<p>Suppose that \( f: \mathbb{R} \to \mathbb{R} \) is a continuous function and satisfies the equation \( f(x)\, f(f(x)) = 1 \) for all \( x \in \mathbb{R} \). Further, if \( f(1000) = 999 \), then which of the following options are necessarily true?</p><ol><li>\( f(500) = \dfrac{1}{500} \)</li><li>\( f(199) = \dfrac{1}{199} \)</li><li>\( f(2000) = \dfrac{1}{2000} \)</li><li>\( f(235) = \dfrac{1}{235} \)</li><li>\( f(1099) = \dfrac{1}{1099} \)</li><li>\( f(x) = \dfrac{1}{x} \; \forall x \in \mathbb{R} - \{0, 1000\} \)</li><li>No such function exists</li></ol><p>Enter the product of the number of all correct options. For example, if correct options are 2 and 3, then enter 6.</p>
Step-by-Step Solution
Key Concept: From f(x)·f(f(x)) = 1, we get f(f(x)) = 1/f(x). Applying f to both sides and using continuity reveals that f can only take values in {x, 1/x} for each x, forcing f(x) = 1/x everywhere except possibly at fixed points.
<p><strong>Step 1:</strong> From the functional equation f(x)·f(f(x)) = 1, we get f(f(x)) = 1/f(x) for all x ∈ ℝ.</p><p><strong>Step 2:</strong> Apply f to both sides: f(f(f(x))) = f(1/f(x)). Also, from f(f(x)) = 1/f(x), apply the functional equation: f(f(x))·f(f(f(x))) = 1, giving (1/f(x))·f(f(f(x))) = 1, so f(f(f(x))) = f(x).</p><p><strong>Step 3:</strong> This means f(1/f(x)) = f(x) for all x. If f is continuous and injective (which follows from the constraint), then 1/f(x) = x, implying f(x) = 1/x.</p><p><strong>Step 4:</strong> Verify f(x) = 1/x satisfies f(x)·f(f(x)) = (1/x)·f(1/x) = (1/x)·x = 1. ✓</p><p><strong>Step 5:</strong> Check consistency with f(1000) = 999: If f(x) = 1/x, then f(1000) = 1/1000 ≠ 999. This is a contradiction.</p><p><strong>Step 6:</strong> Since f(x) = 1/x is the only solution consistent with the functional equation and continuity, but it contradicts f(1000) = 999, no such function exists.</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C