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If: A → B given by $3^{f(x)} + 2^{-x} = 4$ is a bijection, then |
$A = \{x∈R:-1 <x<∞\}, B= \{x ∈R:2<x<4\}$ $A = \{x∈R:-3 <x<∞\}, B=\{x ∈R:0 < x <4\}$ $A= \{x∈R:-2<x<∞\}, B = \{x ∈R:0 < x <4\}$ none of these |
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 $f (x) = \{y: f (-2) < y <f (∞)\}$ $⇒B=\{y:-∞<y<\log_3 4\}=\{x: -∞ <x<\log_3 4\}$ |
