(a.i) Minimum Point Condition:

Step 1: Find the derivative
Using the quotient rule:
\( f'(x) = \frac{\sin x \cdot \frac{1}{2\sqrt{x}} – \sqrt{x} \cos x}{\sin^2 x} \)

Step 2: Set derivative to zero for critical points
\( \frac{\sin x}{2\sqrt{x}} – \sqrt{x} \cos x = 0 \)
\( \frac{\sin x}{2\sqrt{x}} = \sqrt{x} \cos x \)

Step 3: Simplify to required form
\( \frac{\sin x}{\cos x} = 2x \)
\( \tan x = 2x \) ✓

Verification:
The solution \( x \approx 1.17 \) radians (67°) satisfies this equation when checked numerically.

(a.ii) Decreasing Function Interval:

Analysis:
1. The function decreases where \( f'(x) < 0 \)
2. From part (a.i), the minimum occurs at \( x \approx 1.17 \)
3. Testing values:
– At \( x = \frac{\pi}{4} \approx 0.785 \): \( f'(x) < 0 \)
– At \( x = 1.5 \): \( f'(x) > 0 \)

Conclusion:
\( f(x) \) is decreasing for \( 0 < x \leq 1.17 \) radians
(Exact interval: \( 0 < x \leq x_{min} \) where \( x_{min} \) satisfies \( \tan x = 2x \))

(b) Graph Sketch:

Key Features:
1. Asymptotes:
– Vertical asymptote at \( x = \pi \) (denominator → 0)
– Approaches infinity as \( x \to 0^+ \)
2. Minimum Point:
– Located at \( x \approx 1.17 \) (from part a.i)
3. Behavior:
– Concave up throughout domain
– Decreases to minimum, then increases toward \( x = \pi \)

(c) Normal Parallel to y = -x:

Step 1: Find condition for normal
Normal parallel to \( y = -x \) ⇒ slope = -1
Therefore, tangent slope = 1 (since normal slope = -1/m)

Step 2: Solve \( f'(x) = 1 \)
\( \frac{\sin x – 2x \cos x}{2\sqrt{x} \sin^2 x} = 1 \)
Numerical solution gives \( x \approx 1.96 \) radians

Step 3: Find y-coordinate
\( y = f(1.96) = \frac{\sqrt{1.96}}{\sin(1.96)} \approx 1.51 \)

Final Coordinates:
\( (1.96, 1.51) \)

(d) Volume of Revolution:

Disk Method:
\( V = \pi \int_{\pi/6}^{\pi/3} \left( \frac{\sqrt{x}}{\sin x} \right)^2 dx \)
\( = \pi \int_{\pi/6}^{\pi/3} \frac{x}{\sin^2 x} dx \)

Calculation:
Using integration by parts or numerical methods:
\( V \approx 2.68 \) (or \( 0.852\pi \)) cubic units

Verification:
The integral evaluates to approximately 0.852π, which matches the numerical result.