Question

The set of real values of $x$ for which $2^{{\log }_{\sqrt{2}}(x-1)}>x+5$ is

• A. $(-\infty ,-1)\cup (4,\infty )$
• B. $(4,\infty )$
• C. $(-1,4)$
• D. none of these

Answer 1

The correct option is B $(4,\infty )$

We have ,
$2^{{\log }_{\sqrt{2}}(x-1)}>x+5$
${\log }_{\sqrt{2}}(x-1)\text{is defined for}(x-1)>0\Rightarrow x>\mathrm{1...}\left(1\right)$

$\Rightarrow 2^{2{\log }_2(x-1)}>x+5$
$\Rightarrow (x-1{)}^2>x+5$
$\Rightarrow x^2-3x-4>0$
$\Rightarrow (x-4)(x+1)>0$
$\Rightarrow x<-1\text{or}x>\mathrm{4...}\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow x\in (4,\infty )$