x>0时,00,n>0时,m+n > n, f(m+n) = f(m)*f(n) < f(n)
=> x>0时,f(x)单调递减.
f(0) = f(0)*f(0) => f(0) = 0 或 f(0)=1
当f(0) = 0 , m>0 时,f(m+0) = f(m)*f(0) = 0 与题意矛盾
f(0) = 1
当m>0：
f(0) = f(m)*f(-m) = 1 => f(-m) = 1/f(m) => 当x f(1)
f(x^2 + y^2) > f(1)
=> x^2+y^2 < 1 解集为圆的内部
B: f(a*x-y+2) = 1
=> a*x - y + 2 = 0 根据单调性
若A与B的交集是空集
=> 直线a*x - y + 2 = 0 与圆x^2+y^2 = 1至多有一个交点.
=> 作图易得 ：
=> 0= - 根号3