Question

If $f(x)={\log }_{100x}\left(\frac{2{\log }_{10}x+2}{-x}\right);g(x)=\{x\};$ where $\{x\}$ denotes the fractional part of $x$. If the function $(fog)(x)$ exists, then the maximum possible range of $g(x)$ is

• A. $\left(0,10^{-1}\right)$
• B. $\left(0,10^{-2}\right)$
• C. $\left(0,10^{-2}\right)\cup \left(10^{-2},10^{-1}\right)$
• D. None of these

Answer 1

Answer: (C)

$f(x)={\log }_{100x}\left(\frac{2{\log }_{10}x+2}{-x}\right);g(x)=\{x\}$;
For fog $(x)$ to exist.
Range of $f(x)$ must be subset of domain of $f(x)$
Now, range $f(x)=[0,1)$
For domain of $f(x):100x>0,\ne 1$ and
$\frac{2{\log }_{10}x+2}{-x}>0$
$\Rightarrow x>0;x\ne \frac{1}{100}$ and
$\frac{{\log }_{10}x+1}{x}<0$
$\Rightarrow x>0,x\ne \frac{1}{100}{\log }_{10}x<-1$
$\Rightarrow x>0;x\ne \frac{1}{100};x<\frac{1}{10}$
$\Rightarrow x\in \left(0,\frac{1}{10}\right)-\left\{\frac{1}{100}\right\}.$
$\therefore$ Maximum possible range of
$f(x)=\left(0,\frac{1}{10}\right)-\left\{\frac{1}{100}\right\}$
$=\left(0,\frac{1}{100}\right)\cup \left(\frac{1}{100},\frac{1}{10}\right)$
i.e., $\left(0,10^{-2}\right)\cup \left(10^{-2},10^{-1}\right)$.