Question

A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, then f(2a – x) is 

(a) f(x) 

(b) –f(x) 

(c) f(– x) 

(d) f(a) + f(a – x)

Answer 1

Correct option (b) –f(x)

Explanation:

Given f(x – y) = f(x)f(y) – f(a – x) f(a + y)

Put x = 0 = y, we get,

f(0) = f(0)f(0) – (f(a))²

1 = 1 – (f(a))² 

f(a) = 0

Also, f (2a – x) = f(a – (x – a))

= f(a)f(x – a) – f(a – a)f(a + x – a)

= f(a)f(x – a) – f(0)f(x)

= 0 x f(x – a) – 1 x f(x)

= –f(x)