Question

The complete set of real values of $2^{{\log }_{\sqrt{2}}(x-1)}$ is

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

Answer 1

Answer: (D)

none of these

Let $f\left(x\right)=2^{{\log }_{\sqrt{2}}(x-1)}$
Domain of $f\left(x\right)$ is $x\in (1,\infty )$
$f\left(x\right)=2^{{\log }_{\sqrt{2}}(x-1)}$
$\Rightarrow f\left(x\right)=2^{2{\log }_2(x-1)}\Rightarrow f\left(x\right)=2^{{\log }_2{(x-1)}^2}$
$\Rightarrow f\left(x\right)={(x-1)}^2$
For $x>1$
$f\left(x\right)>0$
Therefore, $f\left(x\right)\in (0,\infty )$
Ans: D