If f : A→B given by

Solution:

(d) None of these

$f: A \rightarrow B$

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

$\Rightarrow 3^{f(x)}=4-2^{-x}$

Taking $\log$ on both the sides,

$f(x) \log 3=\log \left(4-2^{-x}\right)$

$\Rightarrow f(x)=\frac{\log \left(4-2^{-x}\right)}{\log 3}$

Logaritmic function will only be defined if $4-2^{-x}>0$

$\Rightarrow 4>2^{-x}$

$\Rightarrow 2^{2}>2^{-x}$

$\Rightarrow 2>-x$

$\Rightarrow-2

$\Rightarrow x \in(-2, \infty)$

That means $A=\{x \in R:-2

As we know that, $f(x)=\frac{\log \left(4-2^{-z}\right)}{\log 3}$

$\Rightarrow f(x)=1$ which does not belong to any of the options.