MCQ
If $f : A → B$ given by $3^{f(x)} + 2^{-x} = 4$ is a bijection, then
  • A
    $\text{A}=\{\text{x}\in\text{R}:-1<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
  • B
    $\text{A}=\{\text{x}\in\text{R}:-3<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
  • C
    $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\},$ $\text{B}=\{\text{x}\in\text{R}:2<\text{x}<4\}$
  • None of these.

Answer

Correct option: D.
None of these.
$\text{f}:\text{A}\rightarrow\text{B}$
$3^\text{f(x)}+2^{-\text{x}}=4$
$\Rightarrow\ 3^{\text{f(x)}}=4-2^{-\text{x}}$
Taking $\log$ on both the sides,
$\text{f(x)}\log3=\log(4-2^{-\text{x}})$
$\Rightarrow\ \text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
Logaritmic function will only be defined if $4-2^{-\text{x}}>0$
$\Rightarrow\ 4>2^{-\text{x}}$
$\Rightarrow\ 2^2>2^{-\text{x}}$
$\Rightarrow\ 2>-\text{x}$
$\Rightarrow-2<\text{x}$
$\Rightarrow\ \text{x}\in(-2,\infty)$
That means $\text{A}=\{\text{x}\in\text{R}:-2<\text{x}<\infty\}$
As we know that, $\text{f(x)}=\frac{\log(4-2^{-\text{x}})}{\log3}$
We take $\text{x}=0\in(-2,\infty)$
$\Rightarrow f(x) = 1$ which does not belong to any of the options.

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