Question:

<p>Let <span class="math">y = \frac{\sin x + 4}{\sin x - 2}</span> where <span class="math">x ∈ [\frac{π}{2}, π]</span></p><p>On the interval <span class="math">[\frac{π}{2}, π]</span>, we have <span class="math">\sin x ∈ [0, 1]</span></p><p>For the numerator: <span class="math">\sin x + 4 ∈ [4, 5]</span></p><p>For the denominator: <span class="math">\sin x - 2 ∈ [-2, -1]</span></p><p>Since the numerator is positive and denominator is negative, <span class="math">y</span> is negative.</p><p>At <span class="math">x = \frac{π}{2}</span> (where <span class="math">\sin x = 1</span>): <span class="math">y = \frac{1 + 4}{1 - 2} = \frac{5}{-1} = -5</span></p><p>At <span class="math">x = π</span> (where <span class="math">\sin x = 0</span>): <span class="math">y = \frac{0 + 4}{0 - 2} = \frac{4}{-2} = -2</span></p><p>As <span class="math">\sin x</span> increases from 0 to 1, <span class="math">y</span> decreases from <span class="math">-2</span> to <span class="math">-5</span></p><p>Therefore: <span class="math">A = [-5, -2]</span></p><p>∴ Answer is (d).</p>