The function $f(x) = kx - \sin x$ is strictly increasing for:

$k > 1$

The correct answer is Option (1) → $k > 1$ ##

$f(x) = kx - \sin x$ is strictly increasing for all $x \in \mathbb{R}$

$\Rightarrow f'(x) > 0, \forall x \in \mathbb{R}$

$\Rightarrow k - \cos x > 0 \quad [-1 \le \cos \theta \le 1]$

$\Rightarrow k - 1 > 0 \Rightarrow k > 1$