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. None of the above
• C. $(4,\infty )$
• D. $(-1,4)$

Answer 1

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

$2^{{\log }_{\sqrt{2}}(x-1)}>x+5$
$\Rightarrow 2^{lo{g}_2(x-1{)}^2}>x+5$
$[\because a^{{\log }_{a}x}=x]$
$\Rightarrow (x-1{)}^2>x+5$
$\Rightarrow x^2-3x-4>0$
$\Rightarrow (x-4)(x+1)>0$
$\Rightarrow x>4\text{or}x<-1$
But for given $log$ to be defined, $x-1>0$
$i.e.,x>1\therefore x>4\Rightarrow x\in (4,\infty )$
Hence, The correct answer is option (b)