For each real <span class="math-inline">x</span>, let f(x) = \(\max\{x^{2}, x^{3}, x^{4}\}\), then f(x) is -
(A) f(x) = <span class="math-inline">x^{4}</span> for <span class="math-inline">x ≤ -1</span>
(B) f(x) = <span class="math-inline">x^{2}</span> for <span class="math-inline">-1 < x ≤ 0</span>
(C) f(\(\frac{1}{2}\)) = <span class="math-inline">\frac{1}{2}</span>
(D) f(\(\frac{1}{2}\)) = <span class="math-inline">\frac{1}{4}</span>
Step-by-Step Solution
Key Concept: To find f(x) = max{x², x³, x⁴}, we need to compare these three functions across different regions of the real line by analyzing where each function dominates the others.
<p><strong>Step 1: Analyze for x ≤ -1</strong></p><p>For x ≤ -1, all values are negative or x = -1. Let x = -2:</p><ul><li>x² = 4</li><li>x³ = -8</li><li>x⁴ = 16</li></ul><p>Here x⁴ > x² > x³. For all x ≤ -1, we have x⁴ ≥ x² because |x| ≥ 1 implies x⁴ ≥ x². Also x⁴ > x³ since x⁴ is positive and x³ is negative for x < 0. Therefore f(x) = x⁴ for x ≤ -1.</p><p><strong>Step 2: Analyze for -1 < x ≤ 0</strong></p><p>For -1 < x < 0: x² is positive, x³ is negative, x⁴ is positive. Since 0 < |x| < 1, we have |x|⁴ < |x|². This means x⁴ < x². So f(x) = x² for -1 < x ≤ 0.</p><p><strong>Step 3: Analyze for 0 < x < 1</strong></p><p>For 0 < x < 1: all three are positive. Since x < 1, we have x² > x³ and x² > x⁴. Therefore f(x) = x² for 0 < x < 1.</p><p><strong>Step 4: Analyze for x ≥ 1</strong></p><p>For x ≥ 1: x⁴ ≥ x² ≥ x³ (since x⁴ = x² · x² ≥ x² when x ≥ 1). Therefore f(x) = x⁴ for x ≥ 1.</p><p><strong>Step 5: Verify boundary cases</strong></p><p>At x = -1: x² = 1, x³ = -1, x⁴ = 1, so max = 1 = x⁴ ✓</p><p>At x = 0: all equal 0 ✓</p><p>At x = 1: x² = 1, x³ = 1, x⁴ = 1, all equal ✓</p><p><strong>Step 6: Check option (C)</strong></p><p>f(1/2) = max{(1/2)², (1/2)³, (1/2)⁴} = max{1/4, 1/8, 1/16} = 1/4 = (1/2)², not 1/2.</p><p><strong>Step 7: Conclusion</strong></p><p>Option (A) states f(x) = x⁴ for x ≤ -1, which is correct based on our analysis.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A