Question

If $f_1\left(x\right)=2^{f_2\left(x\right)}$, where $f_2\left(x\right)={2012}^{f_3\left(x\right)}$, where $f_3\left(x\right)={\left(\frac{1}{2013}\right)}^{f_4\left(x\right)}$, where $f_4\left(x\right)={\log }_{2013}{\log }_{x}2012$, then the range of $f_1\left(x\right)$ is

• A. $(2,\infty )$
• B. $(2012,\infty )$
• C. $(0,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is A $(2,\infty )$

$f_3\left(x\right)=\left[\right(2013{)}^{-1}{]}^{{\log }_{2013}{\log }_{x}2012}$

$=[{\log }_{x}2012{]}^{{\log }_{2013}(2013{)}^{-1}}$

$=[{\log }_{x}2012{]}^{-1}={\log }_{2012}x$

$f_2\left(x\right)={2012}^{{\log }_{2012}x}=x^{{\log }_{2012}2012}$

$=x$

$f_1\left(x\right)=2^x$

where $x$ can be $(0,1)\cup (1,\infty )$

$\therefore f_1\left(x\right)=(1,2)\cup (2,\infty )$