If $f (x) = 2 ℓ (x - 2) - x^{2} + 4 x + 1$ , then find the solution set of the inequality $f (x) \geq 0$ .

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Solution

$f (x) = 2 ln (x - 2) - x^{2} + 4 x + 1 f (x) = 2 \times \frac{1}{x - 2} - 2 x + 4 \geq 0 = \frac{2 - 2 (x - 2) x + 4 (x - 2)}{x - 2} \geq 0 = \frac{2 - 2 x^{2} + 4 x + 4 x - 8}{x - 2} \geq 0 = \frac{- 2 x^{2} + 8 x - 6}{x - 2} \geq 0 = \frac{2 x^{2} - 8 x + 6}{x - 2} \leq 0 = \frac{x^{2} - 4 x + 3}{x - 2} \leq 0 = \frac{(x - 1) (x - 3)}{x - 2} \leq 0 ∴ x b e l o n g t o (- \infty, - 1) \cup (2, 3)$

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