smc

Question

A real-valued function f has the property that for all real numbers a and b, f(a + b) + f(a - b) = 2f(a) f(b). Which one of the following cannot be the value of f(1)?

Accepted Answer

Substituting a = b we get f(2a) + f(0) = 2f(a)^2 Substituting a= 0 we find 2f(0) = 2f(0)^2 f(0) 0, 1. This gives f(2a) = 2f(a)^2 - f(0) 0-1 Plugging in a = 12 implies f(1) -1, so answer choice -2 is impossible.