If `f(x-y),f(x)f(y),a n df(x+y)` are in A.P. for all `x , y ,a n df(0)!=0,` then (a)`f(4)=f(-4)` (b)`f(2)+f(-2)=0` (c)`f^(prime)(4)+f^(prime)(-4)=0` (d)`f^(prime)(2)=f^(prime)(-2)`

Options

A. `f(4)=f(-4)`
B. `f(2)+f(-2)=0`
C. `f'(4)+f'(-4)=0`
D. `f'(2)=f'(-2)`

Explanation

f(x-y),f(x)f(y), and f(x+y) are in A.P Therefore,
f(x+y) + f(x-y) = 2f(x) f(y) for all x,y
Putting x=0, y = 0 in (1), we get
`f(0)+f(0)=f(0)f(0)`
`"or "f(0)=1" "[because f(0)ne0]`
Putting `x=0, y=x,` we get
`f(x)+f(-x)=2f(0)f(x)`
`"or "f(x)=f(-x)" (1)`
`"or "f(4)=f(-4),f(3)=f(-3)`
Differentiating (1) w.r.t. x, `f'(x)+f'(-x)=0`
`"or "f'(4)+f'(-4)=0`