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Question
\[f : A \to \text{B given by } 3^{ f\left( x \right)} + 2^{- x} = 4\] is a bijection, then
Options
\[A = \left\{ x \in R : - 1 < x < \infty \right\}, B = \left\{ x \in R : 2 < x < 4 \right\}\]
\[A = \left\{ x \in R : - 3 < x < \infty \right\}, B = \left\{ x \in R : 2 < x < 4 \right\}\]
\[A = \left\{ x \in R : - 3 < x < \infty \right\}, B = \left\{ x \in R : 2 < x < 4 \right\}\]
None of these
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Solution
(d) None of these
\[f: A \to B\]
\[ 3^{f(x)} + 2^{- x} = 4 \]
\[ \Rightarrow 3^{f(x)} = 4 - 2^{- x} \]
\[\text{Taking log on both the sides} , \]
\[f(x) \log 3 = \log \left( 4 - 2^{- x} \right)\]
\[ \Rightarrow f(x) = \frac{\log \left( 4 - 2^{- x} \right)}{\log 3}\]
\[\text{Logaritmic function will only be defined if} 4 - 2^{- x} > 0\]
\[ \Rightarrow 4 > 2^{- x} \]
\[ \Rightarrow 2^2 > 2^{- x} \]
\[ \Rightarrow 2 > - x\]
\[ \Rightarrow - 2 < x\]
\[ \Rightarrow x \in \left( - 2, \infty \right)\]
\[\text{That means A} = \left\{ x \in R: - 2 < x < \infty \right\}\]
\[\text{As we know that}, f(x) = \frac{\log \left( 4 - 2^{- x} \right)}{\log 3}\]
\[\text{ We take } x = 0 \in \left( - 2, \infty \right)\]
\[ \Rightarrow f(x) = 1 \text{which does not belong to any of the options} . \]
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