒ g′x=sec⁡xsec⁡x+tan⁡x>0 ∀ x ∈-π2,π2
⇒ gx is one - one function
Hence loge⁡gx3 is one - one function.
And gx ∈0, ∞ ∀ x∈-π2,π2
⇒ log⁡gx ∈R Hence fx is an onto function.

Asked in: JEE Advanced 2014 (Paper 1)

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