If `f: A->B` given by `3^(f(x))+2^(-x)=4` is a bijection, then A

Options

A. `A=(x in R: -1 lt x lt oo), B=(x in R: 2 lt x lt 4)`
B. `A=(x in R: -3 lt x lt oo), B=(x in R: 0 lt x lt 4)`
C. `A=(x in R: -2 lt x lt oo), B=(x in R: 0 lt x lt 4)`
D. None of these

Answer

D

Explanation

We have
`3^(f^((x)))+2^(-x)=4Rightarrow 3^(f^((x)))=4-2^(-x) Rightarrow f(x)=log_(3)(4-2^(-x))`
Clearly, f(x) is defined, if
`4-2^(-x)=0`
`Rightarrow 2^(-x) lt 4 Rightarrow 2^(x) gt 2^(-2) Rightarrow x gt -2 Rightarrow A=(x in R: x gt -2)`
Now, `f(x)=log_(3)(4-2^(-x))`
`Rightarrow f'(x) =(1)/((4-2^(-x))log, 3)xx2^(-x)log 2gt 0" for all "x gt -2`
`Rightarrow f(x)"is increasing on"(-2, oo)`
`Rightarrow B="Range of f"(x)={y:f(-2) lt y lt f (oo))`
`Rightarrow B=(y: -oo lt y lt log_(3) 4)=(x: -oo lt x lt log_(3) 4)`