Question

The domain of the function $f\left(x\right)={\log }_2\left(-{\log }_{\frac{1}{2}}\left(1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}\right)-1\right)$ is

• A. $\left(0,1\right)$
• B. $(0,1]$
• C. $[1,\infty )$
• D. $\left(1,\infty \right)$

Answer 1

The correct option is A

$\left(0,1\right)$

Find the domain of the function $f\left(x\right)$:

Consider the given function as,

$f\left(x\right)={\log }_2\left(-{\log }_{\frac{1}{2}}\left(1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}\right)-1\right)$

Defined the above function as,

$$\begin{gathered} \Rightarrow -{\log }_{\frac{1}{2}}\left(1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}\right)-1>0 \\ \Rightarrow {\log }_{\frac{1}{2}}\left(1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}\right)<-1\left[{\text{×}}^{\text{ly}}\text{by}-\right] \\ \Rightarrow 1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}>{\left(\frac{1}{2}\right)}^{-1} \\ \Rightarrow 1+\frac{1}{x^{{\displaystyle \frac{1}{4}}}}>2 \\ \Rightarrow \frac{1}{x^{{\displaystyle \frac{1}{4}}}}>1 \\ \Rightarrow x^{\frac{1}{4}}<1 \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow x^{\frac{1}{4}}={\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}} \\ \Rightarrow x^{\frac{1}{4}}={\left(\sqrt{x}\right)}^{\frac{1}{2}} \\ \Rightarrow x>0 \\ \Rightarrow 0<x^{\frac{1}{4}}<1 \end{gathered}$$

Hence, the domain of the function is $\left(0,1\right)$

Therefore, the correct answer is option (A).