Question

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

• A. $f\left(x\right)$ is an odd funtion
• B. $f\left(x\right)$ is a one-one function
• C. $f\left(x\right)$ is an onto function
• D. $f\left(x\right)$ is an even function

Answer 1

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

The correct options are
A $f\left(x\right)$ is an odd funtion
B $f\left(x\right)$ is a one-one function
C $f\left(x\right)$ is an onto function

$f\left(x\right)=\left(\log \right(\sec x+\tan x){)}^3\forall x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$f(-x)=-f\left(x\right)$, hence $f\left(x\right)$ is odd function
Let $g\left(x\right)=\sec x+\tan x\forall x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$\Rightarrow g^{\prime }\left(x\right)=\sec x(\sec x+\tan x)>0\forall x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$\Rightarrow g\left(x\right)$ is one-one function
Hence $\left({\log }_e\right(g\left(x\right)){)}^3$ is one-one function.
and $g\left(x\right)\in (0,\alpha )\forall x\in (-\frac{\pi }{2},\frac{\pi }{2})$
$\Rightarrow \log \left(g\right(x\left)\right)\in R$. Hence $f\left(x\right)$ is an onto function.