Let a funtion `f` defined on te set of all integers satisfy `f(0)ne 0 ,f(1)=3` and
`f(x).f(y)=f(x+y)+f(x-y)` for all integers x and y. then

Options

A. `f(2)=7`
B. `f(3)=21`
C. `f(4)=47`
D. `f(7)=843`

Answer

A, C, D

Explanation

`f(x+y)+f(x-y)=f(x).f(y)`
given `f(0)ne0`
`f(1)=3`
`x=1,y=0=1`
`f(1)+f(1)=f(1).f(0)`
`implies2f(11)=f(0).f(1)`
`implies2xx3=3f(0)`
`impliesf(0)=2` ..(i)
`x=1,y=1impliesf(2)+f(0)=f(1)f(1)`
`impliesf(2)+2=(3)^(2)` (`becausef(0)=2`)
`impliesf(2)=7` ..(ii)
`x=2,y=1impliesf(3)+f(1)=f(2)f(1)`
`impliesf(3)+3=7xx3`
`impliesf(3)=18` ..(iii)
`x=3,y=1impliesf(4)+f(2)=f(3)f(1)`
`impliesf(4)+7=18xx3`
`impliesf(3)=47` .(iv)
`x=4,y=3impliesf(7)+f(1)=f(4)f(3)`
`impliesf(7)+3=47xx18`
`impliesf(7)=846-3=843`