Question

Let \( f : [0, \frac{\pi}{2}] \to \mathbb{R} \) be defined as \[ f(x) = ax + \beta \sin x, \] where \( a, \beta \in \mathbb{R} \). Let \( f \) have a local minimum at \( x = \frac{\pi}{4} \) with \[ f' \left( \frac{\pi}{4} \right) = -\frac{4}{\sqrt{2}}. \] Then \( 8\sqrt{2a} + 4 \beta \) equals

Hint:

For optimization problems involving local minima, find the critical points using the first derivative and check the second derivative to confirm the type of critical point.

Answer 1

Correct Answer: 4

Step 1: Analyze the function \( f(x) \).
The function is \( f(x) = ax + \beta \sin x \). The derivative of this function is: \[ f'(x) = a + \beta \cos x. \] Since \( f \) has a local minimum at \( x = \frac{\pi}{4} \), we set \( f' \left( \frac{\pi}{4} \right) = 0 \) for the condition of critical points: \[ a + \beta \cos \left( \frac{\pi}{4} \right) = 0. \] Since \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), this gives: \[ a + \beta \cdot \frac{1}{\sqrt{2}} = 0. \] This implies that: \[ a = -\frac{\beta}{\sqrt{2}}. \]
Step 2: Use the second condition.
We are also given that \( f' \left( \frac{\pi}{4} \right) = -\frac{4}{\sqrt{2}} \), which implies that: \[ a + \beta \cdot \frac{1}{\sqrt{2}} = -\frac{4}{\sqrt{2}}. \] Substitute \( a = -\frac{\beta}{\sqrt{2}} \) into this equation: \[ -\frac{\beta}{\sqrt{2}} + \beta \cdot \frac{1}{\sqrt{2}} = -\frac{4}{\sqrt{2}}, \] which simplifies to: \[ \beta = -\frac{8}{\sqrt{2}}. \]
Step 3: Conclusion.
Thus, \( 8\sqrt{2a} + 4\beta = -\frac{8}{\sqrt{2}} \).