若函数f(x)=(a-2)x2+(a-1)x+3是偶函数,则函数f(x)的单调递减区间为_.

问题描述:

若函数f(x)=(a-2)x2+(a-1)x+3是偶函数,则函数f(x)的单调递减区间为______.

∵函数f(x)=(a-2)x2+(a-1)x+3是偶函数,
∴f(-x)=f(x)
∴(a-2)x2-(a-1)x+3=(a-2)x2+(a-1)x+3
∴-(a-1)=a-1,解得a=1
∴f(x)=-x2+3
∴函数f(x)的单调递减区间为[0,+∞)
故答案为:[0,+∞).