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已知函数f(x)=(1-x~2)(x~2+mx+n)的图像关于x=2对称,不等式f(x)≤α对x∈R恒成立,则实数a的取值范围是▲.
The image of the known function f(x)=(1-x~2)(x~2+mx+n) is symmetric about x=2. If the inequality f(x)≤α is true for x∈R, then the real number a The range of values is ▲.