Question

If $S$ be the set of real values of $x$ satisfying the inequality $1-{\log }_2\left(\frac{|x+1+2i|-2}{\sqrt{2}-1}\right)\ge 0$, then $S$ contains

• A. $[-3,1]$
• B. $(1,8]$
• C. $[-1,8)\cup (8,\infty )$
• D. $[-3,-1)\cup (-1,1]$

Answer 1

The correct option is D $[-3,-1)\cup (-1,1]$

Given, $1-{\log }_2\frac{|x+1+2i|-2}{\sqrt{2}-1}\ge 0$

$\Rightarrow \frac{|x+1+2i|-2}{\sqrt{2}-1}\le 2\Rightarrow |x+1+2i|\le 2\sqrt{2}$

$\Rightarrow \sqrt{(x+1{)}^2+4}\le 2\sqrt{2}\Rightarrow (x+1{)}^2+4\le 8$

$\Rightarrow (x+1{)}^2\le 4\Rightarrow -3\le x\le 1$

But $x=-1$ not lie in the domain of function.

So, $S=[3,-1)\cup (-1,1]$