<p>The number of real values of \(x\) for which \(\sin(e^x) = 5^x + 5^{-x}\) is</p>
Step-by-Step Solution
Key Concept: The RHS equals 5^x + 5^(-x) ≥ 2 by AM-GM inequality, while LHS sin(e^x) has maximum value 1. Therefore, no solution exists since the ranges don't overlap.
<p><strong>Step 1:</strong> Identify the range of RHS using AM-GM inequality.</p><p>For positive reals a and b: a + b ≥ 2√(ab)</p><p>5^x + 5^(-x) ≥ 2√(5^x · 5^(-x)) = 2√(1) = 2</p><p><strong>Step 2:</strong> Identify the range of LHS.</p><p>Since sin(e^x) is the sine of any real number, we have:<br/>-1 ≤ sin(e^x) ≤ 1</p><p><strong>Step 3:</strong> Compare ranges.</p><p>LHS: sin(e^x) ∈ [-1, 1]<br/>RHS: 5^x + 5^(-x) ∈ [2, ∞)</p><p>The ranges have no overlap. The minimum value of RHS is 2, while the maximum value of LHS is 1.</p><p><strong>Step 4:</strong> Conclusion.</p><p>No value of x can satisfy the equation since the LHS can never equal the RHS.</p><p>∴ <strong>Answer: 0 (zero real values)</strong></p>
Correct Answer: A