<p>If <i>f</i> : [1, 10] → [1, 10] is a non-decreasing function and <i>g</i> : [1, 10] → [1, 10] is a non-increasing function. Let <i>h</i>(<i>x</i>) = <i>f</i>(<i>g</i>(<i>x</i>)) with <i>h</i>(1) = 1. Then, <i>h</i>(2)</p>
<p>(a) lies in (1, 2)</p>
<p>(b) is more than two</p>
<p>(c) is equal to one</p>
<p>(d) is not defined</p>
Step-by-Step Solution
Key Concept: Since f is non-decreasing and g is non-increasing, the composition h(x) = f(g(x)) is non-decreasing. Combined with the boundary condition h(1) = 1 and the constraint that h maps [1,10] to [1,10], we can determine h(2) precisely.
<p><strong>Step 1: Analyze the composition h(x) = f(g(x))</strong></p><p>Since g is non-increasing on [1,10], for any x₁ < x₂, we have g(x₁) ≥ g(x₂).</p><p>Since f is non-decreasing on [1,10], for g(x₁) ≥ g(x₂), we have f(g(x₁)) ≥ f(g(x₂)).</p><p>Therefore, h(x₁) ≥ h(x₂), meaning <strong>h is non-increasing</strong>.</p><p><strong>Step 2: Apply the given condition h(1) = 1</strong></p><p>Since h is non-increasing and h(1) = 1, for all x ∈ [1,10], we have:</p><p>h(x) ≤ h(1) = 1</p><p><strong>Step 3: Use the codomain constraint</strong></p><p>We're given that h: [1,10] → [1,10], meaning h(x) ∈ [1,10] for all x ∈ [1,10].</p><p>Combined with h(x) ≤ 1 from Step 2, we get:</p><p>h(x) ∈ [1,10] ∩ (-∞, 1] = {1}</p><p>Therefore, h(x) = 1 for all x ∈ [1,10].</p><p><strong>Step 4: Find h(2)</strong></p><p>Since h(x) = 1 for all x in the domain:</p><p>h(2) = 1</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C