题目: 已知函数f(x)=2x, x>0x+1, x≤0,若f(a)+f(1)=0,则实数a=______. 答案: 当a>0时,f(a)=2a,由 f(a)+f(1)=0,可得 2a+2=0,解得a=-1(舍去).当a<0时,f(a)=a+1,由 f(a)+f(1)=0,可得a+1+2=0,解得a=-3,故答案为-3.