If the function f satisfies the relation `f(x+y)+f(x-y)=2f(x)xxf(y), AA x, y, in R and f(0) ne 0`, then

Options

A. `f(x)` is an even function
B. `f(x)` is an odd function
C. If `f(2)=a, " then " f(-2)=a`
D. If `f(4)=b, " then " f(-4)= -b`

Answer

A, C

Explanation

`f(x+y)+f(x-y) =2f(x)*f(y) " (1)" `
Put `x=0." Then " f(y)+f(-y)=2f(0)f(y) " (2)" `
Put `x=y=0. " Then " f(0) +f(0)=2f(0) f(0)`
` :. f(0) =1 ["as "f(0) ne 0]`
` :. f(-y) =f(y) " [from (2)]" `
Hence, the function is even, Then `f(-2)=f(2) =a.`