(a)
$\because\ \ f(x) = x + \cos x$
On differentiating, we get $f^{'}(x) = 1 - \sin x$
$f^{'}(x) > 0$ for all values of $x$
($\because\sin x$ is lying between-1 to $+ 1$)
$\therefore\ \ f(x)$ is always increasing
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.