偶函数f（x）在[0，+∞）上为减函数，不等式f（ax-1）＞f（2+x2）恒成立，则a的取值范围是（ ）
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由题意可得，偶函数f（x）在（-∞，0]上为增函数，
再根据不等式f（ax-1）＞f（2+x2）恒成立可得|ax-1|＜2+x2恒成立．
故有-2-x2＜ax-1＜2+x2，即
恒成立．x2+ax+1＞0 x2-ax+3＞0
∴△=a2-4＜0，且△′=a2-12＜0，
解得a2＜4，即-2＜a＜2，
故选：D．
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| New Kowloon Primary School is going to hold a charity fair (慈善活动) next Sunday. They want to 1 money to buy clothes for poor people. The tickets to the fair are $30 2 . They have already sold 1,000 tickets. The programme of the fair is good. 3 , the pupil team will have a basketball match 4 the teacher team. Then, two singers, Kitty Li and Betty Fong, will 5 at the fair. There will also be a magic show and Chinese dances. After the shows, there will be a 6 . The teachers and parents have donated many things-----books, toys and food. At the 7 of the charity fair there will be a lucky draw (抽奖). Mr. Wong, the headmaster, has donated a tape-recorder as the grand prize, and there will be ten 8 prizes. | ||||
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