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The minimum value of $f (x) = \frac{x^{2} - 2 x + 3}{x^{2} - 4 x + 7}$ is

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$\frac{3 - \sqrt{3}}{3}$

$3 - \frac{1}{\sqrt{3}}$

$2 - \frac{1}{\sqrt{3}}$

$1 + \frac{1}{\sqrt{3}}$

answer is B.

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Detailed Solution

$\frac{x^{2} - 2 x + 3}{x^{2} - 4 x + 7} = y$

$x^{2} - 2 x + 3 = y (x^{2} - 4 x + 7)$

$x^{2} - 2 x + 3 - y (x^{2} - 4 x + 7) = 0$

$x^{2} - 2 x + 3 - y x^{2} + 4 y x - 7 y = 0$

$(1 - y) x^{2} + (4 y - 2) x + (3 - 7 y) = 0$

$D \geq 0$

${(4 y - 2)}^{2} - 4 (3 - 7 y) (1 - y) \geq 0$

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