If: A → B given by $3^{f(x)} + 2^{-x} = 4$ is a bijection, then

none of these

We have,

$3^{f(x)}+2^{-x}=4⇒3^{f(x)} = 4-2^{-x}= f(x) = \log_3 (4-2^{-x})$

Clearly, f(x) is defined, if

$4-2^{-x}>0$

$⇒2^{-x} <4⇒ 2^x >2^{-2}⇒ x>-2⇒ A = \{x ∈R:x>-2\}$

Now, $f(x)=\log_3 (4-2^{-x})$

$⇒f'(x)=\frac{1}{(4-2^{-x})\log_e3}×2^{-x}\log 2>0$ for all $x>-2$

⇒ f(x) is increasing on (-2,∞)

⇒ B = Range of $f (x) = \{y: f (-2) < y <f (∞)\}$

$⇒B=\{y:-∞<y<\log_3 4\}=\{x: -∞ <x<\log_3 4\}$