Question

Find the value of \( a \) for which } \( f(x) = \sqrt{3} \sin x - \cos x - 2ax + 6 \) \(\text{ is decreasing in } \mathbb{R}\).

Hint:

Quick Tip: For a function to be decreasing, its derivative must be non-positive. Check the maximum and minimum values of the derivative expression to find the required conditions on parameters.

Answer 1

To determine when \( f(x) \) is decreasing, we need to find the derivative of \( f(x) \) and analyze where it is negative.
The given function is: \[ f(x) = \sqrt{3} \sin x - \cos x - 2ax + 6 \] First, differentiate \( f(x) \): \[ f'(x) = \frac{d}{dx} \left( \sqrt{3} \sin x - \cos x - 2ax + 6 \right) \] Using the derivative formulas \( \frac{d}{dx}(\sin x) = \cos x \), \( \frac{d}{dx}(\cos x) = -\sin x \), and \( \frac{d}{dx}(x) = 1 \), we get: \[ f'(x) = \sqrt{3} \cos x + \sin x - 2a \] For the function to be decreasing, we require \( f'(x) \leq 0 \) for all \( x \in \mathbb{R} \). Hence, we need: \[ \sqrt{3} \cos x + \sin x - 2a \leq 0 \] Now, since \( \cos x \) and \( \sin x \) are bounded between -1 and 1, the maximum value of \( \sqrt{3} \cos x + \sin x \) occurs when \( \cos x = 1 \) and \( \sin x = 1 \), which gives the maximum value of: \[ \sqrt{3} + 1 \] Therefore, we need: \[ \sqrt{3} + 1 - 2a \leq 0 \] Solving for \( a \): \[ 2a \geq \sqrt{3} + 1 \] \[ a \geq \frac{\sqrt{3} + 1}{2} \] Thus, for the function \( f(x) \) to be decreasing in \( \mathbb{R} \), the value of \( a \) must satisfy: \[ a \geq \frac{\sqrt{3} + 1}{2} \]