<p>If \(f(x) = \left(\dfrac{3}{5}\right)^x + \left(\dfrac{4}{5}\right)^x - 1\), \(x \in \mathbb{R}\), then the equation \(f(x) = 0\) has</p>
Step-by-Step Solution
Key Concept: Analyze the monotonicity of f(x) by examining its derivative to determine how many times the function crosses the x-axis. Since f(x) is a sum of decreasing exponential functions, f'(x) is always negative, making f strictly decreasing.
<p><strong>Step 1:</strong> Check boundary behavior.<br>At x = 0: f(0) = (3/5)⁰ + (4/5)⁰ - 1 = 1 + 1 - 1 = 1 > 0<br>As x → ∞: f(x) → 0 + 0 - 1 = -1 < 0</p><p><strong>Step 2:</strong> Verify monotonicity.<br>f'(x) = (3/5)^x·ln(3/5) + (4/5)^x·ln(4/5)<br>Since ln(3/5) < 0 and ln(4/5) < 0, both terms are negative for all x ∈ ℝ<br>Therefore f'(x) < 0 for all x, so f is strictly decreasing.</p><p><strong>Step 3:</strong> Apply Intermediate Value Theorem.<br>Since f is continuous, strictly decreasing, f(0) = 1 > 0, and f(x) → -1 < 0 as x → ∞, by IVT there exists exactly one x₀ ∈ (0, ∞) where f(x₀) = 0.</p><p>∴ The equation f(x) = 0 has <strong>exactly one real solution</strong></p>
Correct Answer: B