Let $f:\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \rightarrow \mathbb{R}$ be given by \[ f(x)=(\log (\sec x+\tan x))^{3} . \] Then

(A) $f(x)$ is an odd function; (B) $f(x)$ is a one-one function; (C) $f(x)$ is an onto function

(D) $f(x)$ is an even function

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Relations And Functions XII

Let

f : (- \frac{π}{2}, \frac{π}{2}) \to R

be given by

f (x) = (lo g (sec x + tan x))^{3} .

Then

Answer: (A), (B), (C)

First,

f (- x) = (lo g (sec x - tan x))^{3} = (lo g (\frac{1}{s e c x + t a n x}))^{3} = (- lo g (sec x + tan x))^{3} = - f (x)

, so

f

is odd and (A) is correct. Next,

f^{'} (x) = 3 (lo g (sec x + tan x))^{2} \cdot \frac{s e c x t a n x + s e c ^{2} x}{s e c x + t a n x} = 3 (lo g (sec x + tan x))^{2} sec x > 0

for all

x \in (- \frac{π}{2}, \frac{π}{2})

, so

f

is strictly increasing, hence one-one, making (B) correct. As

x \to \frac{π}{2}^{-}

sec x + tan x \to \infty

, so

f (x) \to \infty

. As

x \to - \frac{π}{2}^{+}

f (x) \to - \infty

. Since

f

is continuous, the range is

R

, so

f

is onto and (C) is correct.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options