(3) âµxâ(a,+â)
â´h(x)=f(x)
=2x^2+(x-a)(x-a)
=3x^2-2ax+a^2
h(x)>=1
3x^2-2ax+a^2>=1
3x^2-2ax+a^2-1>=0
Î=4a^2-4*3(a^2-1)
=12-8a^2
=4(3-2a^2)
1ï¼Î<=0
4(3-2a^2)<=0
2a^2-3>=0
a<=-â6/2æè
a>=â6/2
3x^2-2ax+a^2-1>=0ç解é为Rï¼ä½âµxâ(a,+â)
â´è§£é为ï¼xâ(a,+â)
2ï¼Î>0
4(3-2a^2)>0
2a^2-3<0
-â6/2<a<â6/2
âÎ=2â(3-2a^2)
x1=(2a-2â(3-2a^2))/6
=(a-â(3-2a^2))/3
x2=(2a+2â(3-2a^2))/6
=(a+â(3-2a^2))/3
ä¸çå¼ç解é为ï¼x<=(a-â(3-2a^2))/3æè
x>=(a+â(3-2a^2))/3
ä½âµxâ(a,+â)
â´å¿
é¡»å
讨论aä¸x1åx2ç大å°å
³ç³»ï¼å½a<x1æ¶ï¼è§£é为(a,x1]U[x2,+â)ï¼å½x1=<a<=x2æ¶ï¼è§£é为ï¼[x2,+â)ï¼å½a>x2æ¶ï¼è§£é为(a,+â)ã
è¥a<x1
a<(a-â(3-2a^2))/3
3a<a-â(3-2a^2)
â(3-2a^2)<-2a
3-2a^2<4a^2
6a^2>3
a^2>1/2
a<-â2/2æè
a>â2/2
å³å½-â6/2<a<-â2/2æè
â2/2<a<â6/2æ¶ï¼ä¸çå¼ç解éï¼(a,(a-â(3-2a^2))/3]U[(a+â(3+2a^2))/3,+â)
è¥x1=<a<=x2
-â2/2<=a<=â2/2
解éï¼[(a+â(3+2a^2))/3,+â)
è¥a>x2
a>(a+â(3+2a^2))/3
3a>a+â(3+2a^2)
2a>â(3+2a^2)
4a^2>3+2a^2
2a^2>3
a^2>3/2
a<-â6/2æè
a>â6/2ä¸åé¢Î>0å¾å°çâ6/2<a<â6/2çç¾ï¼å æ¤ï¼ä¸ä¼åºç°a>x2çæ
åµã
综述ï¼ä¸çå¼ç解é为ï¼xâ(a,+â)æè
(a,(a-â(3-2a^2))/3]U[(a+â(3+2a^2))/3,+â)æè
[(a+â(3+2a^2))/3,+â)
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