Question:
If $f: A \rightarrow B$ given by $3^{f(x)}+2^{-x}=4$ is a bijection, then
(a) $A=\{x \in R:-1 (b) $A=\{x \in R:-3 (c) $A=\{x \in R:-2 (d) None of these
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.