2026年高三年级综合测试——数学参考答案_2024-2026高三（6-6月题库）_2026年04月高三试卷_2604302026届广东东莞市高三第二次模拟考试（全科）

2026 年高三年级综合测试 数学 参考答案 一、单项选择题 题号 1 2 3 4 5 6 7 8 答案 A C B B D C A D 二、多项选择题 题号 9 10 11 答案 ABD AD ACD 三、填空题 12．1 13． f  x 1（答案不唯一）如 f  x 0， f  x  x， f  x  x3 等 2 64 14． ， （前2分，后3分） 15 15 四、解答题 15. 解：（1）设所求的线性回归方程为yˆ b ˆ xaˆ， 12345 x  3，······················································································· 1分 5 0.50.711.31.5 y  1，··············································································· 2分 5 5 (x x)(y y)(2)(0.5)(1)(0.3)10.320.52.6，······························3分 i i i1 5 (x x)2 411410·················································································· 4分 i i1 5 (x x)(y y) i i b ˆ i1 0.26···················································································5分 5 (x x)2 i i1 所以aˆ  yb ˆ x10.2630.22··········································································· 6分 所以yˆ 0.26x0.22····························································································7分 （2）当x6时，y0.2660.221.78，则Y N  1.78,0.022 ··································8分 由正态分布性质，可知PY 1.75PY 1.76PY  .···································9分 因为P(Y)0.6827，···············································································10分 P(Y) 0.6827 所以PY 0.5 0.5 0.84135.···································11分 2 2 数学参考答案 第1页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}25 因为0.84135 ，···························································································12分 30 所以该月日参与人数超过1.75万人的天数不少于25天.············································13分 16. 解：（1）方法一：因为AB1,AC 2,BC  3, 所以AB2BC2  AC2, 即AB BC,··1分 因为PA平面ABCD,BC平面ABCD, 所以PABC,·············································2分 又因为ABPA A,AB平面PAB,PA平面PAB, 所以BC平面PAB,······················3分 因为PB平面PAB, 所以BC PB,·······································································4分 所以PBA为平面PBC 与平面ABC所成的角,··························································5分 因为PA AB, 所以PBA45, 2 即平面PBC 与平面ABC所成角的余弦值为 ;······················································· 6分 2 方法二：以A为坐标原点，建立空间直角坐标系如下图所示, 3 1 A(0,0,0),B( , ,0),C(0,2,0),P(0,0,1),···································································1分 2 2  3 1  3 3 则PB( , ,1),BC ( , ,0),······································································2分 2 2 2 2  3 1   n    P  B  0   2 x 2 yz0 设平面PBC 的法向量为n(x,y,z), 则  , 即 ,······················3分 nBC 0  3 3  x y 0   2 2  令x 3, 则y1,z2, 所以n( 3,1,2),······························································4分  因为PA底面ABCD, 所以平面ABC的法向量为PA(0,0,1),····································5分     |nPA| 2 2 因此平面PBC与平面ABC所成角的余弦值为|cosn,PA|     ;·····6分 |n||PA| 2 21 2 （2）方法一：取线段PB上的中点E, 因为PA AB, 所以AE  PB,····························7分 由（1）可知BC平面PAB,AE平面PAB, 所以BC  AE,······································8分 又因为PBBC B,PB平面PBC,BC平面PBC, 所以AE平面PBC,·····················9分 因为AE平面ADE, 所以平面ADE平面PBC,····················································10分 延长DA、CB交于点G, 连接GE, 并延长GE交线段PC于点F ,则A,E,F,D四点共面, 过点P作PH FG,交CB延长线于点H , ·······························································11分 数学参考答案 第2页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}CF CG 因为CFG CPH , 所以  ①·································································12分 PF HG 1 CG 因为BEG BPH , 所以BE BG 2 ②······················································· 13分   PE HG HG CF 2BE PF 1 联立①②可得  2, 即  ,·······························································14分 PF PE PC 3 所以存在这样的点E,F满足题意，此时点E位于线段PB上的中点、点F位于线段PC上靠近点P 的三等分点.····································································································· 15分 方法二：取线段PB上的中点E, 因为PA AB, 所以AE  PB,··································· 7分 由（1）可知BC平面PAB,AE平面PAB, 所以BC  AE,······································8分 又因为PBBC B,PB平面PBC,BC平面PBC, 所以AE平面PBC,·····················9分 因为AE平面ADE, 所以平面ADE平面PBC,····················································10分 以A为坐标原点，建立空间直角坐标系如下图所示, 3 1 3 1 1 A(0,0,0),D( , ,0),E( , , ), 2 2 4 4 2 假设存在这样的点F，使得A,E,F,D四点共面，不妨设F(0,22b,b)(其中0b1),·······11分  3 1  3 1 1  则AD( , ,0),AE ( , , ),AF (0,22b,b),··············································12分 2 2 4 4 2    因为存在唯一的有序实数对(x,y)，使得AF  xAE yAD, 3 1 1 3 1 3 3 1 1 1 所以(0,22b,b) x( , , ) y( , ,0)( x y, x y, x),····················13分 4 4 2 2 2 4 2 4 2 2 2 4 2 2 2 解得b ,x ,y ，此时F(0, , ),································································14分 3 3 3 3 3 所以存在这样的点E,F满足题意，此时点E位于线段PB上的中点、点F位于线段PC上靠近点P 的三等分点.····································································································· 15分 方法三：以A为坐标原点，建立空间直角坐标系如下图所示, 数学参考答案 第3页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}3 1 3 1 A(0,0,0),B( , ,0),C(0,2,0),D( , ,0),P(0,0,1), 2 2 2 2 假设存在这样的点E,F， 1 不妨设E( 3a,a,12a)(其中0 a ),F(0,22b,b)(其中0b1),·································7分 2  3 1   则AD( , ,0),AE ( 3a,a,12a),AF (0,22b,b),···········································8分 2 2    3 1   nAD0   x y0 设平面ADE 的法向量为m(x,y,z), 则  , 即 2 2 , nAE 0   3axay(12a)z0 6a  6a 令x 3, 则y3,z , 所以m( 3,3, ),·················································9分 2a1 2a1  因为平面ADE与平面PBC垂直，由（1）可得平面PBC 的法向量为n( 3,1,2),··········10分   1 3 1 1 由mn0，可得a ，此时E( , , ),···························································· 11分 4 4 4 2    又因为A,E,F,D四点共面，所以存在唯一的有序实数对(x,y)，使得AF  xAE yAD, 3 1 1 3 1 3 3 1 1 1 即(0,22b,b) x( , , ) y( , ,0)( x y, x y, x),·······················13分 4 4 2 2 2 4 2 4 2 2 2 4 2 2 2 解得b ,x ,y ，此时F(0, , ),································································14分 3 3 3 3 3 所以存在这样的点E,F满足题意，此时点E位于线段PB上的中点、点F位于线段PC上靠近点P 的三等分点.····································································································· 15分 17. 解：（1）当a0时， f xtanx2x，···························································· 1分 1 所以 f'x 2，······················································································ 2分 cos2x    1 2cosx 1 2cosx  ，···············································································3分 cos2x π π π 当x(0，)时，cosx0，令 f'x0，得 x ，·············································4分 2 4 2 π 令 f'x0，得0x ，··················································································5分 4 数学参考答案 第4页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}π π π 故 f(x)的减区间为(0，)，增区间为( ，)；···························································6分 4 4 2 （2）由 f(x)b0恒成立，即basinx2xtanx，·················································7分 令gxasinx2xtanx，则bgx ································································8分 max 因为gxf(x)， π π π 当a0时，由（1）知，gx在(0，)上单调递增，在( ，)单调递减，······················9分 4 4 2 π π 故此时gx g( ) 1， max 4 2 π 所以，ab的最小值为 1；·············································································10分 2 1 当a0时，g'xacosx2 ，·································································11分 cos2x π 当x(0，)时，易得g'x为减函数，··································································12分 2 g'( π ) 2 a0，x π 时，g'x， 4 2 2 π π 由零点存在性定理得，存在x ( ，)，使得g'x 0，·········································13分 0 4 2 0 π gx在(0,x )上单调递增，在(x，)单调递减，故此时 0 0 2 π 2 π π gx  gx  g( ) a 1 1,··························································· 14分 max 0 4 2 2 2 π π 此时ab 1，综上：ab的最小值为 1.······················································15分 2 2 18. 解：（1）因为T 的两条渐近线为两条坐标轴，对称轴为 y x，顶点坐标分别为  1,1  ， 0   1,1 ，实半轴长为 2 ，······················································································1分  将曲线T 绕原点顺时针旋转 得到曲线T ，则T 是焦点在x轴上的双曲线，渐近线方程为 0 4 y x，实半轴长为 2 ，················································································· 2分   所以T 的左顶点A的坐标为  2,0 ，···································································3分 x2 y2 所以曲线T的标准方程为：  1··································································4分 2 2 数学参考答案 第5页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}（2）（i）方法一：  证明：因为曲线T是曲线T 绕原点顺时针旋转 得到，不妨设曲线T 上的点A,B,C分别对应曲 0 4 1 线T 上的点A',B',C'，曲线T 的方程为 y  ，则A'的坐标为(1,1)，··················· 5分 0 0 x 1 1 设点D'为A'B'C'的垂心，B'(x , ),C'(x , ),D'(x,y)，且x  x .······················6分 1 x 2 x 1 2 1 2 1 1 所以直线A'B'的斜率 x 1 ，·································································7分 k  1  1 x 1 x 1 1 1 1 边A'B'的高所在的直线l 的方程为 yx (xx ) x xx x  ， ①·····8分 1 1 2 x 1 1 2 x 2 2 1 同理：边A'C'的高所在的直线l 的方程为 y x xx x  ，②····························9分 2 2 1 2 x 1  1 x  1 因为垂心D'(x,y)同时在l 、l 上，联立①②，得 x x ，即 y  .···················· 10分 1 2 1 2  x y  x x 1 2  所以A'B'C'的垂心D'在曲线T 上.将曲线T 上的点 A',B',C',D'绕原点顺时针旋转 后得到 0 0 4 曲线T上对应的点分别为A,B,C,D，则ABC的垂心D在曲线T 上.························11分 方法二： 证明：因为 B、C 为曲线T 右支上不同的两点,设 B(x ,y ),C(x ,y ) ，设ABC 的垂心为  1 1 2 2   D x ,y ，有x2  y 2 2，x 2 y 2 2，CD(x x ,y  y )， AB x  2,y ，  0 0 1 1  2 2  0 2 0 2 1 1 BD(x x ,y  y )，AC  x  2,y ，·······················································5分   0 1 0 1 2 2 CDAB(x x )(x  2) y (y  y )0恒成立， ① 0 2 1 1 0 2   BDAC (x x )(x  2) y (y  y )0恒成立， ②·····································6分 0 1 2 2 0 1 ①式两边同时乘以 y ，得 y (x x )(x  2) y2(y  y )0，又因为 y 2  x2 2， 1 1 0 2 1 1 0 2 1 1 所以有y (x x )(x  2)(y  y )0恒成立. ③······································· 7分 1 0 2 1 0 2 数学参考答案 第6页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}同理，②式两边同时乘以y ，化简可得y (x x )(x  2)(y  y )0恒成立 ，④ 2 2 0 1 2 0 1 ③④，得：y (x x )(x  2)(y  y )0恒成立， ⑤··································8分 0 1 2 0 1 2 因为D x ,y 为ABC的垂心， 0 0   所以有ADCB(x  2)(x x ) y (y  y )0恒成立. ⑥································ 9分 0 1 2 0 1 2 ⑤式两边同时乘以 y ，得 y 2(x x ) y (x  2)(y  y )0恒成立，⑦ 0 0 1 2 0 0 1 2 ⑥式两边同时乘以(x  2)，得(x 2 2)(x x ) y (x  2)(y  y )0恒成立 ⑧ 0 0 1 2 0 0 1 2 ⑧⑦，得(x 2  y 2 2)(x x )0恒成立，······················································10分 0 0 1 2 当x x 0时，x2  y2 2恒成立，即ABC的垂心D  x,y  在曲线T 上. 1 2 当x x 0时，ABC的垂心是双曲线的右顶点，综上ABC的垂心总在曲线T 上.···11分 1 2 方法三： 证明：因为B、C为曲线T 右支上不同的两点, x2 y 2 当BC的斜率不存在时,不妨设B(x ,y ),C(x ,y )，则有 1  1 1， 1 1 1 1 2 2   所以BC边上的高所在的直线方程为y0，则y0与双曲线交于点D 2,0 ，············5分   下证D 2,0 为ABC的垂心.         因为DB x  2,y ，DC  x  2,y ，AC  x  2,y ,AB x  2,y ， 1 1 1 1 1 1 1 1 所以DBAC (x  2)(x  2) y 2  x2  y 2 20，所以DB AC, 1 1 1 1 1 DCAB(x  2)(x  2) y 2  x2  y 2 20,所以DC  AB, 1 1 1 1 1   即D 2,0 为ABC的垂心，所以垂心D在曲线T 右支上.········································6分 当BC的斜率存在时，设直线BC的方程： y kxb， x2 y2   1   联立 2 2 ，得 1k2 x2 2kbxb2 20，  y kxb 数学参考答案 第7页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}1k2 0      4k2b2 41k2 b2 2 0  k2 1 2kb 所以有x x  0 ， ，·······································7分  1 2 1k2  b2 22k2  b2 2 x x  0  1 2 1k2 b2 2k2 所以有y y   kx b  kx b   ，···························································8分 1 2 1 2 1k2 1    则BC边上的高所在的直线方程为 y x 2 ，交双曲线右支于点D x ,y ，交左支于点 0 0 k x2 y2   1  2 2   A,联立 ，可得 k2 1x2 2 2x2k2 20，  1  y  x 2   k 2(k2 1) 2(k2 1) 2 2k 所以有 2x  ，即x  ， y  ， 0 k2 1 0 k2 1 0 k2 1  2(k2 1) 2 2k 即D    k2 1 , k2 1    ··················································································9分 下证D为ABC的垂心. 2(k2 1) 2 2k   因为DB(x  ,y  )，AC  x  2,y , 1 k2 1 1 k2 1 2 2 2(k2 1) 2 2k   DC (x  ,y  )，AB x  2,y ， 2 k2 1 2 k2 1 1 1 2(k2 1) 2(k2 1) 2 2ky 所以DBAC  x x  2x  x   y y  2 ③ 1 2 1 k2 1 2 k2 1 1 2 k2 1 b2 2 b2 2k2 因为x x  , y y  ，y kx b，代入③中，可得： 1 2 k2 1 1 2 1k2 2 2 2 2kb 2(k2 1)x 2 2kb 2 2kb 2 2kb DBAC   2x  2   2(x x )  0 k2 1 1 k2 1 k2 1 1 2 k2 1 1k2 所以DB AC，····························································································· 10分 同理可得DCAB0，DC  AB， 2(k2 1) 即D为ABC的垂心，因为k2 1， 0所以垂心D在曲线T 右支上. k2 1 综上所述：ABC的垂心D  x,y  在曲线T上.·······················································11分 数学参考答案 第8页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}方法四： 证明：因为B、C为曲线T 右支上不同的两点, 当BC的斜率不存在时,由双曲线的对称性不妨设B(x ,y ),C(x ,y )， 1 1 1 1 所以BC边上的高所在的直线方程为y0， x  2 AC边上的高所在的直线方程为 y  1  xx   y ， y 1 1 1 y 2 联立可得，x x  1 ，因为点B在双曲线上，有x2  y 2 2，所以x  2 ， 1 x  2 1 1 1   即垂心D的坐标为 2，0 ，显然点D在曲线T 右支上；············································5分   当x  2或x  2 时，不妨设C 2,0 ,则边AC的高所在的直线方程为x x ，边 AB 的 1 2 1 x  2   高所在的直线方程为y  1 x 2 ，联立可得y y ， y 1 1 x2 (y )2   因为 1  1 1，所以D x ,y 在曲线T 上.···················································6分 1 1 2 2 当BC的斜率存在时，设B(x ,y ),C(x ,y )，且x  2,x  2， 1 1 2 2 1 2 x2 y2   1   设直线BC的方程： ykxb，联立 2 2 ，得 1k2 x2 2kbxb2 20，  y kxb 1k2 0      4k2b2 41k2 b2 2 0  k2 1 2kb 所以有x x  0 ， ，·······································7分  1 2 1k2  b2 22k2  b2 2 x x  0  1 2 1k2 数学参考答案 第9页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}b2 2k2 所以有y y   kx b  kx b   ，···························································8分 1 2 1 2 1k2 x  2 所以AB边上的高所在的直线方程为 y  1  xx   y ， ① y 2 2 1 x  2 AC边上的高所在的直线方程为 y  2  xx   y ， ② y 1 1 2   设垂心D的坐标为 x ,y ，联立①、 ②， 0 0     x  2 x  2 x x  2 x x  2 得    2 y  1 y    x 0  1 2 y  2 1 y   y 1  y 2  ，····························9分 2 1 2 1       x y x  2 x y x  2  y y y y 即垂心D的横坐标x  1 1 2 2 2 1 1 2 1 2 0 y (x  2)y (x  2) 1 2 2 1 x (kx b)(x  2)x (kx b)(x  2)k(x x )y y  1 1 2 2 2 1 1 2 1 2 (kx b)(x  2)(kx b)(x  2) 1 2 2 1   (x x )kx x  2k(x x ) 2bky y  1 2 1 2 1 2 1 2 (x x )( 2kb) 1 2      2 2kb k2 1 2 k2 1 所以x 0   1k2  b 2k   k2 1 ···························································10分   1  又因为垂心D x ,y 必然在BC边上的高所在的直线方程 y x 2 上， 0 0 k 2 2k  2  k2 1   2 2 2k  2 2  k2 1 2 所以 y  ，即有x 2  y 2      2， 0 k2 1 0 0   k2 1     k2 1    k2 1 2   2 k2 1 x2 y2 因为k2 1，所以x  0，即垂心D在曲线T ：  1的右支上. 0 k2 1 2 2 综上所述：ABC的垂心D在曲线T 上.·······························································11分 （ii）方法一 证明：由（i）可知，曲线T 上的点A,B,C,D分别对应曲线T 上的点A',B',C',D'，有 0 1 1 1 A' 1,1  ,B'(x , ),C'(x , )，A'B'C'的垂心D' ( ,x x )在曲线T 上，则D'关于原 1 x 2 x x x 1 2 0 1 2 1 2 1 点的对称点E' ( ,x x )也在曲线T 上.·························································12分 x x 1 2 0 1 2 1 由双曲线的对称性，不妨设点B'在直线A'C'的上方，E'在直线A'C'的下方，因为x  x  ， 1 2 x x 1 2 数学参考答案 第10页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}1 1 所以直线 A'B', A'E',C'B',C'E'的斜率都存在.即k  ，k  x x ，k  ， A'B' x A'E' 1 2 C'B' x x 1 1 2 k  x ，····································································································13分 C'E' 1 1 x x k k x 1 2 1x2x 当x 1时，所以tanB'A'E' A'B' A'E'  1  1 2 ， 2 1k k 1 x (1x ) A'B' A'E' 1 x x 1 2 x 1 2 1 1 x  k k 1 x x 1x 2x tanB'C'E' C'E' C'B'  1 2  1 2 ,··········································14分 1k k 1 x (1x ) C'E' C'B' 1 x 1 2 x x 1 1 2 即有tanB'A'E'tanB'C'E'0，所以B'A'E'B'C'E'. 所以A',B',C',E'四点共圆.·················································································15分 1 1 1 当x 1时，E' ( ,x )，所以A'B'(x 1, 1),A'E'( 1,x 1)， 2 x 1 1 x x 1 1 1 1 1 1 所以A'B'A'E'(x 1 1)( x 1)( x 1)(x 1 1)0，所以 A'B' A'E'···············16分 1 1 同理：可得C'B'C'E'，即B'A'E'B'C'E',所以A',B',C',E'四点共圆.  将点A',B',C',E'绕原点顺时针旋转 后得到对应的点分别为A,B,C,E ， 4 则A,B,C,E 四点共圆.·······················································································17分 方法二 证明：由（i）可知，点D在双曲线T 右支上，因为双曲线T 的图像关于原点中心对称，E为D 关于原点的对称点，所以点E在双曲线T 的左支上.   当BC的斜率不存在时,不妨设B(x ,y ),C(x ,y )，此时垂心D的坐标为 2,0 ，则D关于原 1 1 1 1 点的对称点E与点A( 2,0)重合.因为ABC存在外接圆，所以A,B,C,E四点共圆. ····················································································································· 12分 当BC的斜率存在时，设直线BC的方程为 y kxb， 由双曲线的对称性，不妨设点B在直线AC的上方，点E在直线AC的下方，则此时AB,AE,CE 数学参考答案 第11页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}    的斜率都存在，设点D x ,y ，点E x ,y ， 0 0 0 0 x 2 y 2 ky y 1 因为点D为垂心，且 0  0 1，所以k k  0 1，即k  0  ， 2 2 DA BC x  2 DA x  2 k 0 0 y y y 2 又因为k k  0  0  0 1，即k k k ，··················13分 DA EA x  2  2x x 2 2 EA CB 0 0 0 x 2 y 2  0  0 1  2 2 因为 ，所以x 2 x 2  y 2  y 2， x 2 y 2 2 0 2 0 2  2 1   2 2 y  y y  y y 2  y 2 因为k k 1，k k  2 0  2 0  2 0 1，所以k k ，·····14分 DC AB CD EC x x x x x 2 x 2 AB EC 2 0 2 0 2 0 当k k  1时，此时有k k 1，即AB AE,CBCE,所以BAEBCE AB AE EC CB 则A,B,C,E 四点共圆；·····················································································15分 k k k k 当k k  1时，即k k 1，此时有tanBAE  AB AE  AB ， AB AE EC CB 1k k 1k k AB AE AB k k k k tanBCE CE CB  AB ，则tanBAEtanBCE0，·····················16分 1k k 1k k CE CB AB 即BAEBCE，则A,B,C,E四点共圆. 综上所述：A,B,C,E 四点共圆.···········································································17分 方法三 证明：由（i）可知，点D在双曲线T 右支上，因为双曲线T 的图像关于原点中心对称，E为D 关于原点的对称点，所以点E在双曲线T 的左支上.   当BC的斜率不存在时,不妨设B(x ,y ),C(x ,y )，此时垂心D的坐标为 2,0 ，则D关于原 1 1 1 1 点的对称点E与点A( 2,0)重合.因为ABC存在外接圆，所以A,B,C,E四点共圆. ····················································································································· 12分 当BC的斜率存在时，设直线BC的方程为 y kxb， 由双曲线的对称性，不妨设点B在直线AC的上方，点E在直线AC的下方，则此时AB,AE,CE 数学参考答案 第12页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#} 2(k2 1) 2 2k 的斜率都存在，设点D , ，   k2 1 k2 1   2 2k 则E      2 k ( 2 k  2 1 1) , 2 k2  2k 1     ,k EA  2(k k 2 2   1 1 ) k k BC ，························13分   2 k2 1 2 2k  y 因为k  y 1 ，k  k2 1 2 ， AB x  2 CE 2(k2 1) 1  x k2 1 2 2 2k  y y k2 1 2 y 2 2k  y (k2 1) 则k k  1   1  2 AB CE x  2 2(k2 1) x  2 2(k2 1)(k2 1)x 1  x 1 2 k2 1 2     y 2(k2 1)(k2 1)x (x  2)2 2k y (k2 1)  1  2 1  2 (x  2) 2(k2 1)(k2 1)x 1 2 2(k2 1)(y  y )2 2y (k2 1)(x y x y )2 2kx 4k  1 2  2 1 2 2 1 1 (x  2) 2(k2 1)(k2 1)x 1 2 2b 4k 因为 y  y k(x x )2b ，x y x y 2kx x b(x x ) ， 1 2 1 2 1k2 1 2 2 1 1 2 1 2 k2 1 2b 2(k2 1) 2 2k(x x )2 2b 代入得： 1k2 1 2 k k    AB CE (x  2) 2(k2 1)(k2 1)x 1 2   2 2b(k2 1)2k2 (1k2) 即k k   0 AB CE (1k2)(x  2) 2(k2 1)(k2 1)x 1 2 所以k k ··························································································15分 AB CE 当k k  1时，此时有k k 1，即AB AE,CBCE,所以BAEBCE AB AE EC CB 则A,B,C,E 四点共圆；·····················································································16分 k k k k 当k k  1时，即k k 1，此时有tanBAE  AB AE  AB ， AB AE EC CB 1k k 1k k AB AE AB k k k k tanBCE CE CB  AB ，则tanBAEtanBCE0， 1k k 1k k CE CB AB 即BAEBCE，则A,B,C,E四点共圆. 综上所述：A,B,C,E 四点共圆.···········································································17分 19.解：(1)（i）5、7、11、13················································································ 4分 （ii）设m  a 3n a 3n1a 32 a 31a ,其中a  0,1,2  , iN,1in n n1 2 1 0 i 数学参考答案 第13页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}n 则S  m a ，·······························································································5分 i i0 因为3m2a 3n1a 3n a 33a 32 a 312，·······························6分 n n1 2 1 0 n 所以S  3m2 a 2S  m 2，································································7分 i i0 同理，9m4a 3n2 a 3n1a 34 a 33a 32 3130，··················8分 n n1 2 1 0 n 所以S  9m4 a 2S  m 2，所以S  3m2 S  9m4  ．······················9分 i i0 （2）因为4T 2 b2 2mb m2  0，所以4T 2 b2 2mb m2，即(2T )2  b m 2， n n n n n n n n 由数列  b  为正项数列，则T 0，所以2T  b m .···········································10分 n n n n 又因为mN*，m80，则m0，   所以2T 2(b b b )b b m，所以2T mb ， nN* ·················11分 n 1 2 n n n n n 1 则T  mb  n 2 ，两式作差得b  b  n2 ， n1 n1 n 3 n1 又因为T b  m ，所以  b  是以 m 为首项， 1 为公比的等比数列，可得b  m .···· 12分 1 1 3 n 3 2 n 3n 方法一： 因为m80，设m33x 32x 31x 30x （x ，x ，x ，x  0,1,2  ）， 3 2 1 0 0 1 2 3  1  2 m1 m2 b   b      1 3     1 3     3     3   33x 32x 3x x 1 33x 32x 3x x 2  3 2 1 0  3 2 1 0      3   3  x 1 x 2 232x 23x 2x  0  0 3 2 1   3     3   x 1 x 2 当x 0时， 0  0 0； 0   3     3   x 1 x 2 当x 0 1时，   0 3      0 3   1， x 1 x 2 当x 2时， 0  0 2，     0  3   3  x 1 x 2  1  2 所以   0 3      0 3    x 0 ，即   b 1  3      b 1  3   232x 3 23x 2 2x 1 x 0 ；····· 13分 数学参考答案 第14页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#} 1  2 m 3  m 6  b   b        2 3     2 3    32 32    32 32   33x 32x 3  x 1 x  33x 32x 3  x 2 x   3 2 1 0  3 2 1 0  32 32     3  x 1 x  3  x 2 x  23x 2x  1 0  1 0  3 2  32   32  3  x 1 x  3  x 2 x  3x  6x  当x 0时， 1 0  1 0   0    0  0； 1  32   32   32   32  3  x 1 x  3  x 2 x  6x  9x  当x 1时， 1 0  1 0    0    0  1， 1  32   32   32   32  3  x 1 x  3  x 2 x  9x  12x  当x 2时， 1 0  1 0   0    0  2， 1  32   32   32   32  3  x 1 x  3  x 2 x  所以 1 0  1 0   x ，  32   32  1  1  2 即   b 2  3      b 2  3   23x 3 2x 2 x 1 ，····························································14分 同理可得  1  2  1  2   b 3  3      b 3  3   2x 3 x 2 ，   b 4  3      b 4  3    x 3 ，······································15分 1 2 m 2 80 2  1  2 当n5时，b n  3 b n  3  3n  3  243  3 1，所以   b n  3      b n  3   0.······16分   1  2 所以      b i ＋ 3      b i ＋ 3     i1  1  2  1  2  1  2 =  b 1  3      b 1  3      b 2  3      b 2  3      b 4  3      b 4  3   (232x 23x 2x x ) (23x 2x x ) (2x x ) x 3 2 1 0 3 2 1 3 2 3   232 2321  x  2321  x  21  x x 3 2 1 0 33x 32x 3x x m．证毕······································································17分 3 2 1 0 m 方法二：上接b  ， n 3n 对任意实数x，记xk f ，其中k  x  ，0 f 1 1  1  2 当0 f  时，  3x 3k ，  x  x  x kkk 3k     3  3  3 数学参考答案 第15页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}1 2  1  2 当  f  时，  3x 3k 1，  x  x  x k k1 k 3k1     3 3  3  3 2  1  2 当  f 1时，  3x 3k2，  x  x  x k k1  k1 3k2     3  3  3  1  2 故对任意实数x，恒有  3x  x  x  x ，      3  3  1  2 则  x    x   3x  x  .·········································································15分  3  3 m  1  2  m  m 代入xb  ，得 b   b   3b  b   ， n 3n   n 3     n 3   n n  3n1    3n    m  N  1  2 取正整数N 满足3N m，则  3N   0，记H N      b n  3      b n  3     ， n1 N  m  m  m  则H N     3n1     3n      m   3N    m， n1  m    1  2 当N ，  3N   0，故     b n  3      b n  3     m ，得证.······························17分 n1 数学参考答案 第16页（共16页） {#{QQABAYSlwgAwgISACY4qQwleCUqYkIEjJIgshVCQOAxLCJNABIA=}#}