To determine when \( f(x) \) is decreasing, we must first compute its derivative and identify where it is non-positive.
The given function is: \[ f(x) = \sqrt{3} \sin x - \cos x - 2ax + 6 \] Differentiating \( f(x) \) yields: \[ f'(x) = \frac{d}{dx} \left( \sqrt{3} \sin x - \cos x - 2ax + 6 \right) \] Applying standard differentiation rules (\( \frac{d}{dx}(\sin x) = \cos x \), \( \frac{d}{dx}(\cos x) = -\sin x \), and \( \frac{d}{dx}(x) = 1 \)), we obtain: \[ f'(x) = \sqrt{3} \cos x + \sin x - 2a \] For \( f(x) \) to be decreasing across all \( x \in \mathbb{R} \), we require \( f'(x) \leq 0 \). This implies: \[ \sqrt{3} \cos x + \sin x - 2a \leq 0 \] Since \( \cos x \) and \( \sin x \) are bounded within \([-1, 1]\), the maximum value of \( \sqrt{3} \cos x + \sin x \) is \( \sqrt{3} + 1 \), which occurs when \( \cos x = 1 \) and \( \sin x = 1 \). Therefore, the condition becomes: \[ \sqrt{3} + 1 - 2a \leq 0 \] Rearranging to solve for \( a \): \[ 2a \geq \sqrt{3} + 1 \] \[ a \geq \frac{\sqrt{3} + 1}{2} \] Consequently, for \( f(x) \) to be decreasing for all real numbers, \( a \) must satisfy: \[ a \geq \frac{\sqrt{3} + 1}{2} \]