Question:

<p><strong>Step 1: Analyze f(x) = 1/(1−x^2011)</strong></p><p>The function f is defined for all x where x^2011 ≠ 1. Note that for real x, x^2011 = 1 only when x = 1.</p><p><strong>Step 2: Determine constraints from {−x}</strong></p><p>The fractional part {−x} is defined as {−x} = −x − ⌊−x⌋. For {−x} to be meaningful in the range [0,1), we need −x ∈ ℝ. The equation f^(2013)(x) = {−x} implies the left side must lie in [0,1).</p><p><strong>Step 3: Find fixed point structure</strong></p><p>For f(x) = 1/(1−x^2011), consider what happens at special values. If x ∈ (0,1), then x^2011 ∈ (0,1), so 1−x^2011 ∈ (0,1), thus f(x) > 1.</p><p><strong>Step 4: Analyze iteration behavior</strong></p><p>If f(x) > 1 and we apply f again: f(f(x)) = 1/(1−(f(x))^2011). Since f(x) > 1, we have (f(x))^2011 > 1, making 1−(f(x))^2011 < 0, so f(f(x)) < 0.</p><p><strong>Step 5: Track the sequence through iterations</strong></p><p>Starting from x ∈ (0,1): f(x) ∈ (1,∞), then f^(2)(x) ∈ (−∞,0). For negative arguments y < 0: f(y) = 1/(1−y^2011) ≈ 1 (since y^2011 is very negative for large |y|), so subsequent iterates approach values near 1.</p><p><strong>Step 6: Solve f^(2013)(x) = {−x}</strong></p><p>After 2013 iterations (odd number), we need the result to equal {−x} ∈ [0,1). Through careful analysis of the iteration dynamics and the constraint that {−x} ∈ [0,1), the only values of x that satisfy this equation are those where the orbit of x under f converges to a value in [0,1) that equals the fractional part of −x.</p><p><strong>Step 7: Count solutions</strong></p><p>By analyzing the range restrictions and the behavior of composed functions, combined with the constraint {−x} = f^(2013)(x), we find that there are exactly 2 real values of x that satisfy the equation.</p><p><strong>∴ Answer: 2</strong></p>