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The solution set of ${(x - 2)}^{x^{2} - 6 x + 8} > 1$ is

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$(2, \infty)$

$(2, 3) \cup (4, \infty)$

$(4, 5) \cup (5, \infty)$

$(2, 3) \cup (4, 5)$

answer is B.

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Detailed Solution

$Case(i) : x - 2 > 1 Case(ii) : x - 2 < 1$

$\begin{array}{l} x^{2} - 6 x + 8 > 0 (o r) x^{2} - 6 x + 8 < 0 \\ \Rightarrow x \in (4, \infty) \Rightarrow x \in (2, 3) \end{array}$

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