数学答案2025适应性练习数学参考答案（仅供参考）_2025年3月_250317福建2025届高中毕业班适应性练习卷（3月高三省质检）_2025福建高三3月适应性考试数学

2025 届高中毕业班适应性练习卷 数学参考答案及评分细则 评分说明： 1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内 容比照评分标准制定相应的评分细则。 2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度， 可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答 有较严重的错误，就不再给分。 3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。 4．只给整数分数。选择题和填空题不给中间分。 一、选择题：本题考查基础知识和基本技能。每小题5分，满分40分。 1．C 2．D 3．A 4．A 5．B 6．C 7．A 8．B 二、选择题：本题考查基础知识和基本技能。每小题 6 分，满分 18 分。全部选对的得 6 分，部分选对的 得部分分，有选错的得0分。 9．AD 10．ABD 11．ACD 三、填空题：本题考查基础知识和基本技能。每小题5分，满分15分。 1 12．y2 =4x 13．g(x)=2x（写出一个满足条件的函数即可） 14． 16 四、解答题：本题考查基础知识、基本技能、基本思想和基本经验，考查发现问题、提出问题、分析问题 和解决问题的能力。共5小题，满分77分。 15. 本小题主要考查数列的基本概念、等比数列等基础知识，考查运算求解能力、逻辑推理能力等，考查 函数与方程思想、分类与整合思想等，考查数学运算、逻辑推理等核心素养，体现基础性和综合性．满 分13分． 解法一：（1）因为a −S =1，所以当n2时，a −S =1， ·················································· 1分 n+1 n n n−1 所以a −a −S +S =0， ························································································ 2分 n+1 n n n−1 所以a =2a（n2）． ································································································· 4分 n+1 n 当n=1时，a =S +1=2，满足a =2a ，也符合a =2a ． ············································· 5分 2 1 2 1 n+1 n a 因为a ≠0，所以a ≠0， n+1 =2， 1 n a n 所以{a }是以1为首项，2为公比的等比数列， ································································ 6分 n 所以a =2n−1． ·········································································································· 7分 n 数学参考答案及评分细则 第1页（共 19页）n2 （2）由（1）知a =2n−1，所以b = ，······································································· 8分 n n 2n−1 (n+1)2 所以 b n+1 = 2n = (n+1)2 ． ······················································································ 9分 b n2 2n2 n 2n−1 b 令 n+1 >1，解得− 2+11，又因为b >0，所以b >b ； ······································· 11分 n n+1 n b n 同理，当n3且n∈N*时，b 0 ，解得− 2+1b ； ·············································································· 11分 n+1 n 同理，当n3且n∈N*时，b 0)， ································································································ 9分 2x−1 x(2−xln2) 则 f′(x)= ， ····························································································· 10分 2x−1 2 2 所以当x∈(0, )时， f′(x)>0；当x∈( ,+∞)时， f′(x)<0． ln2 ln2 2 2 所以 f (x)在(0, )单调递增，在( ,+∞)单调递减． ···················································· 11分 ln2 ln2 2 9 又2< <3，b = f (n)，b =2，b = ， ln2 n 2 3 4 所以b b >b >， ····················································································· 12分 1 2 3 4 5 9 所以b 的最大值为b = ． ························································································· 13分 n 3 4 16．本小题主要考查直线与平面垂直的判定与性质、平面与平面垂直的判定与性质、直线与平面所成的角、 二面角、点到直线的距离、空间向量和解三角形等基础知识，考查直观想象能力、逻辑推理能力、运 算求解能力等，考查函数与方程思想、数形结合思想、化归与转化思想等，考查直观想象、逻辑推理、 数学运算等核心素养，体现基础性、综合性．满分15分． 解法一：（1）由题可知，∠ADC =45°，CD=2 2 ，AC = AP=2， CD AC 在△ACD中，由正弦定理，得 = . sin∠CAD sin∠ADC 2 2 2 所以 = ，sin∠CAD=1， ·········································································· 2分 sin∠CAD 2 2 所以∠CAD=90°，即AD⊥ AC ， ················································································· 3分 所以AD=2. 又AP=2，PD=2 2 ，所以AP2 + AD2 =PD2，即AD⊥ AP. ··········································· 5分 又因为APAC = A，AP,AC⊂平面PAC ，所以AD⊥平面PAC . ······································ 6分 又因为AD⊂平面ACD，所以平面PAC ⊥平面ACD. ······················································· 7分 数学参考答案及评分细则 第3页（共 19页）（2）过M 作ME⊥ AC于点E，过E作EF ⊥ CD于点F ，连接MF. 因为平面PAC ⊥平面ACD，平面PAC∩平面ACD= AC，ME ⊂平面PAC ， 所以ME ⊥平面ACD. ································································································ 9分 又CD ⊂平面ACD，所以ME ⊥ CD. 又EF ⊥ CD，ME,EF ⊂平面MEF ，EF∩ME =E，所以CD ⊥平面MEF . ······················· 10分 P 又MF ⊂平面MEF ，所以CD⊥MF ，即M 到直线CD的距离为MF. 14 14 M 因为点M 到直线CD的距离为 ，所以MF = . ······················································ 11分 2 2 N 设AE=x，则CE=2−x. B A 在Rt△AEM 中，∠EAM =60°，所以ME= 3x. E C F D 2 在Rt△EFC中，由（1）可知，∠ECF =45°，所以EF = (2−x). 2 又ME ⊥平面ACD，EF ⊂平面ACD，所以ME⊥EF . 14 2 在Rt△EFM 中， MF2 =ME2 +EF2，即( )2 =( 3x)2 +[ (2−x)]2， ····························· 12分 2 2 3 解得x=1或x=− （舍去）.所以AM =2，故M 与点P重合. ··········································· 13分 7 取线段PA的中点N，连接CN ，易知CN ⊥ AP. 由（1）知AD⊥平面PAC ,CN ⊂平面PAC ，所以AD⊥CN . 又ADAP= A，AD,AP⊂平面PAD， 所以CN ⊥平面PAD， 所以∠CPN为直线CM 与平面PAD所成的角. ································································· 14分 z 因为在等边△ABC中，∠CPN =60°，所以直线CM 与平面PAD所成的角为60°. ·P················· 15分 Z 解法二：（1）同解法一.····································································································· 7分 （2）在平面PAC 内作AZ ⊥ AC，如图. B 由（1）知平面PAC ⊥平面ACD，且平面PAC 平面ACD= AC， A 所以AZ ⊥平面ACD. C D    x y 又AC ⊥ AD，分别以AC,AD,AZ为x,y,z轴的正方向建立空间直角坐标系Axyz，如图所示. ····· 8分 则A(0,0,0)，C(2,0,0)，D(0,2,0)，P(1,0, 3).设AM =2m（0≤m≤1），则M(m,0, 3m).     所以AD=(0,2,0)，AP=(1,0, 3)，CD=(−2,2,0)，CM =(m−2,0, 3m). ····························· 9分    CD (−2,2,0) 2−m 所以|CM |2=(m−2)2 +( 3m)2 =4m2 −4m+4，且CM ⋅  =(m−2,0, 3m)⋅ = . |CD| 2 2 2 数学参考答案及评分细则 第4页（共 19页）设点M 到直线CD的距离为d，    CD 所以d2 =|CM |2 −(CM ⋅  )2 ····················································································· 11分 |CD| 2−m 7 =4m2 −4m+4−( )2 = m2 −2m+2. 2 2 7 14 3  依题意得， m2 −2m+2=( )2，解得m=1或m=− （舍去），所以CM =(−1,0, 3). ········ 12分 2 2 7 设平面PAD的法向量n=(x,y,z)，直线CM 与平面PAD所成的角为θ.   AP⋅n=0, x+ 3z=0， 由 得 AD⋅n=0, 2y=0. 取z=1，则平面PAD的一个法向量为n=(− 3,0,1)， ······················································ 13分   |CM ⋅n| 所以sinθ=|cos|=  ············································································· 14分 |CM ||n| 3 = ， 2 所以直线CM 与平面PAD所成的角为60°. ····································································· 15分 P 解法三：（1）取AC的中点Q，连接PQ，QD. 因为△ABC是边长为2的等边三角形， B 所以AC =2，PQ⊥ AC. ···············································································A·············· 1分 Q 在△ACD中，∠ADC =45°，CD=2 2 ， C D 2 由余弦定理，得AC2 =CD2 + AD2 −2CD⋅ADcos∠ADC，22 =（2 2)2 + AD2 −2×2 2AD× ， 2 解得AD=2， ··········································································································· 2分 所以AD2 + AC2 =CD2，所以∠CAD=90°，即AD⊥ AC . ·················································· 3分 1 又因为AQ= AC =1，所以QD2 = AD2 + AQ2 =5. 2 又PQ= 3，PD=2 2 ，所以PQ2 +QD2 =PD2，所以PQ⊥QD. ······································ 4分 又PQ⊥ AC，QDAC =Q，QD,AC ⊂平面ACD， ······················································· 5分 所以PQ⊥平面ACD. ································································································· 6分 又因为PQ⊂平面PAC ，所以平面PAC ⊥平面ACD. ······················································· 7分 （2）取CD的中点R，连接QR，所以QR∥AD，又AD⊥ AC ，所以QR⊥AC.    结合（1）可知，PQ,AC,QR两两互相垂直，分别以QC,QR,QP为x,y,z轴的正方向建立空间直角坐 数学参考答案及评分细则 第5页（共 19页）标系Qxyz，如图所示. ································································································ 8分 依题意得，P（0,0, 3），A(-1,0,0)，C(1,0,0)，D(-1,2,0)， z     所以AP=(1,0, 3)，CA=(−2,0,0)，CD=(−2,2,0)，AD=(0,2,0)． ···································· 9分 P   因为M 是线段PA上的点，可设AM =λAP=(λ,0, 3λ) (0≤λ≤1)，    所以CM =CA+ AM =(λ−2,0, 3λ). B 14 A 因为点M 到直线CD的距离为 ， Q 2    CM ⋅CD 14 x C R D 所以 |CM |2 −(  )2 = ， ·············································································· 11分 |CD| 2 y 4−2λ 14 即 (λ−2)2 +3λ2 −( )2 = ， 2 2 2 3 解得λ=1或λ=− （舍去）． 7  所以CM =（-1,0, 3）． ······························································································· 12分 设平面PAD的法向量n=(x,y,z)，直线CM 与平面PAD所成的角为θ.   AP⋅n=0, x+ 3z=0, 由 得 AD⋅n=0, 2y=0. 取z=−1，则平面PAD的一个法向量为n=( 3,0,−1)． ···················································· 13分   |CM ⋅n| 所以sinθ=|cos|=  ············································································· 14分 |CM ||n| 3 = ． 2 所以直线CM 与平面PAD所成的角为60°． ··································································· 15分 1 解法四：（1）分别取AC,CD的中点为Q,R，连接PQ，QR，PR，则QR∥AD，且QR= AD． 2 因为△ABC是边长为2的等边三角形，所以AC =2，PQ⊥AC，PQ= 3． ························· １分 在△ACD中，∠ADC =45°，CD=2 2 ， 2 由余弦定理，得AC2 =CD2 + AD2 −2CD⋅ADcos∠ADC，22 =（2 2)2 + AD2 −2×2 2AD× ， P 2 解得AD=2， ··········································································································· 2分 所以QR=1. B A Q 数学参考答案及评分细则 第6页（共 19页） C R D因为AC2 + AD2 =CD2，所以AD⊥ AC ， ······································································· 3分 所以QR⊥AC.又PQ⊥AC，所以∠PQR为二面角P−AC−D的平面角． ······························· 5分 CP2 +CD2 −PD2 4+(2 2)2 −(2 2)2 1 因为在△PCD中，由余弦定理，得cos∠PCD= = = ， 2CP⋅CD 2×2×2 2 2 2 所以在△PCR中，由余弦定理，得 1 PR= CP2 +CR2 −2CP⋅CRcos∠PCR = 22+( 2)2 −2×2× 2× =2． 2 2 所以在△PQR中，PQ2 +QR2 =PR2，所以∠PQR=90°，即二面角P−AC−D是直二面角． ···· 6分 所以平面PAC⊥平面ACD. ··························································································· 7分 （2）同解法三．······································································································· 15分 17．本小题以人工智能为背景，主要考查条件概率、概率的基本性质、互斥事件、对立事件、独立事件等 基础知识，考查数学建模能力、运算求解能力、逻辑推理能力等，考查统计与概率思想、分类与整合 思想、化归与转化思想等，考查数学抽象、数学建模、逻辑推理和数学运算等核心素养，体现基础性、 综合性和应用性．满分15分． 解法一：（1）设“甲平台向该用户推送A”为事件M ，“甲平台向该用户推送B”为事件N，则“甲平台 没有向该用户推送A”为事件M ，“甲平台没有向该用户推送B”为事件N， ························ 1分 依题意得，P(M)=0.7，P(N)=0.5，P(MN)=0.3，所以P(M)=1−P(M)=1−0.7=0.3． ······· 2分 因为N =MN∪MN ， 所以P(MN)=P(N)−P(MN)=0.5−0.3=0.2． ································································· 4分 P(MN) 所以P(N|M)= ······························································································· 6分 P(M) 0.2 2 = = ． 0.3 3 2 所以在甲平台没有向该用户推送A的条件下，它向该用户推送B的概率为 ． ······················ 7分 3 （2）设“乙平台向该用户推送C”为事件L，“这两个平台至少向该用户推送A、B、C中的一种” 为事件R，则P(L)=0.6，所以P(L)=1−P(L)=1−0.6=0.4． ············································· 8分 依题意，“甲平台向该用户推送A或B”为事件M∪N，“甲平台既不向该用户推送A，也不向该用户 推送B”为事件MN . 因为P(M)=0.7，P(N)=0.5，P(MN)=0.3，由概率的基本性质可得： P(M∪N)=P(M)+P(N)−P(MN) ················································································· 9分 数学参考答案及评分细则 第7页（共 19页）=0.7+0.5−0.3=0.9． ················································································· 10分 所以P(MN)=1−P(M∪N) ·························································································· 11分 =1−0.9=0.1．·························································································· 12分 依题意，R与MNL互为对立事件，且MN 与 L 相互独立， ··············································· 13分 所以P(R)=1−P(MNL)=1−P(MN)P(L) ········································································ 14分 =1−0.1×0.4=0.96． 所以这两个平台至少向该用户推送A、B、C中的一种的概率为0.96． ································· 15分 解法二：（1）同解法一． ··································································································· 7分 （2）设“乙平台向该用户推送C”为事件L，“这两个平台至少向该用户推送A、B、C中的一种” 为事件R． 因为P(L)=0.6，所以P(L)=1−P(L)=1−0.6=0.4． ························································· 8分 依题意，“甲平台向该用户推送A或B”为事件M∪N． 因为P(M)=0.7，P(N)=0.5，P(MN)=0.3，由概率的基本性质可得： P(M∪N)=P(M)+P(N)−P(MN) ················································································· 9分 =0.7+0.5−0.3=0.9． ·················································································· 10分 依题意，L与(M∪N)L 互斥，M∪N与L相互独立，······················································· 11分 所以P(R)=P(L∪(M∪N)L) ························································································· 12分 =P(L)+P((M∪N)L) ···················································································· 13分 =P(L)+P(M∪N)P(L) ·················································································· 14分 =0.6+0.9×0.4=0.96． 所以这两个平台至少向该用户推送A、B、C中的一种的概率为0.96． ································· 15分 解法三：（1）同解法一． ··································································································· 7分 （2）设“乙平台向该用户推送C”为事件L，“这两个平台至少向该用户推送A、B、C中的一种” 为事件R． 因为P(L)=0.6，所以P(L)=1−P(L)=1−0.6=0.4． ························································· 8分 数学参考答案及评分细则 第8页（共 19页）依题意，R=MNL∪MNL∪MNL∪MNL∪MNL∪MNL∪MNL， ············································ 10分 且MNL、MNL、MNL、MNL、MNL、MNL、MNL互斥，MN 与L、MN 与L、MN 与L、MN 与L、MN 与L、MN 与L、MN 与L均相互独立． ························································ 12分 又因为P(M)=0.7，P(N)=0.5，P(MN)=0.3， 所以P(R)=P(MNL∪MNL∪MNL∪MNL∪MNL∪MNL∪MNL) =P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL)+P(MNL) ····················· 13分 =P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L)+P(MN)P(L) ······················································································································· 14分 =(1−0.7−0.5+0.3)×0.6+(0.5−0.3)×(1−0.6)+(0.7−0.3)×(1−0.6)+(0.7−0.3)×0.6 +(0.5−0.3)×0.6+0.3×(1−0.6)+0.3×0.6 =0.96． 所以这两个平台至少向该用户推送A、B、C中的一种的概率为0.96． ································· 15分 18．本小题主要考查导数及其应用、函数的基本性质、三角函数的图象与性质等基础知识，考查运算求解 能力、逻辑推理能力、直观想象能力等，考查化归与转化思想、函数与方程思想、数形结合思想，考 查数学抽象、逻辑推理、直观想象、数学运算等核心素养，体现基础性和综合性．满分17分． π π π 解法一：（1）由 f(x)=( −x)sinx+ax−cosx得， f′(x)=−sinx+( −x)cosx+a+sinx=( −x)cosx+a， 2 2 2 ························································································································ 2分 π π π 所以 f′(0)=a+ ．依题意，得a+ = ，所以a=0， ····················································· 4分 2 2 2 π 此时 f′(x)=( −x)cosx， 2 π π 3π π 所以当x∈(0, )时， f′(x)>0；当x∈( , )时， f′(x)>0，且 f′( )=0． 2 2 2 2 3π π 所以当x∈(0, )时， f′(x)0，当且仅当x= 时， f′(x)=0； ·········································· 6分 2 2 3π 当x∈( ,2π)时， f′(x)<0． ······················································································ 7分 2 3π 3π 故 f(x)在(0, )单调递增，在( ,2π)单调递减． ···························································· 8分 2 2 π （2）由 f(x)=( −x)sinx+ax−cosx得， 2 数学参考答案及评分细则 第9页（共 19页）π π f(π−x)=(x− )sin(π−x)+a(π−x)−cos(π−x)=(x− )sinx+a(π−x)+cosx， 2 2 所以 f(π−x)+ f(x)=πa， ··························································································· 9分 π πa 所以曲线y= f(x)关于点P( , )对称. ········································································ 10分 2 2 因为 f(0)=−1， f(−π)=−πa+1，取A(0,−1)，B(−π,−πa+1)， ········································· 12分 πa πa +1 +πa−1 则直线AP的斜率k = 2 =a+ 2 ，直线BP的斜率k = 2 =a− 2 ， AP π π BP 3π 3π 2 2 则∀a∈R，k ≠k ，所以A,B,P不共线. ····································································· 14分 AP BP 设A,B关于点P的对称点分别为C,D，则|PA|=|PC|，|PB|=|PD|， 所以四边形ABCD为平行四边形. ················································································· 16分 π πa 因为曲线y= f(x)关于点P( , )对称，所以C,D也在曲线y= f(x)上， 2 2 所以曲线y= f(x)上存在四个点A,B,C,D，使得四边形ABCD为平行四边形. ······················· 17分 解法二：（1）同解法一.····································································································· 8分 （2）曲线y = f(x)上存在四个点，使得以这四点为顶点的四边形是平行四边形． ·················· 9分 证明如下： π π 3π 3π 取A(− ,−π− a)，B(0,−1)，C( ,π+ a)，D(π,πa+1)， ··········································· 12分 2 2 2 2 π π 3π 3π 因为 f(− )=−π− a， f(0)=−1， f( )=π+ a， f(π)=πa+1， 2 2 2 2 所以A,B,C,D四点都在曲线y= f(x)上. ········································································ 13分  π π  π π 因为AB=( ,π+ a−1)，DC =( ,π+ a−1)， 2 2 2 2   所以AB=DC， ········································································································ 15分   因为AC =(2π,2π+2πa)，BD=(π,πa+2)，2π(πa+2)−π(2π+2πa)=4π−2π2 ≠0，   所以AC与 BD 不共线， 所以四边形ABCD为平行四边形． 所以曲线y= f(x)上存在四个点A,B,C,D，使得四边形ABCD为平行四边形． ······················ 17分 解法三：（1）同解法一.····································································································· 8分 （2）曲线y = f(x)上存在四个点，使得以这四点为顶点的四边形是平行四边形． ·················· 9分 数学参考答案及评分细则 第10页（共 19页）证明如下： π π 3π 3π 取A(− ,−π− a)，B(0,−1)，C( ,π+ a)，D(π,πa+1)， ··········································· 12分 2 2 2 2 π π 3π 3π 因为 f(− )=−π− a， f(0)=−1， f( )=π+ a， f(π)=πa+1， 2 2 2 2 所以A,B,C,D四点都在曲线y= f(x)上. ········································································ 13分 2πa+2π πa+2 2 因为k = =a+1，k = =a+ ， AC 2π BD π π 所以∀a∈R，k ≠k ，所以A,B,C,D四点不共线． ······················································ 15分 AC BD π πa 因为线段AC与线段BD的中点都是P( , )，所以|AP|=|PC|，|BP|=|PD|， 2 2 所以四边形ABCD为平行四边形． 所以曲线y= f(x)上存在四个点A,B,C,D，使得四边形ABCD为平行四边形． ······················ 17分 19．本小题主要考查双曲线的标准方程与简单几何性质、直线与圆锥曲线的位置关系、点与圆的位置关系 等基础知识，考查逻辑推理能力、直观想象能力、运算求解能力和创新能力等，考查函数与方程思想、 化归与转化思想、分类与整合思想、数形结合思想和特殊与一般思想等，考查逻辑推理、直观想象、 数学运算等核心素养，体现基础性、综合性与创新性．满分17分． 解法一：（1）依题意，可设P(x,y)(y≠0)，则Q(x,0)．························································· 1分 因为A(−1,0),B(1,0)，Q在线段AB外，所以|AQ|⋅|BQ|=|x+1||x−1|=x2 −1． ····················· 2分 y2 y 又因为|PQ|2=3|AQ|⋅|BQ|，所以y2 =3(x2 −1)，即x2 − =1． 3 P y2 故C 的方程为x2 − =1(y≠0)． ················································································ 3分 3 A O B Q x （2）选择②作为条件． ······························································································ 4分 （i）设M(x,y ),N(x ,y )，则T(x,-y )． 1 1 2 2 1 1 显然l的斜率不为零，否则，有x =-x,y = y ，l:y= y ， 2 1 2 1 1 −y y y2 3x2 −3 此时k k = 1 ⋅ 2 = 1 = 1 =3，与直线TB和NB的斜率之积为6，矛盾． TB NB x −1 x −1 x2 −1 x2 −1 1 2 1 1 ························································································································ 5分 故可设l：x=my+t， x=my+t,  由 y2 得 ( 3m2 −1 ) y2 +6mty+3t2 −3=0， ·························································· 6分 x2 − =1  3 数学参考答案及评分细则 第11页（共 19页）依题意，3m2 −1≠0且∆=36m2t2 −4 ( 3m2 −1 )( 3t2 −3 ) =12 ( 3m2 +t2 −1 ) >0， 3 6mt 3t2 −3 所以m≠± 且3m2 −1+t2 >0，y + y =− ,y y = ． ································ 7分 3 1 2 3m2 −1 1 2 3m2 −1 x=my+t, 由   y2 得 ( 3m2 −1 ) x2 +2tx−3m2 −t2 =0，所以x +x =− 2t ,x x =− 3m2 +t2 ， x2 − =1 1 2 3m2 −1 1 2 3m2 −1  3 y y 因为直线TB和NB的斜率之积为6，所以− 1 ⋅ 2 =6， ············································ 8分 x −1 x −1 1 2 3 ( t2 −1 ) y y 3m2 −1 t+1 即 1 2 =−6， =−6， =2，解得t =3． x x −(x +x )+1 3m2 +t2 2t t−1 1 2 1 2 − + +1 3m2 −1 3m2 −1 此时∆=12(3m2 +8)>0恒成立，所以l：x=my+3，过定点(3,0)． ·································· 10分 y M 6 3m2 +9 24 （ii）由（i）知，x +x =− ，x x =− ，y y = ． 1 2 3m2 −1 1 2 3m2 −1 1 2 3m2 −1 3 3 ①当3m2 −1<0，即− 0，所以M,N 均在C 的右支，如图． ·············· 11分 1 2 3 3   此时BM×BN =(x -1)(x -1)+y y =xx -(x +x )+1+y y 1 2 1 2 1 2 1 2 1 2 A O B x 3m2+9 6 24 20 =- + +1+ = <0， 3m2-1 3m2-1 3m2-1 3m2-1 所以ÐMBN是钝角，△BMN 是钝角三角形．·································································· 13N分 3 3 3m2 +9 ②当3m2 −1>0，即m> 或m<− 时，x x =− <0， 3 3 1 2 3m2 −1 所以M,N 分别在C 的两支．不妨设M 在C 的右支，则x >1，如图． ································ 14分 1   设R(3,0)，则MB×MR=(1-x )(3-x )+y2 =x2-4x +3+3x2-3=4x (x -1)>0， ··········· 16分 1 1 1 1 1 1 1 1 y π 所以ÐBMRÎ(0, )． 2 π A O B R x 因为l过点R，所以ÐBMN =π-ÐBMRÎ( ,π)， M 2 N 所以ÐBMN是钝角，△BMN 是钝角三角形． 综上可知，△BMN 不可能是锐角三角形． ····································································· 17分 数学参考答案及评分细则 第12页（共 19页）y M 解法二：（1）同解法一． ··································································································· 3分 （2）选择②作为条件． ······························································································· 4分 （i）同解法一． ······································································································· 10分 （ii）由（i）知，直线l过定点（3,0）． A O B x ①当M,N 均在C的右支，不妨设M 在x轴的上方，如图． 设直线MB,NB的倾斜角分别为,．则tan0，tan0，MBN=π． N 因为直线TB和NB的斜率之积为6，所以直线MB和NB的斜率之积为-6，即tantan6， tantan tantan 所以tanMBN tan(π)tan()  0， 1tantan 5 所以MBN 是钝角，△MBN 是钝角三角形． ·································································· 13分 ②当M,N 分别在C 的两支时，不妨设M 在C 的右支，则x >1，如图． ····························· 14分 1   设R(3,0)，则MB×MR=(1-x )(3-x )+y2 =x2-4x +3+3x2-3=4x (x -1)>0， ··········· 16分 1 1 1 1 1 1 1 1 y π 所以ÐBMRÎ(0, )． 2 π 因为l过点R，所以ÐBMN =π-ÐBMRÎ( ,π)， A O B R x 2 M N 所以ÐBMN是钝角，△BMN 是钝角三角形． 综上可知，△BMN 不可能是锐角三角形． ····································································· 17分 解法三：（1）同解法一． ··································································································· 3分 （2）选择②作为条件． ······························································································· 4分 y （i）设M(x,y ),N(x ,y ),T(x ,y )，则x =x,y =-y ． 1 1 2 2 T T T 1 T 1 当l的斜率不存在时，x =x,y =-y ，l:x=x ． 2 1 2 1 1 T −y y y2 3x2 −3 3(x +1) A O B R x 所以k k = 1 ⋅ 2 = 1 = 1 = 1 ， M TB NB x −1 x −1 (x −1)2 (x −1)2 x −1 1 2 1 1 1 N 3(x +1) 依题意 1 =6，解得x =3．此时，l:x=3，过点R(3,0)． ······································· 5分 x −1 1 1 下面我们证明，当l的斜率存在时，M,N,R共线． 显然直线TB斜率存在且不为零，可设直线TB：x=m y+1(m ¹0)， 1 1 x=m y+1, 由   y 1 2 得 ( 3m2 −1 ) y2 +6m y=0，解得y =- 6m 1 ，x =- 3m 1 2+1 ，   x2 − 3 =1 1 1 T 3m 1 2-1 T 3m 1 2-1 数学参考答案及评分细则 第13页（共 19页）6m 3m2+1 所以y = 1 ，x =- 1 ． 1 3m2-1 1 3m2-1 1 1 6m 1 y 3m2-1 3m 所以k = 1 = 1 =- 1 ． ································································ 7分 MR x -3 3m2 +1 6m2-1 1 - 1 -3 1 3m2-1 1 6m 3m2+1 同理可设直线NB：x=m y+1(m ¹0)，得y =- 2 ，x =- 2 ， 2 2 2 3m2-1 2 3m2-1 2 2 6m - 2 y 3m2-1 3m 所以k = 2 = 2 = 2 ． ··································································· 9分 NR x -3 3m2 +1 6m2-1 2 - 2 -3 2 3m2-1 2 1 1 1 因为直线TB和NB的斜率之积为6，所以 ⋅ =6，即m = ， m m 2 6m 1 2 1 3m 3m 所以k = 2 =- 1 =k ，所以M,N,R共线，即l过点R(3,0)． NR 6m2-1 6m2-1 MR 2 1 综上，l过定点(3,0)． ······························································································· 10分 x=my+3,  （ii）依题意，可设l：x=my+3，由 y2 得 ( 3m2 −1 ) y2 +18my+24=0， x2 − =1  3 依题意，3m2 −1≠0且∆=(18m)2 −4×24 ( 3m2 −1 ) =12 ( 3m2 +8 ) >0，即m≠± 3 ， 3 18m 24 此时y + y =− ,y y = ． ······································································· 11分 1 2 3m2 −1 1 2 3m2 −1 3 3 ①当3m2 −1<0，即− 0，即m> 或m<− 时，y y = >0， 3 3 1 2 3m2 −1 N 数学参考答案及评分细则 第14页（共 19页）所以M,N 分别在C 的两支．不妨设M 在C 的右支，则x >1，如图． ································ 14分 1   所以MB×MR=(1-x )(3-x )+y2 =x2-4x +3+3x2-3=4x (x -1)>0， ······················· 16分 1 1 1 1 1 1 1 1 y π 所以ÐBMRÎ(0, )． 2 T π 因为l过点R，所以ÐBMN =π-ÐBMRÎ( ,π)， A O B R x 2 M N 所以ÐBMN是钝角，△BMN 是钝角三角形． 综上可知，△BMN 不可能是锐角三角形． ····································································· 17分 解法四：（1）同解法一． ··································································································· 3分 （2）选择③作为条件． ······························································································ 4分 （i）设M(x,y ),N(x ,y )，则T(x,-y )． 1 1 2 2 1 1 显然l的斜率不为零，否则x =-x,y = y ，l:y= y ， 2 1 2 1 1 −y 2y 因为直线TB和NA的斜率之商为2，所以 1 = 2 ， x −1 x +1 1 2 −y 2y 从而有 1 = 1 ，解得y =0，此时，l与C 不存在公共点，与题设矛盾． ···················· 5分 x −1 −x +1 1 1 1 故可设l：x=my+t， x=my+t,  由 y2 得 ( 3m2 −1 ) y2 +6mty+3t2 −3=0， ·························································· 6分 x2 − =1  3 依题意，3m2 −1≠0且∆=36m2t2 −4 ( 3m2 −1 )( 3t2 −3 ) =12 ( 3m2 +t2 −1 ) >0， 3 6mt 3t2 −3 所以m≠± 且3m2 −1+t2 >0，y + y =− ,y y = ． ································ 7分 3 1 2 3m2 −1 1 2 3m2 −1 x=my+t, 由   y2 得 ( 3m2 −1 ) x2 +2tx−3m2 −t2 =0，所以x +x =− 2t ,x x =− 3m2 +t2 ， x2 − =1 1 2 3m2 −1 1 2 3m2 −1  3 −y 2y 因为直线TB和NA的斜率之商为2，所以 1 = 2 ． ·················································· 8分 x −1 x +1 1 2 y y 因为点M 在C上，所以y2 =3(x2 −1)，即 1 ⋅ 1 =3， 1 1 x −1 x +1 1 1 数学参考答案及评分细则 第15页（共 19页）3 ( t2 −1 ) 所以− 2y 2 ⋅ y 1 =3，即 2y 1 y 2 =−3， 2⋅ 3m2 −1 =−3， 2(t−1) =1， x +1 x +1 x x +x +x +1 3m2 +t2 2t t+1 2 1 1 2 1 2 − − +1 3m2 −1 3m2 −1 解得t =3．此时∆=12(3m2 +8)>0恒成立，所以l：x=my+3，过定点(3,0)． ·················· 10分 6 3m2 +9 24 （ii）由（i）知，x +x =− ，x x =− ，y y = ． y 1 2 3m2 −1 1 2 3m2 −1 1 2 3m2 −1 M 3 3 ①当3m2 −1<0，即− 0，所以M,N 均在C 的右支，如图． ·············· 11分 1 2 3 3   此时BM×BN =(x -1)(x -1)+y y =xx -(x +x )+1+y y 1 2 1 2 1 2 1 2 1 2 3m2+9 6 24 20 =- + +1+ = <0， A O B R x 3m2-1 3m2-1 3m2-1 3m2-1 所以ÐMBN是钝角，△BMN 是钝角三角形．·································································· 13分 3 3 3m2 +9 N ②当3m2 −1>0，即m> 或m<− 时，x x =− <0， 3 3 1 2 3m2 −1 所以M,N 分别在C 的两支，不妨设M 在C 的右支，则x >1，如图． ································ 14分 1 设R(3,0)，F(2,0)，则F(2,0)是以BR为直径的圆的圆心， MF 2 =(x -2)2 +y2 =x2-4x +4+y2 =x2-4x +4+3x2-3=4x (x -1)+1>1， ············ 16分 1 1 1 1 1 1 1 1 1 1 π y 所以M 在以BR为直径的圆外，所以ÐBMRÎ(0, )． 2 T π 因为l过点R(3,0)，所以ÐBMN =π-ÐBMRÎ( ,π)， F 2 A O B R x M 所以ÐBMN是钝角，△BMN 是钝角三角形． N 综上可知，△BMN 不可能是锐角三角形． ····································································· 17分 解法五：（1）同解法一． ··································································································· 3分 （2）选择③作为条件． ······························································································· 4分 （i）设M(x,y ),N(x ,y ),T(x ,y )，则x =x,y =-y ． 1 1 2 2 T T T 1 T 1 当l的斜率不存在时，x =x,y =-y ，l:x=x ． 2 1 2 1 1 −y 2y 因为直线TB和NA的斜率之商为2，所以 1 = 2 ， x −1 x +1 1 2 数学参考答案及评分细则 第16页（共 19页）−y −2y 即 1 = 1 ，解得x =3，此时，直线l:x=3，过点R(3,0)． ······································ 5分 x −1 x +1 1 1 1 下面我们证明，当l的斜率存在时，M,N,R共线． 显然直线TB斜率存在且不为零，可设直线TB：x=m y+1(m ¹0)， 1 1 x=m y+1, 由   y 1 2 得 ( 3m2 −1 ) y2 +6m y=0，解得y =- 6m 1 ，x =- 3m 1 2+1 ，   x2 − 3 =1 1 1 T 3m 1 2-1 T 3m 1 2-1 6m 3m2+1 所以y = 1 ，x =- 1 ． 1 3m2-1 1 3m2-1 1 1 6m 1 y 3m2-1 3m 所以k = 1 = 1 =- 1 ． ································································ 7分 MR x -3 3m2 +1 6m2-1 1 - 1 -3 1 3m2-1 1 x=m y−1,  2 同理可设直线NA：x=m y-1(m ¹0)，由 y2 得 ( 3m2 −1 ) y2 −6m y=0， 2 2 x2 − =1 2 2  3 6m 2 6m 3m2+1 y 3m2-1 3m 所以y = 2 ，x =m y -1= 2 ，k = 2 = 2 =- 2 ． ············· 9分 2 3m2-1 2 2 2 3m2-1 NR x -3 3m2 +1 3m2-2 2 2 2 2 -3 2 3m2-1 2 1 2 因为直线TB和NA的斜率之商为2，所以 = ，即m =2m ． m m 2 1 1 2 3m 3m 所以k =- 2 =- 1 =k ，所以M,N,R共线，即l过点R(3,0)． NR 3m2-2 6m2-1 MR 2 1 综上，l过定点(3,0)． ······························································································· 10分 （ii）同解法二． ······································································································ 17分 解法六：（1）同解法一． ··································································································· 3分 （2）选择②作为条件． ······························································································· 4分 （i）设M(x,y ),N(x ,y )，则T(x,-y )． 1 1 2 2 1 1 当l的斜率不存在时，x =x,y =-y ，l:x=x ． 2 1 2 1 1 −y y y2 3x2 −3 3(x +1) 所以k k = 1 ⋅ 2 = 1 = 1 = 1 ． TB NB x −1 x −1 (x −1)2 (x −1)2 x −1 1 2 1 1 1 3(x +1) 又因为直线TB和NB的斜率之积为6，所以 1 =6，解得x =3． x −1 1 1 数学参考答案及评分细则 第17页（共 19页）此时，直线l:x=3，过点(3,0)． ·················································································· 5分 当l的斜率存在时，设l：y=kx+m， y=kx+m,  由 y2 得 ( 3−k2) x2 −2kmx−m2 −3=0， x2 − =1  3 依题意，3−k2 ≠0且∆=4k2m2 −4 ( 3−k2)( −m2 −3 ) =12 ( m2 +3−k2) >0， 2km m2 +3 所以k ≠± 3且m2 +3−k2 >0，x +x = ,x x =− ． ······································ 7分 1 2 3−k2 1 2 3−k2 y y 因为直线TB和NB的斜率之积为6，所以− 1 ⋅ 2 =6， ············································ 8分 x −1 x −1 1 2 ( m2 +3 ) k2 2k2m2 (kx +m)(kx +m) k2x x +km(x +x )+m2 − + +m2 即 1 2 =−6， 1 2 1 2 =−6， 3−k2 3−k2 =−6， x x −(x +x )+1 x x −(x +x )+1 m2 +3 2km 1 2 1 2 1 2 1 2 − − +1 3−k2 3−k2 3(m−k) 所以 =−6，所以m=−3k ，所以l：y=kx-3k，过点(3,0)． −m−k 综上，l过定点(3,0)． ······························································································· 10分 6k2 9k2 +3 （ii）由（i）知，m=−3k ，故m2 +3−k2 =8k2 +3>0恒成立，且x +x = ，x x = ． 1 2 k2 −3 1 2 k2 −3 y M ①当k2 −3>0，即k > 3或k <− 3时，x x >0，所以M,N 均在C 的右支，如图． ············ 11分 1 2   此时BM×BN =(x -1)(x -1)+y y =xx -(x +x )+1+k2(x -3)(x -3) 1 2 1 2 1 2 1 2 1 2 =xx -(x +x )+1+k2xx -3k2(x +x )+9k2 1 2 1 2 1 2 1 2 = 9k2+3 - 6k2 +1+ k2(9k2+3) -3k2× 6k2 +9k2 A O B x k2-3 k2-3 k2-3 k2-3 9k2+3 6k2 k2-3 k2(9k2+3) 18k4 9k2(k2-3) = - + + - + k2-3 k2-3 k2-3 k2-3 k2-3 k2-3 N -20k2 = <0， k2-3 所以ÐMBN是钝角，△BMN 是钝角三角形．·································································· 13分 ②当k2 −3<0，即− 31，如图． ································· 14分 1 数学参考答案及评分细则 第18页（共 19页）  设R(3,0)，则MB×MR=(1-x )(3-x )+y2 =x2-4x +3+3x2-3=4x (x -1)>0， ··········· 16分 1 1 1 1 1 1 1 1 π y 所以ÐBMRÎ(0, )． 2 T π 因为l过定点R(3,0)，所以ÐBMN =π-ÐBMRÎ( ,π)， 2 A O B R x M 所以ÐBMN是钝角，△BMN 是钝角三角形． N 综上可知，△BMN 不可能是锐角三角形． ····································································· 17分 数学参考答案及评分细则 第19页（共 19页）