Question

Range of the function defined by $9^x+3^{f\left(x\right)}=3$ is :

• A. $(-\infty ,1)$
• B. $(-1,1)$
• C. $(-\infty ,\frac{1}{2})$
• D. $(-\infty ,1)\cup (1,\infty )$

Answer 1

The correct option is A $(-\infty ,1)$

$3^{f\left(x\right)}=3-9^x$
$\Rightarrow f\left(x\right)={\log }_3(3-9^x)$
Domain: $3-9^x>0$ for $f\left(x\right)$ to be defined
$\Rightarrow 3^1>3^{2x}$
$\Rightarrow 1>2x$
$\Rightarrow x<\frac{1}{2}$
In the domain, $x\in (-\infty ,\frac{1}{2})$
$9^x\in (0,3)$
$\Rightarrow 3-9^x\in (0,3)$
$\Rightarrow {\log }_3(3-9^x)\in (-\infty ,1)$
$\therefore$Range of the function is $(-\infty ,1)$