if f(x+y)+f(x-y)=2f(x)f(y) for all x,y ε R and f(0)≠0 then prove that f(x) is an even function.
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2 Answers
106Setting x=y=0, we get
2f(0)=2f2(0)
So, f(0) = 1 [given f(0)≠0]
Set x=y=x
f(2x)+1 = 2f2(x) ... (i)
Set y= -x, we have
1+f(2x) = 2f(x) .. (ii)
From 1 and 2
2f(x) = 2f2(x)
=> f(x)=0 OR f(x)=1
Either case f(x) is even
1after, f(0)=1,
u can directly have
x=0, so
1*f(-y)+1*f(y)=2*1*f(y)
or f(-y)=f(y)
simple way out.
so f(x) is even
cheers!!
