Explanation:

Let’s break down the function:

Given: \( x = \frac{\sin^2A + \sin A + 1}{\sin A} \) for \( 0 < A \leq \frac{\pi}{2} \)

Let’s analyze how x varies as A changes:

- Substitute \( y = \sin A \), where \( 0 < y \leq 1 \)

- Now, \( x = \frac{y^2 + y + 1}{y} = y + 1 + \frac{1}{y} \)

- As y increases from 0 to 1, \( x = y + 1 + \frac{1}{y} \)

- The minimum for \( y + \frac{1}{y} \) for \( 0 < y \leq 1 \) is at \( y = 1 \)

- For \( y = 1 \) (\( \sin A = 1 \implies A = \frac{\pi}{2} \)), x is minimized.

Now check all options:

- Option 1: \( A = \frac{\pi}{6} \): \( \sin A = \frac{1}{2} \), x = 0.5 + 1 + 2 = 3.5

- Option 2: \( A = \frac{\pi}{4} \): \( \sin A = \frac{\sqrt{2}}{2} \), x ˜ 0.7071 + 1 + 1.4142 ˜ 3.12

- Option 3: \( A = \frac{\pi}{3} \): \( \sin A = \frac{\sqrt{3}}{2} \), x ˜ 0.8660 + 1 + 1.1547 ˜ 3.02

- Option 4: \( A = \frac{\pi}{2} \): \( \sin A = 1 \), x = 1 + 1 + 1 = 3

Summary of Options:

- Option 1: 3.5

- Option 2: 3.12

- Option 3: 3.02

- Option 4: 3 (Minimum value)

Correct Option:

- Option 4: \( \frac{\pi}{2} \) is the correct and minimum point.