If $f$ is an even function and g is an odd function, then the function $fog$ is

We have, $fog(-x)=f\left[g\right(-x\left)\right]$

$\begin{array}{l}=\mathrm{f}[-\mathrm{g}(\mathrm{x}\left)\right](\because \mathrm{g}\text{ is odd })\\ =\mathrm{f}\left[\mathrm{g}\right(\mathrm{x}\left)\right](\because \mathrm{f}\text{ is even })\\ =\mathrm{fog}\left(\mathrm{x}\right)\mathrm{\forall }\mathrm{x}\in \mathrm{R}\end{array}$

$\therefore fog$ is an even function