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若不等式|1-a|+|2a-1|〈=2恒成立,则实数a的取值范围是?
人气:498 ℃ 时间:2020-01-29 05:21:07
解答
答:
当a<=1/2时:1-a+(1-2a)=2-3a<=2,0<=a<=1/2
当1/2<=a<=1时:1-a+2a-1=a<=2恒成立
当a>=1时:a-1+2a-1=3a-2<=2,1<=a<=4/3
综上所述:0<=a<=4/3
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