<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: The range is found by tracking how the innermost function sin(x) ∈ [-1,1] transforms through each composition, then applying the outermost sine to the resulting interval.
<p><strong>Step 1:</strong> Find the range of the innermost function.</p><p>Since x ∈ ℝ, we have sin(x) ∈ [-1, 1]</p><p><strong>Step 2:</strong> Apply the next layer.</p><p>Therefore (π/2)sin(x) ∈ [-π/2, π/2]</p><p>So sin((π/2)sin(x)) ∈ [sin(-π/2), sin(π/2)] = [-1, 1]</p><p><strong>Step 3:</strong> Apply the outermost layer.</p><p>Now (π/6)sin((π/2)sin(x)) ∈ [-π/6, π/6]</p><p><strong>Step 4:</strong> Find the final range.</p><p>Since sin is an increasing function on [-π/6, π/6], we have:</p><p>f(x) = sin((π/6)sin((π/2)sin(x))) ∈ [sin(-π/6), sin(π/6)]</p><p>f(x) ∈ [-1/2, 1/2]</p><p>∴ Answer: C (Range is [-1/2, 1/2])</p>
Correct Answer: C