2Tn?1?22?2?23?3?24?...??n?1??2n?n?2n?1
123nn?1两式相减得?Tn?2?2?2?...?2?n?2??1?n??2n?1?2
n?1所以Tn?2??n?1??2.
?2??2n?1?2?1?n?2n?1?2n?1?2?n?2n?1
18.【答案】(1)见解析;(2)1021 3
【解析】证明:(1)过点PH?BC交BC于H点,连接QH,可知QHPDC,PHPEC,可知面PHDP面EFDC,则PQP面FECD.
(2)连接AE,AC,作DM?AB交AB于M点,DM?AD2?AM2?52?22?21, 可知多面体ABCDEF分为两部分,四棱锥A?DCEF,三棱锥E?ABC
114VA?DCEF??DM?SDCEF??21?22?21,
333
111VE?ABC??EC?S?ABC??2??21?6?221,可知该几何体的体积为
332V?VA?DCEF?VE?ABC?
41021?221?21. 33
M
y?19. 【答案】(1)$3433x?. (2) 225【解析】(1)根据给出的数据可知x?1?2?3?4?520?25?30?30?25?3,y??26,
55$?可知b?xyii?15i?15i?5x?y2??x2i?5x(1?20?2?25?3?30?4?30?5?25)?5?26?3153??,且经过点
(12?22?32?42?52)?5?32102343$$?43,则回归直线方程为$?3?a,?ay?x?,故所求的回归直线的?x,y?,可知26?32222方程为y$?3432x?2. (2)根据分层抽样可知,1班选2名分为A1,A2,3班选3名分为B1,B2,B3
所有的情况为:(A1,A2,B1),(A1,A2,B2),(A1,A2,B3),(A1,B1,B2),(A1,B1,B3),(A1,B2,B3), (A2,B1,B2),(A2,B1,B3),(A2,B2,B3),(B1,B2,B3),共10种情况
其中1班1名,3班2
名的有(A1,B1,B2),(A1,B1,B3),(A1,B2,B3)(A2,B1,B2),(A2,B1,B3),(A2,B2,B3)共有6种,所求的概率为p?610?35. 【答案】(1)x2y220.4?2?1(2)见解析
??2?2???a2?12b2?1?a?2【解析】(1)由题意,得??c?2得??b?2?a2 ?222??a?b?c?c?2??故椭圆的方程为x2y24?2?1 ?x2y2(2)①由题意知y,由???4?2?1得?x22?2x?16?8y20?00?2y0x?8x00?0 ?x0x?2y0y?4?0因为点M?x2?2y24,则x2?2x220,y0?在椭圆上,所以x00?0x?x0?0,即?x?x0??0,得x?x0,y?y0.
所以直线l与椭圆C有且只有一个公共点,即点M. ②由(1)知A??2,0?,B?2,0?
过点A且与x轴垂直的直线的方程为x??2,
,
?x?2?结合方程x0x?2y0y?4?0,得点P??2,0?
y0??x0?2?0x?2 直线PB的斜率k?y0??0?2?24y0则直线PB的方程为y??x0?2?x?2?. 4y0y0??因为MN?AB于点N,所以N?x0,0?,线段MN的中点坐标为?x0,?
2??2x0?24?x0?x0?2?? 令x?x0,得y??4y04y022??x02y0y??0, 因为x?2y?4,所以y?4y04y022020y0??所以直线PB经过线段MN的中点?x0,?.
2??21.【答案】(1)见解析;(2)(??,?1710].
【解析】(1)函数f(x)?lnx?ax2?3x的定义域为(0,??),
f'(x)?1?2ax?3,f'(1)?1?2a?3?0,a?1, x212x2?3x?1?0, 可知f(x)?lnx?x?3x,f'(x)??2x?3?xx1?1??1?x1?1,x2?,可知在x??0,?时,f'(x)?0,函数f(x)单调递增,在x??,1?时,f'(x)?0,
2?2??2?函数f(x)单调递减,在(1,??)单调递增,可知函数f(x)的极小值为f(1)?ln1?1?3??2,1135?1?极大值为f???ln????ln2?.
2424?2?m(x2?x1)mm(2)f(x1)?f(x2)?可以变形为f(x1)?f(x2)??,可得
x2x1x1x2f(x1)?mmm?f(x2)?,可知函数f(x)?在[1,10]上单调递减 x1x2xh(x)?f(x)?h'(x)?mm?lnx?x2?3x?, xx1m?2x?3?2?0,可得 xx32m??2x3?3x2?x,设F(x)??2x?3x?x,
1?1?F'(x)??6x?6x?1??6?x????0,可知函数F(x)在[1,10]单调递减,
2?2?22F(x)min?F(10)??2?103?3?102?10??1710,可知
m??1710,可知参数m的取值范围为(??,?1710]. 22.【答案】(1)4p;(2)见解析
22【解析】(1)由?sin??2pcos??p?0?,x??cos?,y??sin?,得y?2px?p?0?,
?p?其交点F?,0?.
?2?又弦BC所在直线的倾斜角为??x???其参数方程为??y???3π, 4p2?t22(t为参数). 2t22将它代入y?2px?p?0?中,整理得t2?22pt?2p2?0, ??16p2?0,
设B,C对应的参数分别为t1,t2,则t1?t2??22p,t1t2??2p2,
?BC?t1?t2??t1?t2?2?4t1t2?4p.
p??x??tcos?2(2)根据题意可设弦BC所在直线的倾斜角为?,则直线BC的参数方程为???y?tsin?2(t为参数),代入y?2px?p?0?,整理得t2sin2??2ptcos??p2?0.
??4p2cos2??4p2sin2??4p2?0,
2pcos??p2设B,C对应的参数分别为t1,t2,则t1?t2?,t1t2?2.
sin2?sin?p2. 则FB?FC?t1?t2?t1t2?2sin?1pcos?QM为BC的中点,?MF?t1?t2?,
2sin2??MN?MF?tan??2pcos?p?tan??,
sin2?sin?p22?MN?2,MN?FB?FC.
sin?
1323. 【答案】(1)xx??1或x?1(2)??t?
22??【解析】(1)当a?b?1,c?3时,函数y?log2[f(x)?2c]?log2?x?1?2x?1?3?, ∴2x?1?x?1?3?0,
当x??1时,?(2x?1)?(x?1)?3?0,可得x??1; 当?1?x?1时,可知 2?2x?1?x?1?3?0,解得x??1,可知无解; 1时 22x?1?x?1?3?0,x?1,可知x?1
当x?故函数的定义域为xx??1或x?1.
???3x?a?b?c,x??a?b?(2) f?x??x?a?2x?b?c???x?a?b?c,?a?x?,根据函数的解析式可知
2?b?3x?a?b?c,x???2??当x?bbb9时,取得最小值为a??c,Q2a?b?2c?9,?a??c?, 22229?2t2?2t?3,?4t2?4t?3?0,(2t?3)(2t?1)?0,解得2f(x)?2t2?2t?3恒成立,可知
1313??t?,故参数t的取值范围为??t?. 2222