Explanation

(a)
Given, $f(x) = \cos{x + \cos{\sqrt{2}x}}$
$\Rightarrow \ \ \ \ \ \ \ f^{'}(x) = - \sin x - \sqrt{2}\sin\sqrt{2}x$
$\Rightarrow \ \ \ \ f^{''}(x) = - \cos x - 2\cos\sqrt{2}x$
For extremum value, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ - \sin x - \sqrt{2}\sin\sqrt{2}x = 0\ \ \ \ \ \Rightarrow \ \ \ \ x = 0$
$\therefore$ At $x = 0,\ f^{''}(x) < 0$, maxima
$\therefore\ \ f(x)$ Is maximum only once.