Answer:
Option A,B,C
Explanation:
Plan
(i) f '(x)>0, $\forall x \in(a,b)$ , then f(x) is an increasing function in (a,b) and thus f(x) is one-one function in (a,b)
(ii) if range of f(x)= codomain of f(x) , then f(x) is an onto function.
(iii) A function f(x) is said to be odd function , if f(-x)= -f(x) $\forall x \in R$, i.e, f(-x)+f(x)=0, $\forall x \in R$,
$f(x)=[ln(\sec x+\tan x)]^{3}$
$f'(x)=\frac{3 [ln(\sec x+\tan x)]^{2}(\sec x \tan x+\sec^{2}x)}{(\sec x+\tan x)}$
$f'(x)=3\sec x[ln(\sec x+\tan x)]^{2}>0,$
$\forall x \in(-\frac{\pi}{2},\frac{\pi}{2})$
f(x) is an increasing function.
$\therefore$ f(x) is an one-one function.
$(\sec x+\tan x)=\tan (\frac{\pi}{4}+\frac{x}{2})$
as $x \epsilon (-\frac{\pi}{2},\frac{\pi}{2})$ then
$0< \tan (\frac{\pi}{4}+\frac{x}{2})< \infty$
0 < sec x +tan x < $\infty$
$\Rightarrow$ -$\infty$ < (ln (sec x+tan x) <$\infty$
-$\infty < [ln(sec x+tan x)]^{3} < \infty$
$\Rightarrow$ - $\infty$ < f(x0< $\infty$
Range of f(x) is R and thus f(x) is an onto function.
$f(-x)= [ln(\sec x-\tan x)]^{3}$
= $\left[ln\left(\frac{1}{\sec x+\tan x}\right)\right]^{3}$
$f(-x)=-[ln(\sec x-\tan x)]^{3}$
f(x)+f(-x)=0
$\Rightarrow$ f(x) is an odd function.
