Question:

<p><strong>Step 1: Construct the quadratic function</strong></p><p>Since f(x) = 0 is a quadratic with roots at x = -π and x = π:</p><p>$$f(x) = a(x + π)(x - π) = a(x^2 - π^2)$$</p><p><strong>Step 2: Find the coefficient 'a'</strong></p><p>Using the condition f(3π/2) = -3π/4:</p><p>$$f\left(\frac{3π}{2}\right) = a\left(\left(\frac{3π}{2}\right)^2 - π^2\right) = a\left(\frac{9π^2}{4} - π^2\right)$$</p><p>$$= a\left(\frac{9π^2 - 4π^2}{4}\right) = a\left(\frac{5π^2}{4}\right) = -\frac{3π}{4}$$</p><p>$$a = -\frac{3π}{4} \cdot \frac{4}{5π^2} = -\frac{3}{5π}$$</p><p><strong>Step 3: Write the complete function</strong></p><p>$$f(x) = -\frac{3}{5π}(x^2 - π^2)$$</p><p><strong>Step 4: Evaluate the limit</strong></p><p>As x → -π, both numerator and denominator → 0 (indeterminate form 0/0), so apply L'Hôpital's rule:</p><p>$$\lim_{x \to -π} \frac{f(x)}{\sin(\sin x)} = \lim_{x \to -π} \frac{f'(x)}{\frac{d}{dx}[\sin(\sin x)]}$$</p><p>$$f'(x) = -\frac{3}{5π} \cdot 2x = -\frac{6x}{5π}$$</p><p>$$\frac{d}{dx}[\sin(\sin x)] = \cos(\sin x) \cdot \cos x$$</p><p>$$\lim_{x \to -π} \frac{-\frac{6x}{5π}}{\cos(\sin x) \cdot \cos x} = \frac{-\frac{6(-π)}{5π}}{\cos(\sin(-π)) \cdot \cos(-π)}$$</p><p>$$= \frac{\frac{6π}{5π}}{\cos(0) \cdot (-1)} = \frac{\frac{6}{5}}{1 \cdot (-1)} = -\frac{6}{5}$$</p><p>$$∴ \text{Answer: Not specified}$$</p>