Question

If $f(x)=\log _{100 x}\left(\frac{2 \log _{10} x+2}{-x}\right) ; g(x)=\{x\} ;$ where $\{x\}$ denotes the fractional part of $x$. If the function $(f o g)(x)$ 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)

$f(x)=\log _{100 x}\left(\frac{2 \log _{10} x+2}{-x}\right) ; g(x)=\{x\}$;
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 $f(x): 100 x > 0, \neq 1$ and
$\frac{2 \log _{10} x+2}{-x} > 0$
$\Rightarrow x > 0 ; x \neq \frac{1}{100}$ and
$\frac{\log _{10} x+1}{x} < 0$
$\Rightarrow x > 0, x \neq \frac{1}{100} \log _{10} x < -1$
$\Rightarrow x > 0 ; x \neq \frac{1}{100} ; x < \frac{1}{10}$
$ \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., $\left(0,10^{-2}\right) \cup\left(10^{-2}, 10^{-1}\right)$.