Q16: a for which $a(x - \cos x)$ is strictly decreasing in $\mathbb{R}$

$a$ is a constant. The derivative of $x$ is 1 and the derivative of $\cos x$ is $-\sin x$⁠, so

The constant $a$ stays in front: $f'(x) = a(1 + \sin x)$⁠.

Since $\sin x \ge -1$⁠, we get $1 + \sin x \ge 0$⁠. It equals 0 only where $\sin x = -1$⁠, that is at $x = 2n\pi - \frac{\pi}{2}$⁠, single points spread apart.

A negative number times a positive number is negative. A function whose derivative is negative everywhere except at isolated points is still strictly decreasing.

If $a = 0$⁠, then $f(x) = 0$ is constant, and a constant function is not strictly decreasing. If $a > 0$⁠, then $f'(x) \ge 0$⁠, so f is increasing.

So option (a), $\{0\}$⁠, and option (d), all real numbers, are wrong.