STD 11 - basic of algoritham — JEE STD 11 Science — Question

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., $x \in [ - 1,\,1 - \sqrt 3 )\,\, \cup (1 + \sqrt 3 ,\,3)$.