Problem:
A real-valued function f has the property that for all real numbers a and b, f(a+b)+f(aβb)=2f(a)f(b)
Which one of the following cannot be the value of f(1)?
Answer Choices:
A. 0
B. 1
C. β1
D. 2
E. β2
Solution:
First note that if f(x)=0 for all x, then the defining identity is met, and f(1)=0, eliminating option (A). Now assume that f is not identically zero. Setting b=0 and choosing a so that f(a)ξ =0, the identity reduces to f(a)+f(a)=2f(a)f(0), yielding f(0)=1. Setting a=b=21β and using f(0)=1 gives f(1)+1=2f(21β)f(21β). If f(1)=β2, then β2+1=2[f(21β)]2, which is impossible. So f(1) cannot equal (E)β2β , and the correct answer is (E).
For completeness, it remains to show why options (C), (B) and (D) are attainable as f(1). Note that the cosine function satisfies the defining identity, as does the hyperbolic cosine function, and more generally so do cos(kx) and cosh(kx). Letting k equal Ο,2Οβ, and 0 in cos(kx), respectively, shows that options (C), (A), and (B) are possible values of f(1). The range of the continuous function f(x)=cosh(kx) is the interval [1,β), which includes 2 . Therefore there must be a value of k that makes f(1)=2, eliminating option (D).
Note: While not required for this problem, it is possible to determine a specific x satisfying coshx= 2. Observe that