一道数学题
若函数f(x)=(k-2)x^2+(k-1)x+2是偶函数,则f(x)的单调递增区间是?
参考答案:f(x)=f(-x)
(k-2)x^2+(k-1)x+2=(k-2)x^2-(k-1)x+2
即k-1=-k+1 k=1
f(x)=-x^2+2
单调递增区间(-∞,0]
若函数f(x)=(k-2)x^2+(k-1)x+2是偶函数,则f(x)的单调递增区间是?
参考答案:f(x)=f(-x)
(k-2)x^2+(k-1)x+2=(k-2)x^2-(k-1)x+2
即k-1=-k+1 k=1
f(x)=-x^2+2
单调递增区间(-∞,0]