题目内容

1.已知函数f(x)=x2-2(a+2)x+a2,g(x)=-x2+2(a-2)x-a2+8.设函数H1(x)=$\left\{\begin{array}{l}{f(x),f(x)≥g(x)}\\{g(x),f(x)<g(x)}\end{array}\right.$,H2(x)=$\left\{\begin{array}{l}{g(x),f(x)≥g(x)}\\{f(x),f(x)<g(x)}\end{array}\right.$,记H1(x)的最小值为A,H2(x)的最大值为B,则A-B(  )
A.16B.-16C.a2+2a-16D.a2-2a-16

分析 作差f(x)-g(x)=2x2-4ax+2a2-8=2(x-a-2)(x-a+2),从而化简H1(x)与H2(x),从而结合二次函数的性质求最值,从而解得.

解答 解:f(x)-g(x)=2x2-4ax+2a2-8=2(x-a-2)(x-a+2),
故当x≥a+2或x≤a-2时,f(x)≥g(x);
当a-2<x<a+2时,f(x)<g(x),
∵H1(x)=$\left\{\begin{array}{l}{f(x),f(x)≥g(x)}\\{g(x),f(x)<g(x)}\end{array}\right.$,H2(x)=$\left\{\begin{array}{l}{g(x),f(x)≥g(x)}\\{f(x),f(x)<g(x)}\end{array}\right.$,
∴H1(x)=$\left\{\begin{array}{l}{{x}^{2}-2(a+2)x+{a}^{2},x≥a+2或x≤a-2}\\{-{x}^{2}+2(a-2)x-{a}^{2}+8,a-2<x<a+2}\end{array}\right.$,
H2(x)=$\left\{\begin{array}{l}{-{x}^{2}+2(a-2)x-{a}^{2}+8,a-2≤x≤a+2}\\{{x}^{2}-2(a+2)x+{a}^{2},x>a+2或x<a-2}\end{array}\right.$,
结合二次函数的性质可知,
A=H1(a+2)=(a+2)2-2(a+2)(a+2)+a2=-4a-4,
B=H1(a-2)=-(a-2)2+2(a-2)(a-2)-a2+8=-4a+12,
故A-B=-4a-4-(-4a+12)=-16,
故选B.

点评 本题考查了分段函数的应用及二次函数的应用,同时考查了分类讨论的思想应用.

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