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 function
• 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)

$f\left(x\right)$ is an onto function

$f\left(x\right)=\left(\log \right(\sec x+\tan x){)}^3$
$f\left(x\right)={\left({\log }^{}(\frac{1}{\cos x}+\frac{\sin x}{\cos x})\right)}^3$

$f\left(x\right)={\left({\log }^{}\left(\frac{1+\sin x}{\cos x}\right)\right)}^3$

$f\left(x\right)={\left({\log }^{}\left(\frac{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}\right)\right)}^3$

$f\left(x\right)={\left({\log }^{}\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)\right)}^3$

$f\left(x\right)={\left({\log }^{}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)\right)}^3$

$f\left(x\right)={\left({\log }^{}\left(\tan (\frac{x}{2}+\frac{\pi }{4})\right)\right)}^3$


when $x\in (-\frac{\pi }{2},\frac{\pi }{2}),$
$\tan (\frac{x}{2}+\frac{\pi }{4})\in (0,\infty )$
and from the above graph, clearly $y=\log x$ is one one for $x\in (0,\infty )$ as the graph is continuously increasing.
Hence, $f\left(x\right)$ is a one-one function.

As for $x\in (0,\infty ),\log x\in (-\infty ,\infty )$
Similarly $f\left(x\right)\in (-\infty ,\infty )$
Hence, $f\left(x\right)$ is an onto function.
Now,
$f\left(x\right)=\left(\log \right(\sec x+\tan x){)}^3$
$\Rightarrow f(-x)=\left(\log \right(\sec x-\tan x){)}^3$
$\Rightarrow f(-x)={\left(\log \left(\frac{1}{(\sec x+\tan x)}\right)\right)}^3$
$f(-x)=-\left(\log \right(\sec x+\tan x){)}^3=-f(x)$
Hence, $f\left(x\right)$ is an odd function.