若函数f(x)=kx2+(k-1)x+2是偶函数,则f(x)的递减区间是

若函数f(x)=kx2+(k-1)x+2是偶函数,则f(x)的递减区间是
若函数f(x)=kx2+(k-1)x+2是偶函数,则f(x)的递减区间是

若函数f(x)=kx2+(k-1)x+2是偶函数,则f(x)的递减区间是
f(x)=kx2+(k-1)x+2是偶函数,据f(x)=f(-x)得,K=0或1,K=0时不符题意,舍去
则f(x)=X*X+2,递减区间是（负无穷,0]

过程：（负无穷，0]
f(x)=f(-x)=kx^2+(k+1)x+2
=k(-x)^2+(k-1)(-x)+2
=-[kx^2+(k-1)x]+2
因为k在(x)中证明取值为R，但是k-1不等于0，所以1排除，且k=0也不行。则递减区间是（负务求，0]