Question

Let $f\left(x\right)=\log |\sec x+\tan x|,g\left(x\right)=x^6{\sin }^2{x}^4.$ If $fog$ and $gof$ are defined, then which of the following option's is correct

• A. $f\left(x\right)$ is even function
• B. $fog$ is even function
• C. $g\left(x\right)$ is odd function
• D. $gof$ is odd function

Answer 1

The correct option is B $fog$ is even function

Given: $f\left(x\right)=\log |\sec x+\tan x|$
$\Rightarrow f(-x)=\log |\sec (-x)+\tan (-x)|$
$\Rightarrow f(-x)=\log |\sec x-\tan x|$
$\Rightarrow f(-x)=\log \left|\frac{1}{\sec x+\tan x}\right|(\because {\sec }^{2}x-{\tan }^{2}x=1)$
$\Rightarrow f(-x)=-\log |\sec x+\tan x|$
$\Rightarrow f(-x)=-f\left(x\right)$
$\Rightarrow f\left(x\right)$ is odd function.
and $g\left(x\right)=x^6{\sin }^2{x}^4$
$\Rightarrow g(-x)=(-x{)}^6{\sin }^2(-x{)}^4=g\left(x\right)$
$\Rightarrow g\left(x\right)$ is even function.

As we know that composition of two functions (which are either odd or even) will be even if atleast one of them is even. so, both $fog$ and $gof$ are even functions.