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Question Number 10981 by ridwan balatif last updated on 05/Mar/17

Commented by prakash jain last updated on 06/Mar/17

$(x^{2} - 2 x + 4) = x^{2} - 2 x + 1 + 3 = {(x - 1)}^{2} + 3 ⩾ 3 > 0$ $∣ 3 x - 6 ∣⩾ 0$ $\frac{∣ 3 x - 6 ∣}{x^{2} - 2 x + 4} ⩾ 0 (= 0 only at x = 2)$
	Answered by Mechas88 last updated on 05/Mar/17

	$\frac{0}{-}$ $\frac{0}{+}$ $\frac{+}{-}$ $\frac{-}{+}$ $∣ 3 x - 6 ∣= 0$ $x = 2$ $x^{2} - 2 x + 4 < 0$ $(i x + \sqrt{3} - i) (- i x + \sqrt{3} + i) < 0$ $i x < i - \sqrt{3}$ $x < \frac{i - \sqrt{3}}{i}$ $- i x < - i - \sqrt{3}$ $x > \frac{i + \sqrt{3}}{i}$ $x > \frac{i - \sqrt{3}}{i}$ $x < \frac{i + \sqrt{3}}{i}$ $3 x - 6 > 0$ $x > 2$ $3 x - 6 < 0$ $x < 2$ $R t a : A$
	Commented by prakash jain last updated on 06/Mar/17

	$The question has ∣ 3 x - 6 ∣ which is always$ $+ ve even for x < 2 .$
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