The number of integral value(s) of $x$ , that disobey the inequality ${log}_{2} | 4 - 5 x | > 2$ , is

0

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1

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2

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infinitely many

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Solution

The correct option is C $2$
${log}_{2} | 4 - 5 x | > 2$
$\Rightarrow | 4 - 5 x | > 4$
$\Rightarrow 4 - 5 x < - 4$ or $4 - 5 x > 4$
$\Rightarrow 5 x > 8$ or $- 5 x > 0$
$\Rightarrow x > \frac{8}{5}$ or $x < 0$
$\Rightarrow x \in (- \infty, 0) \cup (\frac{8}{5}, \infty)$
$x = 0$ and $x = 1$ are not in the solution set.
Hence, $0, 1$ disobey the given inequality.

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