Let `f:R to R ` such that `f(x+y)+f(x-y)=2f(x)f(y)` for all ` x,y in R`. Then,

Options

A. f(x) an even function , if `f(0) ne 0`
B. f(x) is an odd function, if `f(0) ne 0`
C. f(x) an even function , if `f(0)=0`
D. f(x) is an odd function , if `f(0)=0`

Answer

A, D

Explanation

We have ,
`f(x+y)+f(x+y)=2f(x)` for all ` x,y in R" "` ….(i)
`implies 2f(0)=2{f(0)}" "` [Replacing x, y be zero ]
`implies 2f(0)(f(0)-1)=0`
`implies f(0)=0 or f(0)=1`
CASE `" "`When f(0)=0
Replacing x by 0 in (i) , we get
`f(y)+f(-y)=2(0) f(y)` for all `y in R`
`implies f(-y)=-f(y)` for all ` y in R `
`implies f(x) ` is an odd functions .
CASE II - When `f(0)=1`
Replacing x by 0 in (i), we get
`f(y)+f(-y)=2f(0)f(y)` for all `y in R`
`implies f(-y)=f(y)` for all ` y in R `.
`implies f(x)` is an even function.