Question

若函数f(x)=（k-2)x2+(k-1)x+3是偶函数,则f(X)的递减区间是?

Answer 1

f(x)=f(-x)

(k-2)x²+(k-1)x+3=(k-2)x²-(k-1)x+3

(k-1)x=0

所以k-1=0

k=1

f(x)=-x²+3

开口向下

所以对称轴x=0右边递减

所以是(0,+∞)