Question

$f:A\to B$ is a function satisfying $3^{f\left(x\right)}+2^{-x}=4$. Then the domain and range of $f$ are

• A. $A=\{x\in R:-1<x<\infty \}$;
  $R\left(f\right)=\{x\in R:2<x<4\}$
• B. $A=\{x\in R:-3<x<\infty \}$;
  $R\left(f\right)=\{x\in R:0<x<{\log }_{3}4\}$
• C. $A=\{x\in R:-2<x<\infty \}$;
  $R\left(f\right)=\{x\in R:-\infty <x<{\log }_{3}4\}$
• D. $A=\{x\in R:-2<x<\infty \}$;
  $R\left(f\right)=\{x\in R:-\infty <x<{\log }_{3}41\}$

Answer 1

The correct option is C $A=\{x\in R:-2<x<\infty \}$;

$R\left(f\right)=\{x\in R:-\infty <x<{\log }_{3}4\}$
$3^{f\left(x\right)}+2^{-x}=4$
$\Rightarrow 3^{f\left(x\right)}=4-2^{-x}$
$\Rightarrow f\left(x\right)={\log }_3(4-2^{-x})$

For domain
$4-2^{-x}>0$
$\Rightarrow 2^{-x}<2^2$
$\Rightarrow -x<2[\because \text{For}a>1,a^x<a^y\Rightarrow x<y]$
$\Rightarrow x>-2$
$\therefore$ Domain $=\{x\in R:-2<x<\infty \}$


Now, $-2<x<\infty$
$\Rightarrow -\infty <-x<2$
$\Rightarrow 0<2^{-x}<4$
$\Rightarrow 0<4-2^{-x}<4$
Since, ${\log }_{a}x$ is strictly increasing for $a>1$, we get
$-\infty <{\log }_3(4-2^{-x})<{\log }_{3}4$
$\therefore R\left(f\right)=\{x\in R:-\infty <x<{\log }_{3}4\}$