<p>Let \(f(x) = \sin\!\left(\dfrac{\pi}{6}\sin\!\left(\dfrac{\pi}{2}\sin x\right)\right)\) for all \(x \in R\). Then the range of \(f(x)\) is:</p>
<p>(a) \((-0.25,\, 0.5)\)</p>
<p>(b) \((-1,\, 1)\)</p>
<p>(c) \([-0.5,\, 0.5]\)</p>
<p>(d) \((-0.25,\, 0.25)\)</p>
Step-by-Step Solution
Key Concept: To find the range, work from the innermost function outward by determining the range at each composition level: first find the range of sin(x), then apply subsequent transformations, and finally identify which values the outermost sine can actually achieve.
<p><strong>Step 1:</strong> Find the range of the innermost function.</p><p>For <strong>sin(x)</strong>: the range is <strong>[-1, 1]</strong></p><p><strong>Step 2:</strong> Find the range of (π/2)·sin(x).</p><p>When sin(x) ∈ [-1, 1], we have (π/2)·sin(x) ∈ <strong>[-π/2, π/2]</strong></p><p><strong>Step 3:</strong> Find the range of sin((π/2)·sin(x)).</p><p>Since the argument lies in [-π/2, π/2], we have sin((π/2)·sin(x)) ∈ <strong>[-1, 1]</strong></p><p><strong>Step 4:</strong> Find the range of (π/6)·sin((π/2)·sin(x)).</p><p>When sin((π/2)·sin(x)) ∈ [-1, 1], we have (π/6)·sin((π/2)·sin(x)) ∈ <strong>[-π/6, π/6]</strong></p><p><strong>Step 5:</strong> Find the range of the outermost function sin((π/6)·sin((π/2)·sin(x))).</p><p>Since the argument lies in [-π/6, π/6], we have:</p><p>sin((π/6)·sin((π/2)·sin(x))) ∈ <strong>[-sin(π/6), sin(π/6)] = [-1/2, 1/2]</strong></p><p>∴ Answer: <strong>[-1/2, 1/2]</strong> (Option C)</p>
Correct Answer: C