<p>The number of real solutions for <em>x</em> from <span>\(2\cos\dfrac{x}{2} = 2^x + 2^{-x}\)</span> is ______.</p>
Step-by-Step Solution
Key Concept: Recognize that 2^x + 2^(-x) ≥ 2 by AM-GM inequality (with equality only at x=0), while 2cos(x/2) ≤ 2. Solutions exist only where both sides equal 2, which forces x=0 and cos(0)=1.
<p><strong>Step 1:</strong> Apply AM-GM inequality to the right side.</p><p>By AM-GM: 2^x + 2^(-x) ≥ 2√(2^x · 2^(-x)) = 2√1 = 2</p><p>Equality holds if and only if 2^x = 2^(-x), which gives x = 0.</p><p><strong>Step 2:</strong> Analyze the left side constraints.</p><p>Since -1 ≤ cos(x/2) ≤ 1, we have 2cos(x/2) ≤ 2</p><p><strong>Step 3:</strong> Find intersection of constraints.</p><p>For a solution: 2cos(x/2) = 2^x + 2^(-x)</p><p>Left side: ≤ 2, Right side: ≥ 2</p><p>Both sides equal 2 is the ONLY possibility.</p><p><strong>Step 4:</strong> Verify x = 0.</p><p>At x = 0: 2cos(0) = 2(1) = 2 and 2^0 + 2^0 = 1 + 1 = 2 ✓</p><p><strong>∴ Answer: 1</strong></p>
Correct Answer: 1