Question

Let $F\left(x\right)={\left[\frac{g\left(x\right)-g(-x)}{f\left(x\right)+f(-x)}\right]}^m$ such that $m=2n,n\in N$ and $f(-x)\ne -f\left(x\right);g(-x)\ne -g\left(x\right)$ then $F\left(x\right)$ is

• A. an odd function
• B. an even function
• C. both even and odd function
• D. neither odd nor even function

Answer 1

The correct option is B an even function

$F\left(x\right)={\left[\frac{g\left(x\right)-g(-x)}{f\left(x\right)+f(-x)}\right]}^m$
$\Rightarrow F(-x)={\left[\frac{g(-x)-g\left(x\right)}{f(-x)+f\left(x\right)}\right]}^m$
$\Rightarrow F(-x)=(-1{)}^m{\left[\frac{g\left(x\right)-g(-x)}{f\left(x\right)+f(-x)}\right]}^m$
$\Rightarrow F(-x)={\left[\frac{g\left(x\right)-g(-x)}{f\left(x\right)+f(-x)}\right]}^m=F\left(x\right)(\because m=2n)$
$\therefore F\left(x\right)$ is an even function.