Question

The number of integers lying in the domain of the function $f\left(x\right)=\sqrt{{\log }_{0.5}\left(\frac{5-2x}{x}\right)}$ is $-$

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

We have $f\left(x\right)=\sqrt{{\log }_{0.5}\left(\frac{5-2x}{x}\right)}$

$\sqrt{{\log }_{0.5}\left(\frac{5-2x}{x}\right)}$ is defined when ${\log }_{0.5}\left(\frac{5-2x}{x}\right)\ge 0$

$\Rightarrow \frac{5-2x}{x}\le {\left(\frac{1}{2}\right)}^0$

$\Rightarrow \frac{5-2x}{x}\le 1$

$\Rightarrow 5-2x\le x$

$\Rightarrow 5\le 3x$

$\Rightarrow 3x\ge 5$

$\Rightarrow x\ge \frac{5}{3}$

$\therefore x\in [\frac{5}{3},\infty )$

Consider $\frac{5-2x}{x}>0$

$\Rightarrow 5-2x>0$

$\Rightarrow 2x<5$

$\Rightarrow x<\frac{5}{2}$

$\therefore x\in (-\infty ,\frac{5}{2})$

Solution is $x\in (-\infty ,\frac{5}{2})\cap [\frac{5}{3},\infty )=[\frac{5}{3},\frac{5}{2})$

$\therefore x=2$ lies between $\frac{5}{3}$ and $\frac{5}{2}$

Hence only one integer lies between $\frac{5}{3}$ and $\frac{5}{2}$