Consider a differentiable `f:R to R` for which `f(1)=2 and f(x+y)=2^(x)f(y)+4^(y)f(x) AA x , y in R.`
The minimum value of `f(x)` is

Options

A. `1`
B. `-(1)/(2)`
C. `-(1)/(4)`
D. none of these

Answer

C

Explanation

`f(x+y)=2^(x)f(y)+4^(y)f(x)" (i)"`
Interchanging x and y, we get
`f(x+y)=2^(y)f(x)+4^(x)f(y)" (ii)"`
`rArr" "2^(x)f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)`
`rArr" "(f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k`
`rArr" "f(x)=k(4^(x)-2^(x))`
`rArr" "f(1)=k(4-2)=2`
`rArr" "k=1.`
Hence, `f(x)=4^(x)-2^(x).`
`f(4)=4^(4)-2^(4)=240`
`f(x)=(2^(x))^(2)-2^(x)=(2^(x)-1//2)^(2)-1//4`
Thus, f(x) has least value as `-1//4.`