The solution set of the inequality log_{10}(x | vedclass

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Question

(C) Given the inequality $\log_{10}(x^2 - 2x - 2) \le 0$. For the logarithm to be defined,the argument must be positive: $x^2 - 2x - 2 > 0$. Solving t

Vedclass · Accepted Answer

$[-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3]$

Vedclass · Answer

(C) Given the inequality $\log_{10}(x^2 - 2x - 2) \le 0$. For the logarithm to be defined,the argument must be positive: $x^2 - 2x - 2 > 0$. Solving this,we find the roots of $x^2 - 2x - 2 = 0$ are $x = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$. Thus,$x  1 + \sqrt{3}$. Now,from the inequality $\log_{10}(x^2 - 2x - 2) \le 0$,we have: $x^2 - 2x - 2 \le 10^0 = 1$. $x^2 - 2x - 3 \le 0$. $(x - 3)(x + 1) \le 0$. This implies $-1 \le x \le 3$. Combining the domain condition and the inequality result: $x \in [-1, 1 - \sqrt{3}) \cup (1 + \sqrt{3}, 3]$. Therefore,the correct option is $C$.

The solution set of the inequality $\log_{10}(x^2 - 2x - 2) \le 0$ is

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