上海市育才中学高三数学期中考试试卷

19、(本大题满分14分) 111a200解：(1)设AB?a,AP?x,则?x?AQsin60??a?a?sin60,?AQ?2222xPQ?a4a2x??2x?cos600?22x4x2a4a2a2a2x???2??2224x22a22?a?0.707a22当且仅当x?2a时取到2(2)若PQ为圆弧。设AP?x,则PQ?33a2则x?,x?4?2?3x,S扇形?1?11x?x??a?a?sin600232233?a,?PQ?4?333?a?4?0.673a4?23 方案丙：作BC的中垂线AD,分别在AB、AC和AD上截取点P、Q、E,使得AP?AQ?AE，4111133则?PAE?30,2??x?x???a?a?,x?a?0.658a22222243PQ?4?cos750?0.681a20

20、(本大题满分16分) 解：(1)反函数f?1(x)?log2(x?1),(x??1)(2)log2(x?1)?log4(3x?1),?D?[0,1]3x?1(x?1)2(3)(理科)H(x)?log4(3x?1)?log2(x?1)?log4??log43x?1(x?1)2(x?1)2t?1记u?,t?3x?1?[1,4],x?,3x?13t?1(?1)2(t?2)2144843u???(t?)??,当且仅当t?时等号成立，而t9t9t99t8t?2?[1,4],?t?[1,2]单调递减，t?[2,4]单调递增，u(1)?u(4)?5,?u?[，1]，93?H(x)?[0,log23?]2(文科)H(x)?log4(3x?1)??log4(3?2),单调递减x?1122)?a在[0，1]有解x?1 1log2(x?1),x?[0,1]2H(x)min?H(1)?0,H(x)max?H(0)??y?H(x)与y?a有公共点即log4(3?1?a?[0,]2

21、(本题满分18分) 解(1)?f(x)?0仅有唯一的x值满足，???0，?a?0或4，?a?0,?a?4?s1(n?1)??3(n?1)Sn?n2?4n,an?????an?2n?5?Sn?Sn?1(n?2)?2n?5(n?2)(2)bn:b1?2?1?5,b2?2?2?5,b3?2?4?5,....bn?2?2n?1?5Tn?b1?b2?b3?...?bn?2(1?2?4?...?2n?1)?5n1?2n?2?5n?2(2n?1)?5n?2n?1?5n?21?2nn1(3)(文科)cn???(2n?5)(2n?3)4n2?16n?154n?15?16n11515c1?,c2??2,当n?3时，4n?单调递增，且4n??16?0,3nn?(cn)min?c2??2;?(cn)max?c3?1?cn?cn?1?2n?3(理科)?,?cn?2?cn?2,c1?1,c2?4c?c?2n?5n?2?n?1cn:135479...6810...k(k?1)n2?3n当n为偶数，n?2k,Hn?5?9?13?...?5k?4?22当n为奇数，n?2k?1,Hn?1?(7?11?15?...)(k?1)(k?2)n2?3n?2?1?7(k?1)?4?22?n2?3n,n?2k(k?1,2,3...)??2?Hn??2?n?3n?2,n?2k?1(k?1,2,3...)?2?当n?2k与n?2k?1时，分别比较Hn与Sn大小(作差比较)当1?n?10时，Hn?Sn当n?11时，Hn?Sn