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 value of f(4) is

Options

A. 160
B. 240
C. 200
D. None of these

Answer

A

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)" `
`implies 2^(x) f(y)+4^(y)f(x)=2^(y)f(x)+4^(x)f(y)`
`implies (f(x))/(4^(x)-2^(x))=(f(y))/(4^(y)-2^(y))=k`
`implies f(x)=k(4^(x)-2^(x))`
`f(1)=2`
`implies 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 `-1//4`
Also `4^(x)-2^(x)=2`
`implies (2^(x))^(2)-2^(x)-2=0`
`implies (2^(x)-2)(2^(x)+1)=0`
`implies2^(x)-2=0`
`implies x=1`