摘要:知∈[1.2].-a≤-3又x∈[-2,2]∴f(x)max=max{f}而f=16-4a2<0∴f(x)max=f(-2)=-8+4a+2a2+m ---------------10分又∵f(x)≤1在[-2.2]上恒成立∴f(x)max≤1即-8+4a+2a2+m≤1即m≤9-4a-2a2,在a∈[3.6]上恒成立∵9-4a-2a2的最小值为-87
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