数学试题卷答案(省质检-三明卷)_2024-2026高三（6-6月题库）_2026年04月高三试卷_260410福建省2026届高中毕业班适应性练习（省质检）

三明市 2026 届高中毕业班适应性练习（四月） 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、单项选择题 题号 1 2 3 4 5 6 7 8 答案 A C D C C B C D 二、多项选择题 题号 9 10 11 答案 AD BCD ACD 三、填空题 12． 3 13．30 14．4 3 四、解答题 15.本小题主要考查函数的奇偶性、函数的零点、三角恒等变换、等差数列求和等基础知识，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，考查逻辑推理、数学运算等核 心素养，体现基础性.满分13分. 解法一：（1）因为 f(x)为奇函数，所以 f(x)f(x)，····························································1分 即sin(2x)sin(x)[sin(2x)sin(x)] 恒成立. 得sin(x)sin(x)0 恒成立，···············································································2分 所以sinxcossincosxsincosxsinxcos0 恒成立，················································3分 所以sincosx0恒成立，·····························································································4分 所以sin0，············································································································5分 解得kπ,kZ.········································································································6分 数学参考答案及评分细则 第1页（共 15页）π π （2）因为 ，所以 f(x)sin2xsin(x ) ， 2 2 π 令 f(x)0，则sin2xsin(x ) ，················································································· 8分 2 π π 所以2xx 2k π,k Z或2xx π+2k π,k Z ，··················································10分 1 1 2 2 2 2 π π 2 解得x 2k π,k 或x  k π,k ，·································································11分 1 1 2 2 2 6 3 3 2 1 令a (2n )，b ( n )，则b a b ,kN*， n 2 n 3 2 3k2 k 3k1 所以S  x x x (a a a a a )(b b b )，··································12分 20 1 2 20 1 2 3 4 5 1 2 15 5(a a ) 15(b b ) 45π 435π 所以S  1 5  1 15    95π．························································13分 20 2 2 2 6 解法二：（1）因为 f(x)为R上的奇函数，所以 f(0)0，··························································2分 所以sin0，············································································································3分 解得kπ,kZ， ·····································································································4分 经检验， f(x)sin2xsin(xkπ),k是奇函数， 所以kπ,kZ.········································································································6分 π （2）因为 ，所以 f(x)sin2xcosx，·····································································7分 2 令 f(x)0，则sin2xcosx0，····················································································9分 所以cosx(2sinx1)0，·····························································································10分 1 所以cosx0或sinx ， 2 π π 5π 解得x k π,k 或x 2k π,k 或x 2k π,k ，······································11分 1 1 2 2 3 3 2 6 6 1 11 7 令a (n )π，b (2n )π，c (2n )π， n n n 2 6 6 数学参考答案及评分细则 第2页（共 15页）则(2k2)πb a c a 2kπ,kN*， k 2k1 k 2k 所以S  x x x (a a a )(b b b )(c c c )， 20 1 2 20 1 2 10 1 2 5 1 2 5 10(a a ) 5(b b ) 5(c c ) 所以S  1 10  1 5  1 5  95π．···························································13分 20 2 2 2 解法三：（1）同解法一．····································································································· 6分 π π （2）因为 ，所以 f(x)sin2xsin(x ) ，因为 f(x2π) f(x)， 2 2 所以2π是 f(x)的一个周期，··························································································7分 π 当0x2π时，令 f(x)0，则sin2xsin(x )，····························································9分 2 π π 5π 3π 解得x , , , ，·································································································10分 6 2 6 2 π π 5π 3π 所以 f(x)在区间(0,2π]的零点之和为    3π．················································11分 6 2 6 2 令a  x x x x ， n 4n3 4n2 4n1 4n 则{a }是以3π为首项，8π为公差的等差数列，································································12分 n 所以S  x x x a a a a a 20 1 2 20 1 2 3 4 5 5(a a ) 5(3π35π)  1 5   95π．······································································13分 2 2 16.本小题主要考查导数的几何意义、导数的应用等基础知识，考查逻辑推理能力、运算求解能力等，考 查函数与方程思想、化归与转化思想、分类与整合思想等，考查逻辑推理、数学运算等核心素养，体 现基础性．满分15分． a 解法一：（1）函数 f(x)的定义域为(0,)， f(x)x ．···················································2分 x 2 当a2时，因为 f(x)x ，所以 f(1)1，···························································3分 x 1 又 f(1) ，·······································································································4分 2 1 所以曲线y f(x)在点(1, f(1))处的切线方程为y (x1)， 2 数学参考答案及评分细则 第3页（共 15页）即2x2y30．································································································7分 1 1 1 1 1 2 （2）(i)当a0时， f(ea) (ea)2alnea  ea 10不符合题意，舍去；························9分 2 2 1 (ii)当a0时， f(x) x2 0显然成立；··································································11分 2 (iii)当a0时，令 f(x)0，得0 x a ，令 f(x)0，得x a； 所以 f(x)在(0, a)单调递减，在( a,)单调递增．···················································13分 1 所以 f(x)  f( a) aaln a 0，解得0ae．················································14分 min 2 综上所述，a的取值范围为[0,e)．················································································15分 解法二：（1）同解法一．······························································································7分 1 （2）由已知，得 x2 alnx0． 2 x2 (i)当0x1时，可得a ．·············································································8分 2lnx x2 因为0x1，所以 0，·················································································9分 2lnx x2 又因为x0时， 0， 2lnx 所以a0；······································································································· 10分 1 (ii)当x1时， x2 alnx0恒成立，所以aR；······················································11分 2 x2 (iii)当x1时，可得a ． 2lnx 1 2xlnxx2 令g(x) x2 (x1)，g(x) x  x(2lnx1) ，············································12分 2lnx 2(lnx)2 2(lnx)2 1 当1 xe2时，g(x)0，g(x)单调递减； 1 当xe2时，g(x)0，g(x)单调递增；···································································13分 1 所以g(x) g(e2)e，所以ae．······································································· 14分 min 综上所述，a的取值范围为[0,e)．·········································································· 15分 17.本小题主要考查椭圆的定义、直线与椭圆的位置关系、三点共线等基础知识，考查逻辑推理能力、直 观想象能力、运算求解能力等，考查函数与方程思想、数形结合思想、分类与整合思想、转化与化归 数学参考答案及评分细则 第4页（共 15页）思想等，考查逻辑推理、直观想象、数学运算等核心素养，体现基础性与综合性．满分15分． 3 解法一：（1）当MF  x轴时，|MF | ， 2 2 2 9 5 所以|MF | |MF |2 |FF |2  4  ，································································1分 1 2 1 2 4 2 5 3 所以2a|MF ||MF |  4，···········································································3分 1 2 2 2 从而a2，b2 3，·····························································································5分 x2 y2 故C的方程为  1．·····················································································6分 4 3 （2）设M(x ,y )(y 0)，P(4,y )，Q(4,y )，·························································7分 0 0 0 P Q x 2 y 2 则 0  0 1，即3x 2 4y 2 120．····································································· 8分 4 3 0 0 又F(1,0),F (1,0)， 1 2     所以FM (x 1,y )，FP(3,y )，FM (x 1,y )，FQ(3,y ). ······························10分 1 0 0 1 P 2 0 0 2 Q 因为PF MF ，QF MF ， 1 1 2 2     所以FM FP3(x 1) y y 0，FM FQ3(x 1) y y 0，··································12分 1 1 0 0 P 2 2 0 0 Q 6 6x 两式相加、减，得y  y  ，y y  0 ，·····························································13分 P Q y P Q y 0 0   又因为PM (x 4,y  y )，QM (x 4,y y ) ， 0 0 P 0 0 Q (x 4)(y y )(x 4)(y y )x  y y  8y 4  y  y   6x 0 2 8y 0 2 24 0 ，········· 14分 0 0 Q 0 0 P 0 P Q 0 P Q y 0   所以PM∥QM ，故P,M,Q三点共线. ···········································································15分 3 解法二：（1）当MF  x轴时，|MF | ， 2 2 2 3 3 所以M(1, )或M(1, )，······················································································ 1分 2 2 数学参考答案及评分细则 第5页（共 15页）9 所以 1  4 1①，······························································································2分 a2 b2 又a2 b2 1②，··································································································4分 由①②，解得a2 4，b2 3，················································································ 5分 x2 y2 故C的方程为  1．·····················································································6分 4 3 x 2 y 2 3 （2）设M(x ,y )(y 0)，则 0  0 1，即y 2 3 x 2．··········································7分 0 0 0 4 3 0 4 0 y y (i)当直线MF，MF 斜率均存在时，k  0 ，k  0 ， 1 2 MF1 x 1 MF2 x 1 0 0 y y 所以直线PF :x 0 y1，QF :x 0 y1，···················································9分 1 x 1 2 x 1 0 0  y x 0 y1, 3(x 1) 由 x 1 得P(4, 0 )，·····································································10分  0 y x4 0  y x 0 y1, 3(x 1) 由 x 1 得Q(4, 0 )， ···································································11分  0 y x4 0  3(x 1)  3(x 1) 所以PM (x 4,y  0 )，QM (x 4,y  0 )， 0 0 y 0 0 y 0 0 3 6x 2 8(3 x 2)24 因为(x 4)[y  3(x 0 1) ](x 4)[y  3(x 0 1) ] 6x 0 28y 0 224  0 4 0 0 ， 0 0 y 0 0 y y y 0 0 0 0   所以PM∥QM ，故P,M,Q三点共线. ····································································· 12分 (ii)当直线MF或MF 斜率不存在时，根据对称性，不妨设MF 斜率不存在，且y 0， 1 2 2 0 3 3 3 1 1 此时点M(1, )，Q(4,0)，k  ，故直线PF :x y1，从而P(4,4)，则k  ，k  ， 2 MF1 4 1 4 MQ 2 MP 2 所以P,M,Q三点共线.························································································· 14分 综上，P,M,Q三点共线. ·····················································································15分 18.本小题主要考查随机变量的分布列、数学期望、条件概率与全概率公式等基础知识，考查数学建模能 力、运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学运算、逻辑推理、数据分 析、数学建模等核心素养等.体现基础性，应用性.满分17分. 数学参考答案及评分细则 第6页（共 15页）解：（1）由题可知，k(1)2 kk k(1)1，·························································· 1分 1 化简可得k  ， ···················································································2分 2 23 1 4 当 时，k  ， 2 9 16 则E(X)k2k 3k(1)k(52) ， 9 16 即顾客一次性购买文创盲盒数量的平均值为 .····························································4分 9 （2）（i）设事件A “一次性购买i 个文创盲盒”（i0,1,2,3），事件B“顾客为幸运客户”， i ·······················································································································5分 则P(A )（k 1）2，P(A)k，P(A )k，P(A )k(1). 0 1 2 3 1 依题意，得P(B|A )0，P(B|A) ，···································································· 6分 0 1 3 因为每个盲盒是否为封面款相互独立， 1 1 1 2 1 7 所以P(B|A )( )2  ，P(B|A )( )3 C1 ( )2  ，··········································8分 2 3 9 3 3 3 3 3 27 又由题意知，B A BABA BA B，且A B,AB,A B,A B两两互斥，···························9分 0 1 2 3 0 1 2 3 3 3 1 1 7 2k(5) 所以PBPAB[PAPB|A]0 k k k(1) ，·············11分 i i i 3 9 27 27 i0 i0 1 2(5) 由（1）得，k  ，代入化简可得P(B) ， 2 23 27(2 23) 2(5) 所以 f() ，(0，1).·····································································12分 27(2 23) （ii）设事件C “一次性购买的文创盲盒全部是封面款”， 1 依题意，得P(C|A)( )i,i 1,2,3，········································································13分 i 3 且C  ACACAC，AC,AC,AC两两互斥， 1 2 3 1 2 3 3 3 4k(12) 所以PCPAC[PAPC|A] ，·············································14分 i i i 27 i1 i1 2k(5) 由（i）得，PB ， 27 所以幸运客户中，一次性购买的文创盲盒全部是封面款的概率为 数学参考答案及评分细则 第7页（共 15页）P(BC) P(C) 2(12) P(C|B)   ，·······································································16分 P(B) P(B) 5 1 2(12) 1 1 由题意P(C|B) ，可得  ，解得 ， 2 5 2 7 1 又因为01，所以(0, ].············································································ 17分 7 19.本小题主要考查空间点、线、面位置关系，直线与平面所成角，二面角，平面轨迹方程等基础知识；考查运算求解能力，直观想象能力，逻辑推理能力等；考查函数与方程思想，数形结合思想，化归与 转化思想等；考查数学运算，逻辑推理，直观想象等核心素养．体现综合性和创新性．满分17分． 解法一：（1） 因为PA平面，AB,AD,所以PA AB，PA AD．···································1分 不妨设ABa,ADb,APc，且a≤b≤c， 因为AB AD ，所以 BD2 a2 b2， PB2 a2 c2， PD2 b2 c2， 所以BD≤PB≤PD，所以PBD为 PBD的最大内角．·················································· 2分 PB2 △BD2 PD2 a2 由余弦定理，得cosPBD  0，···········································3分 2PBBD PBBD π 所以PBD(0, )，所以 PBD是锐角三角形．·························································· 4分 2 （2）（i）因为PA ，Q △ 在CP上，且CQBCQD， 由对称性知B,D在同一个轨迹上，且轨迹关于AC对称，   故以A为原点，AC,AP分别为x轴和z轴的正方向建立如图所示的空间直角坐标系Axyz． 设Bx ,y ,0，Dx ,y ,0，因为AP AC3，所以P0,0,3，C3,0,0． 1 1 2 2 因为Q是线段CP上靠近C的三等分点，  2 1 故AQ AC AP2,0,1，即Q2,0,1， ······························································· 5分 3 3   故QC 1,0,1，QBx 2,y ,1， 1 1   QCQB x 1 cosCQB    1 ， QC QB 2 x 22  y2 1 1 1 数学参考答案及评分细则 第8页（共 15页）x 1 1 1  依题意得 2 x 22  y2 1 2 ，化简得x 1 2  y 1 2 3，··················································6分 1 1 且x 10，即x 1，故x≥ 3，又点B不在直线AC上，故x  3 ， 1 1 1 1 同理，x2  y2 3，且x  3，················································································7分 2 2 2 故在坐标平面xAy中，B,D是双曲线x2  y2 3右支上的动点，且B,D在x轴的两侧，如图． π 因为x2  y2 3的两条渐近线分别为y x和yx，它们的夹角为 ， 2 π 所以0BAD ．······························································································8分 2 因为平面PAB平面PADPA，PA AB，PA AD， 所以BAD是二面角B APD的平面角，所以二面角B APD为锐角．··························9分 （ii）因为△PBD不是任何一个长方体的截面，所以△PBD是直角三角形或钝角三角形．······ 10分 证明如下： 若 PBD为锐角三角形，有 PB2 PD2 BD2 0 ， PB2 BD2 PD2 0 ， BD2 PD2 PB2 0 ， △ PB2 BD2 PD2 PD2 BD2 PB2 PB2 PD2 BD2 可令a ，b ，c ， 2 2 2 则存在以ABa,ADb,APc为共点棱的长方体， PBD为该长方体的截面． 由（1）知，若 PBD是长方体的截面，则 PBD是锐角 △ 三角形， 所以 PBD不是 △ 任何一个长方体的截面等价 △ 于 PBD是直角三角形或钝角三角形．···············11分 由（△ i）知，0BAD π ，所以  A  B    A  D  0 ，△又因为PA AB，PA AD， 2 所以  P  B    P  D    P  A    A  B    P  A    A  D    P  A  2   A  B    A  D  0，故0BPD π ．···························12分 2 因为PA，所以PBA,PDA分别是直线PB,PD与所成的角， 即PBA=,PDA=， PA2 9 不妨设AB≤AD，则≥，且PB≤PD，所以，tan2  ，····················13分 AB2 x2 y2 1 1 数学参考答案及评分细则 第9页（共 15页）π 且PBD≥ PDB． 2 作QM BD于M，因为平面QBD ，平面QBDBD，QM 平面QBD， 所以QM ，又PA，所以PA∥QM ． 因为Q是线段CP上靠近C的三等分点，所以M 是线段AC上靠近C的三等分点， 所以M2,0,0，即直线BD过M2,0,0，···································································14分 π   所以PBDPBM≥ ，所以BM BP(2x ,y ,0)(x ,y ,3)x2 2x  y2≤0，··············15分 2 1 1 1 1 1 1 1 这样，问题等价于在平面直角坐标系xAy中，B(x ,y ),D(x ,y ) 在双曲线x2  y2 3的右支上，直线BD 1 1 2 2 9 过点M(2,0)，AB AD，x2 2x  y2≤0，求tan2 的最小值． 1 1 1 x2  y2 1 1 y 如图，不妨设点B在第四象限，则y 0，x 2．因为B,D都在双曲线的右支，故k k  1 1， 1 1 BD BM 2x 1 即y 2x 0，所以y2 （2x）2，又x2 2x  y2≤0，x2  y2 3， 1 1 1 1 1 1 1 1 1  7 故    x 1 23（2x 1 ）2, 解得    x 1  4 ， 即 7 x≤ 1 7 ，·····································16分 x 1 22x 1 x 1 23≤0, 1 7 ≤x≤ 1+ 7 , 4 1 2  2 1 2 9 9 9 3 7 3 所以 tan2 ≥ ≥  ， x2  y2 2x 1 7 2 1 1 1 1 7 π 当x  ，即PBD 时，等号成立． 1 2 2 故 tan2的最小值为3 73．·················································································17分 2 解法二：（1）因为PA平面，AB,AD，所以PA AB，PA AD．··································1分    又因为AB AD ，故可以A为原点，AB,AD,AP分别为x轴，y轴和z轴的正方向，建立如图所示的 空间直角坐标系Axyz．························································································2分 数学参考答案及评分细则 第10页（共 15页）设ABa,ADb,APc，所以Ba,0,0,D0,b,0,P0,0,c，在△PBD中，   BDBP(a,b,0)(a,0,c)a2 0，所以PBD为锐角，   DBDP(a,b,0)(0,b,c)b2 0，所以PDB为锐角，   PBPD(a,0,c)(0,b,c)c2 0，所以BPD为锐角， 所以△PBD是锐角三角形．·····················································································4分 （2）（i）同解法一．·····························································································9分 （ii）因为△PBD不是任何一个长方体的截面，所以△PBD是直角三角形或钝角三角形．······ 10分 证明如下： 若 PBD为锐角三角形，有 PB2 PD2 BD2 0 ， PB2 BD2 PD2 0 ， BD2 PD2 PB2 0 ， △ PB2 BD2 PD2 PD2 BD2 PB2 PB2 PD2 BD2 可令a ，b ，c ， 2 2 2 则存在以ABa,ADb,APc为共点棱的长方体， PBD为该长方体的截面． 由（1）知，若 PBD是长方体的截面，则 PBD是锐角 △ 三角形， 所以 PBD不是 △ 任何一个长方体的截面等价 △ 于 PBD是直角三角形或钝角三角形．···············11分 △ △ 作QM BD于M，因为平面QBD ，平面QBDBD，QM 平面QBD， 所以QM ，又PA，所以PA∥QM ． 因为Q是线段CP上靠近C的三等分点，所以M 是线段AC上靠近C的三等分点， 数学参考答案及评分细则 第11页（共 15页）所以M2,0,0，即直线BD过M2,0,0．···································································12分 在平面直角坐标系xAy中，设直线BD的方程为xty2， 联立   x2  y2 3, 得  t2 1  y2 4ty10， xty2  4t   t2 10,   y 1  y 2  t2 1 , 依题意，有   4t24  t21  0, 且  y y  1 .  1 2 t2 1 因为y 1 y 2 0，所以 t2 1 ．    因为PBx ,y ,3,PDx ,y ,3,BDx x ,y  y ,0， 1 1 2 2 2 1 2 1   所以PBPDx x  y y 9 ty 2  ty 2  y y 9 1 2 1 2 1 2 1 2 6  t2 2    t2 1  y y 2ty  y 13 0 ，····························································13分 1 2 1 2 t2 1   BPBD x x x  y y  y  x2  y2 x x  y y ， 1 1 2 1 1 2 1 1 1 2 1 2   同理DPDBx2  y2  x x  y y ， 2 2 1 2 1 2   不妨设x2  y2≤x2  y2，则必有BPBD x2  y2 x x  y y ≤0． 1 1 2 2 1 1 1 2 1 2 3t2 3 6 因为x2  y2 (x x  y y )x2  y2 [(t2 1)y y 2t(y  y )4]x2  y2  2x2  ， 1 1 1 2 1 2 1 1 1 2 1 2 1 1 t2 1 1 t2 1 x 2 因为x ty 2且y 0，所以t  1 ，代入上式得到 1 1 1 y 1 6 6 x2  y2 x x  y y 2x2  2x2  1 1 1 2 1 2 1 t2 1 1 x 2 2  1  1  y 1    2x2  6y 1 2 2x2  6 x 1 2 3 ··························································· 14分 1 x 22 y2 1 x 22   x23  1 1 1 1  1 7 1 7 2  4x 1 3 10x 1 2 9  22x 1 3    x 1  2       x 1  2    ，   4x 7 4x 7 1 1  1 7 1 7 22x 3 x  x   所以 1   1 2     1 2   ， ≤0 4x 7 1 7 1 7 又因为x 1  3，所以x 1   4 , 2    ．·····································································15分 数学参考答案及评分细则 第12页（共 15页）因为PA，所以PBA,PDA分别是直线PB,PD与所成的角，即PBA,PDA， 因为x2  y2≤x2  y2，所以AB≤AD，所以≥，所以，······································16分 1 1 2 2 PA2 9 9 3 7 3 tan2tan2   ≥ ， AB2 x2  y2 2x2 3 2 1 1 1 1 7 π 当x = ，即PBD 时，等号成立． 1 2 2 3 73 故 tan2的最小值为 ．·················································································17分 2 解法三：（1）因为PA平面，AB,AD，所以PA AB，PA AD．··································1分 又因为AB AD ，所以在△PBD中，  B  P    B  D    B  A    A  P    B  A    A  D    B  A  2 0，所以PBD为锐角，···········································2分  D  B    D  P    D  A    A  B    D  A    A  P    D  A  2 0，所以PDB为锐角，··········································3分  P  B    P  D    P  A    A  B    P  A    A  D    P  A  2 0，所以BPD为锐角， 所以△PBD是锐角三角形．·····················································································4分 （2）（i）同解法一．·····························································································9分 （ii）因为△PBD不是任何一个长方体的截面，所以△PBD是直角三角形或钝角三角形．······ 10分 证明如下： 若 PBD为锐角三角形，有 PB2 PD2 BD2 0 ， PB2 BD2 PD2 0 ， BD2 PD2 PB2 0 ， △ PB2 BD2 PD2 PD2 BD2 PB2 PB2 PD2 BD2 可令a ，b ，c ， 2 2 2 则存在以ABa,ADb,APc为共点棱的长方体， PBD为该长方体的截面． 由（1）知，若 PBD是长方体的截面，则 PBD是锐角 △ 三角形， 所以 PBD不是 △ 任何一个长方体的截面等价 △ 于 PBD是直角三角形或钝角三角形．···············11分 由（△ i）知，0BAD π ，所以  A  B    A  D  0 ，△又因为PA AB，PA AD， 2 所以  P  B    P  D    P  A    A  B    P  A    A  D    P  A  2   A  B    A  D  0，故0BPD π ．···························12分 2 因为PA，所以PBA,PDA分别是直线PB,PD与所成的角，即PBA,PDA， PA2 9 不妨设AB≤AD，则≥，且PB≤PD，所以，tan2  ，···················13分 AB2 x2 y2 1 1 π 且PBD≥ PDB． 2 数学参考答案及评分细则 第13页（共 15页）作QM BD于M，因为平面QBD，平面QBDBD，QM 平面QBD， 所以QM ，又PA，所以PA∥QM ． 因为Q是线段CP上靠近C的三等分点，所以M 是线段AC上靠近C的三等分点， 所以M2,0,0，即直线BD过M2,0,0，···································································14分 π        所以PBDPBM≥ ，所以BABM  BPPA BM BPBM≤0．······························· 15分 2 这样，问题等价于在平面直角坐标系xAy中，B(x ,y ),D(x ,y ) 在双曲线x2  y2 3的右支上，直线BD 1 1 2 2 过点M(2,0)，  B  A    B  M  ≤0 ，求tan2 x2  9 y2 的最小值． 1 1   如图，不妨设点B在第四象限，因为 BABM≤0 ，所以点B在以AM 为直径的圆N内（含边界），记 圆N与双曲线在第四象限的交点为B ，则 AB≤ AB ． 0 0 因为AB 在渐近线yx的上方，故k 1，而k k 1，故k 1，即直线B M 与双曲 0 AB 0 AB 0 B 0 M B 0 M 0 线右支有两个交点B 0 ,D 0 ，符合条件．所以当点B位于点B 0 时， AB 最大，则 tan2最小．·····16分  (x1)2  y2 1, 1 7 1 7 联立 ，得 2x2 2x30 ，解得x 或x （舍去）， x2 y2 3 2 2 1 7 π 9 3 7 3 故当x  ，即PBD 时， tan2的最小值为  ． 1 2 2 1 7 2 数学参考答案及评分细则 第14页（共 15页）3 73 故 tan2的最小值为 ．·······················································································17分 2 数学参考答案及评分细则 第15页（共 15页）