The values of $x$ for which the logarithmic function $f (x) = {log}_{| - x^{2} + 2 x - 4 |} (x (3 - x))$ is defined, is

(0, 3)

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(0, 3) - {1}

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R^{+} - {1}

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(- \infty, 0) \cup (3, \infty)

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Solution

The correct option is A $(0, 3)$
$(1) : x (3 - x) > 0 \Rightarrow x (x - 3) < 0 \Rightarrow x \in (0, 3)$
$(2) : | x^{2} - 2 x + 4 | > 0 \Rightarrow x \in R (3) : | x^{2} - 2 x + 4 | \neq 1$
$\Rightarrow | (x - 1)^{2} + 3 | \neq 1$
The minimum value of $(x - 1)^{2} + 3$ is $3.$
Hence, $| (x - 1)^{2} + 3 | \neq 1 \forall x \in R$

$(1) \cap (2) \cap (3) ∴ x \in (0, 3)$

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