Question:

<p><strong>Step 1:</strong> Rewrite the given equation by dividing by sin x:</p><p>$$\frac{dy}{dx} + \frac{2\cos x}{\sin x}y = \frac{8}{\sin x}$$</p><p><strong>Step 2:</strong> Identify P(x) = 2cos x/sin x. The integrating factor is:</p><p>$$\mu(x) = e^{\int \frac{2\cos x}{\sin x}dx} = e^{2\ln|\sin x|} = \sin^2 x$$</p><p><strong>Step 3:</strong> Multiply the equation by μ(x) = sin²x:</p><p>$$\sin^2 x\frac{dy}{dx} + 2y\cos x\sin x = 8\sin x$$</p><p>The left side is $\frac{d}{dx}[y\sin^2 x]$:</p><p>$$\frac{d}{dx}[y\sin^2 x] = 8\sin x$$</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>$$y\sin^2 x = \int 8\sin x\,dx = -8\cos x + C$$</p><p><strong>Step 5:</strong> Apply initial condition f(π/2) = 8:</p><p>$$8 \cdot 1 = -8\cos(\pi/2) + C = 0 + C \implies C = 8$$</p><p><strong>Step 6:</strong> General solution:</p><p>$$y = \frac{8 - 8\cos x}{\sin^2 x} = \frac{8(1-\cos x)}{\sin^2 x}$$</p><p><strong>Step 7:</strong> Simplify using $1 - \cos x = 2\sin^2(x/2)$ and $\sin^2 x = 4\sin^2(x/2)\cos^2(x/2)$:</p><p>$$y = \frac{8 \cdot 2\sin^2(x/2)}{4\sin^2(x/2)\cos^2(x/2)} = \frac{4}{\cos^2(x/2)} = 4\sec^2(x/2)$$</p><p><strong>Step 8:</strong> Find minimum. Since sec²(x/2) ≥ 1 with equality when x/2 = 0 (i.e., x = 0):</p><p>$$f_\text{min} = 4 \cdot 1 = 4$$</p><p>∴ Answer: <strong>A (minimum value = 4)</strong></p>