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Basic of Logarithms
medium
Solution set of inequality ${\log _{10}}({x^2} - 2x - 2) \le 0$ is
A
$[ - 1,\,1 - \sqrt 3 ]$
B
$[1 + \sqrt 3 ,3]$
C
$[ - 1,\,1 - \sqrt 3 ) \cup (1 + \sqrt 3 \,,\,3]$
D
None of these
Solution
(c) ${\log _{10}}({x^2} – 2x – 2) \le 0$…..$(i)$
For logarithm to be defined,
${x^2} – 2x – 2 > 0$
$ \Rightarrow $ ${(x – 1)^2} > 3$
==> $x – 1 < – \sqrt 3 $ or$x – 1 > \sqrt 3 $
==> $x < 1 – \sqrt 3 $ or $x > 1 + \sqrt 3 $
i.e., $x < – (\sqrt 3 – 1)$ or $x > (\sqrt 3 + 1)$
Now from $(i),$ ${x^2} – 2x – 2 \le 1$
==> ${x^2} – 2x – 3 \le 0$
==> $(x – 3)\,(x + 1) \le 0$ $ \Rightarrow $ $ – 1 \le x \le 3$
$\therefore $ $x \in [ – 1,\, – (\sqrt 3 – 1)\,[\, \cup \,]\,\,\sqrt 3 + 1,\,3]$.
i.e., $x \in [ – 1,\,1 – \sqrt 3 )\,\, \cup (1 + \sqrt 3 ,\,3)$.
Standard 11
Mathematics
