If `f(x-y), f(x) f(y), and f(x+y)` are in A.P. for all x, y, and `f(0) ne 0,` then

Options

A. `f(1)+f(-1)=0`
B. `f(2)+f(-2)=0`
C. `f'(3)+f'(-3)=0`
D. `f'(2)=f'(-2)`

Answer

C

Explanation

It is given that `f(x-y),f(x)f(y)" and "f(x+y)` are in A.P for all x,y`inR.`
`:." "2f(x)f(y)=f(x-y)+f(x+y)" for all "x,yinR" "...(i)`
Putting `x=y=0,` we get
`2{f(0)}^(2)=2f(0)`
`implies" "f(0){f(0)-1}=0impliesf(0)=1" "[becausef(0)ne0]`
Putting x=0 and y=x in (i), we get
`{:(" "2f(0)f(x)=f(-x)+f(x)," for all "x inR),(implies2f(x)=f(-x)+f(x)," for all "x inR),(impliesf(-x)=f(x)," for all "x inR):}`
`implies" "f(x)" is an even function"`
`implies" "f'(x)" is an odd function"`
`:." "f(-1)=f(1)" and "f'(-3)=-f'(3)`
`implies" "f(1)=f(-1)" and "f'(3)+f'(-3)=0.`