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=}#}