福建省泉州市2025届高中毕业班适应性练习卷数学答案_2025年4月_250427福建省泉州市2025届高中毕业班适应性练习卷（泉州四检）（全科）

保密★使用前 泉州市 2025 届高中毕业班适应性练习卷参考答案 2025.04 高 三 数 学 一、选择题：选择题：本题共8小题，每小题5分，共40分。在每小题给出的四个选项中，只有一项是符合题 目要求的。 1~8：CDCA BBDC 二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的选项中，有多项符合题目要求。全部选对 的得6分，有选错的得0分，部分选对的得部分分。 9．AB 10．BCD 11．ACD 三、填空题：本题共3小题，每小题5分，共15分。 7 4 2 2 12．0.7（ 也可） 13． 14． ． 10 9 π 四、解答题：本题共5小题，共77分。解答应写出文字说明，证明过程或演算步骤。 15.（13分） 某旅行社推出“文化古城游”旅游路线后，为了解游客的满意度，对该路线的游客进行随机抽样调查，得到 如下满意度评分的频率分布直方图． （1）估计“文化古城游”路线游客满意度评分的众数和第80百分位数；（同一组中的数据用该区间的中点 值为代表） （2）现从参与“文化古城游”的游客中随机抽取3人，设这3人中满意度评分不低于90分的人数为X ，求 X 的分布列和数学期望．（以样本数据中游客的满意度评分位于各区间的频率作为游客的满意度评分位于该区 间的概率） 【命题意图】本题主要考查频率分步直方图、样本的数字特征、二项分布等基础知识；考查数据处理能力、 运算求解能力等；考查数形结合思想、化归与转化思想、或然与必然思想等；体现基础性与应用性，导向对发展 数学运算、数据分析、数学建模等核心素养的关注． 【试题解析】 9095 （1）估计“文化古迹游”路线游客满意度评分的众数为 92.5，···································1分 2 高三数学试题 第 1 页（共 13 页）满意度评分在 70,95内的频率为（0.0080.0120.0200.0400.070）50.750.8， 故第80百分位数位于[95,100]，·····················································································2分 0.80.75 所以第80百分位数为95 596．（列式1分，计算1分，只写答案扣1分）·········4分 0.25 （2）解法一： 3 由表可知，“文化古迹游”线路游客满意度评分不低于90分的概率为（0.0700.050）50.6 ，5分 5 依题意，X 的可能取值为0,1,2,3，·················································································6分 3 8 则PX 0(1 )3  ，·························································································7分 5 125 3 3 36 PX 1C1 (1 )2  ，······················································································8分 3 5 5 125 3 3 54 PX 2C2( ）2（ 1 ） ，·················································································9分 3 5 5 125 3 27 PX 3（ ）3 ，······························································································10分 5 125 所以X 的分布列为： X 0 1 2 3 27 P 8 36 54 125 125 125 125 ·······························································································································11分 8 36 54 27 9 所以EX0 1 2 3  ．（列式1分，计算1分）·························13分 125 125 125 125 5 解法二： 设事件A“抽取一名‘文化古迹游’线路游客的满意度评分不低于90分”， 3 由表可知，P(A)（0.0700.050）50.6 ．·································································5分 5 3 依题意，X 表示事件A发生的次数，则X ~B(3, )，·························································7分 5 3 3 于是，X 的分布列为P(X k)Ck ( )k（ 1 )3k，k 0,1,2,3．·········································9分 3 5 5 即： X 0 1 2 3 27 P 8 36 54 125 125 125 125 （需要计算出每个X 取值的概率，否则扣2分）······························································11分 3 9 所以E(X)3  ．（答案也可以写为1.8）································································13分 5 5 高三数学试题 第 2 页（共 13 页）16.（15分） 我们把公差不为零的等差数列称为一阶等差数列．若{a a }是一阶等差数列，则称{a }为二阶等差数列． n1 n n （1）若a n2 n1，判断{a }是否为二阶等差数列，并说明理由； n n （2）若{a }是二阶等差数列，且a 1,a 4,a 9． n 1 2 3 （ⅰ）求{a }的通项公式； n  a  （ⅱ）求数列 n 的前n项和S ． 4a 1 n n 【命题意图】本题主要考查数列的基本概念、等差数列的定义与通项公式、数列求和等基础知识；考查运算 求解能力、推理论证能力等；考查函数与方程思想、化归与转化思想等；体现基础性和综合性，导向对发展数学 抽象、数学运算、逻辑推理等核心素养的关注． 【试题解析】 （1）因为a n2 n1，a (n1)2 (n1)1n2 3n3，·········································1分 n n1 所以a a 2n2，·································································································2分 n1 n 所以(a n2 a n1 )(a n1 a n )20，nN，（不等于0没写不扣分）································3分 所以{a a }是一阶等差数列，····················································································4分 n1 n 所以{a }是二阶等差数列．（有判断是二阶等差数列得1分）··············································5分 n （2）（ⅰ）因为{a }是二阶等差数列，所以{a a }是一阶等差数列． n n1 n 因为a 1,a 4,a 9，所以a a 3,a a 5， 1 2 3 2 1 3 2 所以{a a }是首项为3，公差为2的等差数列，····························································6分 n1 n 所以a n1 a n 2n1，nN．·····················································································7分 当n≥2时， a (a a )(a a )(a a )a n n n1 n1 n2 2 1 1 (2n1)(2n3)31··························································································8分 n(2n11)  n2．···································································································9分 2 当n1时，a 1也符合上式，（没检验扣1分）·····························································10分 1 所以{a }的通项公式是a n2．···················································································11分 n n a n2 （ⅱ）因为 n  4a 1 4n2 1 n 1 1  (1 )················································································· 12分 4 4n2 1 1 1 1 1   (  )，···································································· 13分 4 8 2n1 2n1 n 1 1 1 1 1 1 所以S   [(1 )(  )(  )]····················································· 14分 n 4 8 3 3 5 2n1 2n1 n 1 1 n n n2 n   (1 )    ．（后面这三个答案皆可）···················15分 4 8 2n1 4 4(2n1) 4n2 高三数学试题 第 3 页（共 13 页）17.（15分） 已知函数 f(x)ln(x1)aex 1． （1）当a1时，求 f(x)的单调区间； （2）若 f(x)≤lna，求a的取值范围． 【命题意图】本题主要考查导数的运算，利用导数研究函数的单调性、不等式恒成立等基础知识；考查运算 求解能力、推理论证能力等；考查划归与转化思想、函数与方程思想、数形结合思想等；体现综合性，导向对数 学运算、逻辑推理、数学抽象、直观想象等核心素养的关注． 【试题解析】 （1）当a1时， f(x)ln(x1)ex1，其定义域为(1,)，（写出定义域1分）············ 1分 1 则 f(x) ex，································································································ 2分 x1 1 1 令g(x) ex，则g(x) ex 0， x1 (x1)2 1 所以 f(x) ex在区间(1,)内单调递减，（没证明直接写出单调递减扣1分）········· 3分 x1 又 f(0)0，·········································································································· 4分 所以当x(1,0)， f(x)0；当x(0,)， f(x)0． 所以 f(x)的单调递增区间为(1,0)，单调递减区间为(0,)．········································· 5分 （备注：没写定义域且单调递增区间写成(,0)，扣2分） （2）解法一： 1 1a(x1)ex 由题意可知a0， f(x) aex  ,(x1)，·········································· 6分 x1 x1 令F(x)1a(x1)ex，则F(x)a(x2)ex 0，故F(x)在定义域上单调递减．················ 7分 注意到F(1)10，且x→,F(x)→， 1 故存在唯一的x (1,)，使得F(x )0，即aex0  ，·········································· 8分 0 0 x 1 0 故当x(1,x )，F(x)0，即 f(x)0；当x(x ,)，F(x)0，即 f(x)0． 0 0 所以 f(x)在区间(1,x )内单调递增；在区间(x ,)内单调递减， 0 0 所以当xx 时， f(x)取到最大值 f(x )ln(x 1)aex0 1，·········································· 9分 0 0 0 故由题意等价于ln(x 1)aex0 1≤lna， 0 1 注意到aex0  注意到等号两边皆为正数，故取自然对数得x lnaln(x 1) x 1 0 0 0 即lnax ln(x 1)，····························································································10分 0 0 1 所以题意等价于2ln(x 1)(x 1) ≤0，····························································11分 0 0 x 1 0 1 设tx 1，则题意等价于2lntt ≤0， 0 t 高三数学试题 第 4 页（共 13 页）1 2 1 (t1)2 设函数G(t)2lntt ,t0，其导函数为G(t) 1  0 ， t t t2 t2 故G(t)在区间(0,)单调递增，·················································································12分 又因为G(1)0，所以题意等价于0t≤1．···································································13分 因为lnax ln(x 1)lntt1，令h(t)lntt1(0t≤1)， 0 0 1 1t 则h'(t) 1 ＜0，所以h(t)为关于t的单调递减函数，值域为[0,) t t 所以当0t≤1时，可得lna的取值范围为[0,)，························································14分 故所求的a的取值范围为[1,)．···············································································15分 解法二： 由题意得 f(x)≤lna恒成立，即ln(x1)1≤lnaaex，其中a0,x1，························6分 从而可得，题意等价于ln(x1)x1≤lnaaex x，···················································7分 注意到x10，所以上式等价于ln(x1)eln(x1)≤exlna (xlna)，·······························8分 令t ln(x1),t  xlna, 1 2 故可得，题意等价于t et1≤t et2 ．·········································································9分 1 2 考察函数F(t)tet，其导函数为F(t)1et 0， 故F(t)在区间R单调递增，····················································································10分 因此题意等价于F(t )t et1≤t et2  F(t )，即t≤t ，·············································11分 1 1 2 2 1 2 从而题意等价于ln(x1)≤xlna，即ln(x1)x≤lna，··············································12分 考察函数G(x)ln(x1)x,x1， 1 x 其导函数为G(x) 1 ,(x1)， x1 x1 故可得在区间x(1,0)，G(x)0,G(x)单调递增； 而在区间x(0,)，G(x)0,G(x)单调递减，··························································13分 所以当x0时，G(x)取到最大值G(0)，···································································14分 从而可得题意的充要条件为lna≥G(0)0，解得a≥1． 故所求a的取值范围是[1,)．···············································································15分 解法三： 由 f(x)≤lna的必要性可知，当x0时， f(x)≤lna，整理得a1≤lna，即lnaa≥1， ···························································································································· 6分 设F(a)lnaa为关于a的单调递增函数，·································································· 7分 且注意到F(1)1，故解得a≥1．················································································ 8分 下面证明充分性：当a≥1时， f(x)≤lna， 即证：当a≥1时，ln(x1)aex 1≤lna,(x1)，·························································9分 即证lnaexa1ln(x1)≥0． 高三数学试题 第 5 页（共 13 页）将a视为自变量，设函数G(a)lnaexa1ln(x1)，················································10分 1 导函数G(a) ex， a 因为a≥1且ex 0，故G(a)0，所以G(a)为单调递增，··············································· 11分 所以G(a)≥G(1)ex 1ln(x1)． 故只需要证明：当x1时，ex 1ln(x1)≥0，·························································12分 考察函数H(x)ex x1，H(x)ex 1， 可得当x(1,0)，H(x)0，H(x)单调递减；当x(0,)，H(x)0，H(x)单调递增； 所以H(x)≥H(0)0，······························································································13分 即ex≥x1，即x≥ln(x1)． 所以G(1)ex 1ln(x1)≥xln(x1)≥0，充分性得证．···············································14分 综上，所求a的取值范围是[1,)．············································································15分 18.（17分） 如图，在四棱锥PABCD中，底面ABCD为正方形，平面ABCD平面ABP，PAB60，AD1,AP2， 点F 在线段PD上（F 与P,D不重合）． （1）若平面AFB平面PCDl，证明：l ∥平面ABCD； （2）当△AFB的面积最小时，求二面角PAF B的正弦值； （3）在（2）的条件下，若点F,F,,F 是线段PF 的n1等分点，分别过点F,F ,,F 在四棱锥上作平行 1 2 n 1 2 n 1 n 7 3 于平面AFB的截面，记相应的截面面积为S (i1,2,3,,n)，证明： S  ． i n i 48 i1 n(n1)(2n1) （参考公式：12 22 32 n2  ） 6 【命题意图】本题主要考查线面平行判定定理和性质定理、二面角的求解，以及数列求和与不等式证明等基 础知识；考查推理论证能力、运算求解能力、空间想象能力等；考查数形结合思想、化归与转化思想等；体现基 础性、综合性与创新性；导向对直观想象、数学运算、逻辑推理等核心素养的关注． 【试题解析】 （1）证明：因为CD∥AB，AB平面ABF ，CD平面ABF ，·········································· 1分 则CD∥平面ABF ．······································································································2分 又因为CD平面PCD，且平面ABF平面PCDl，所以CD∥l．········································3分 高三数学试题 第 6 页（共 13 页）因为CD平面ABCD，l平面ABCD，所以l∥平面ABCD．（漏掉部分条件酌情扣1-2分）···4分 （2）（ⅰ）解法一： 因为四边形ABCD为正方形，则BCAB， 因为平面ABCD平面ABP，平面ABCDI平面ABP AB,BC 平面ABCD， 所以BC平面ABP．···································································································5分 在△ABP中，AB1,AP2,PAB60°，由余弦定理得BP2 AB2AP22ABAPcos60°3， 所以BP 3，则AB2BP2 AP2，即ABBP．································································6分 由（1）得l∥平面ABCD，不妨设l交PC于点G，连结BG,FG 即平面ABFI平面PCDFG,平面ABFI平面PBC BG， 如图１，则CD∥FG∥AB，所以点F到AB的距离等于点G到AB的距离． 因为ABBC,ABBP,BPIBCB,BP,BC 平面PBC ，所以AB平面PBC ， 又因为BG平面PBC ，所以ABBG． 故只需BG最小时，则△AFB的面积最小，当BGPC时，BG最小．····································7分 因为BC平面ABP，所以BCBP，所以PC2， 3 在Rt△BCP中，BGPC，由BGPCBCBP，可得BG ， 2 1 1 所以在Rt△BGC中，CG BC2BG2  ，即CG CP．·················································· 8分 2 4 以B为原点，分别以BP，BA，BC所在的直线为x,y,z轴，建立如图所示的空间直角坐标系Bxyz． 则P( 3,0,0),A(0,1,0),C(0,0,1),D(0,1,1)． 由AB平面PBC ,PC 平面PBC ，所以ABPC． 高三数学试题 第 7 页（共 13 页）又因为BGPC，ABBGB，AB,BG平面AFGB，所以PC平面AFGB，························9分 uuur 则PC ( 3,0,1)为平面ABF的一个法向量．··································································10分 uuur uuur 设n (x,y,z)为平面PAD的一个法向量，AD(0,0,1)，AP( 3,1,0)， uuur  nAD0 z0 由 uuur ，得 ，取x1,y 3，得n(1, 3,0)，·········································11分 nAP0  3xy0 uuur uuur nPC 3 则cosn,PC uuur  ，···················································································12分 |n||PC| 4 uuur 13 设二面角PAFB的平面角为，则sin 1cos2n,PC ， 4 13 所以二面角PAFB的正弦值为 ．·········································································13分 4 解法二： 因为四边形ABCD为正方形，则BCAB， 因为平面ABCD平面ABP，平面ABCDI平面ABPAB，BC 平面ABCD， 所以BC平面ABP．···································································································5分 在△ABP中，AB1,AP2,PAB60°，由余弦定理得BP2 AB2AP22ABAPcos60°3， 所以BP 3，则AB2BP2 AP2，即ABBP．································································6分 以B为原点，分别以BP,BA,BC所在的直线为x,y,z轴，建立如图所示的空间直角坐标系Bxyz． 则P( 3,0,0),A(0,1,0),C(0,0,1),D(0,1,1)． uur uuur uuur uuur BA(0,1,0),PD（ 3,1,1），因为点F在PD上，不妨设PF PD,(01)， uuur uur uuur uur uuur 则BF BPPF BPPD（ 3,0,0）（ 3,1,1）（ 3 3,,）， uuur uur uuur (BFBA)2 3 3 点F到AB的距离为 |BF|2  uur  42 63  4( )2  ，·····························7分 |BA|2 4 4 3 当 时，点F到AB的距离最小，即△AFB的面积最小．·················································8分 4 uuur 3 3 3 此时，BF=（ , , ）．······························································································9分 4 4 4 设n (x,y ,z )为平面ABF的一个法向量， 1 1 1 1 高三数学试题 第 8 页（共 13 页）uur y 0  n BA0  1 由 n 1 B uu F ur 0 ，得  3 x  3 y  3 z 0 ，取x 1  3,z 1 1，得n 1 ( 3,0,1)．·······················10分 1  4 1 4 1 4 1 设n (x,y,z)为平面PAD的一个法向量， 2 uuur uuur AD(0,0,1),AP( 3,1,0)， uuur  nAD0  z0 由 uuur ，得 ，取x1,y 3，得n (1, 3,0)．········································11分 nAP0  3xy0 2 n n 3 则cosn,n  1 2  ，··················································································12分 1 2 |n ||n | 4 1 2 13 设二面角PAFB的平面角为，则sin 1cos2n,n  ， 1 2 4 13 所以二面角PAFB的正弦值为 ．·········································································13分 4 （ⅱ） 3 3 由（ⅰ）知FG∥AB，FG CD ，AB1， 4 4 因为ABBG，所以四边形ABGF为直角梯形． 3 1 3 3 7 3 又BG ，所以S  (1+ )  ．··························································14分 2 梯形ABGF 2 4 2 16 由题意，不妨设F,F ,,F 是距离点 由近及远的n个等分点，（这句不写不扣分） 1 2 n P 设过点F截四棱锥PABCD的截面交PC于点G，交BP于点B，交AP于点A， i i i i 则截面ABGF∥平面ABGF，i1,2,,n． i i i i 则 PG i  i ，所以 S 梯形AiBiGiFi ( i )2． PG n1 S n1 梯形ABGF i 7 3 i2 即S ( )2S  ．···································································15分 梯形AiBiGiFi n1 梯形ABGF 16 (n1)2 1 n 1 7 3 1222n2 7 3(2n1) 所以 S     ·····························································16分 n i n 16 (n1)2 96 (n1) i1 7 3 1 7 3 = (2 )＜ ．············································17分 96 n1 48 高三数学试题 第 9 页（共 13 页）19.（17分） x2 y2 2 已知椭圆E：  1(ab0)的离心率为 ，且过点( 2,1)． a2 b2 2 （1）求E的方程； （2）若直线与椭圆交于两点，当以这两点和椭圆的中心为顶点的三角形面积达到最大值时，称该直线为椭圆 的“好直线”． （ⅰ）设O为坐标原点，若斜率存在的直线l是E的“好直线”，l与E交于A,B两点，求△AOB的面积； （ⅱ）已知四边形MNPQ为平行四边形，若直线MN,NP,PQ,QM均为E的“好直线”，且不与y轴平行，求 四边形MNPQ的面积的最小值． 【命题意图】本题主要考查椭圆基本性质、直线与椭圆的位置关系等基础知识；考查运算求解能力、推理论 证能力等；考查化归与转化思想、数形结合思想、函数与方程思想等；体现综合性与创新性，导向对直观想象、 逻辑推理、数学运算、数学抽象等核心素养的关注． 【试题解析】  a2 b2 c2  c 2 (1)依题意得，  ，（离心率1分，点代入1分）·················································2分 a 2  2 1   1 a2 b2 a2 4 解得  b2 2，因此椭圆E的方程为 x2  y2 1．·································································3分  4 2  c2 2 （2）（ⅰ）解法一： 设直线l:ykxm，l与E的交点为A(x ,y ),B(x ,y )． 1 1 2 2 ykxm  联立x2 y2 ，消去y得，(12k2)x2 4kmx2m2 40 ，···········································4分   1  4 2 (4km)24(2m24)(12k2)8(4k2m22)0 时， 4mk 2m2 4 由韦达定理，x x  ,x x  ，································································5分 1 2 12k2 1 2 12k2 因此 AB  x x 2 y y 2  1k2 (x x )2 4x x 1 2 1 2 1 2 1 2 4mk 2m2 4 2 2(1k2)(2+4k2 m2)  1k2 ( )2 4  ，·······························6分 12k2 12k2 12k2 |m| 原点O到直线l的距离d  ，·············································································7分 1k2 1 1 2 2(1k2)(24k2 m2) |m| 2(24k2 m2)m2 因此S  |AB|d     ·············8分 △AOB 2 2 12k2 1k2 12k2 高三数学试题 第 10 页（共 13 页）2[(24k2 m2)m2] ≤  2．··································································9分 2(12k2) 当且仅当24k2 m2 m2，即m2 2k2 1时等号成立.所以△AOB的面积为 2．·············· 10分 解法二： 设A(x ,y ),B(x ,y )，O为原点，直线OB的方程为y xx y0，····································4分 1 1 2 2 2 2 |x y x y | 点A到OB的距离d  1 2 2 1 ，·············································································5分 x2  y2 2 2 1 1 |x y x y | 1 因此S △AOB  2 |OB|d  2 x 2 2  y 2 2  1 x 2 2  y 2 2 1  2 |x 1 y 2 x 2 y 1 |． 2 2 1 （用面积公式S  |OB||OA| 1cos2∠AOB证明也可，直接用公式扣3分）·············6分 △AOB 2 x2 y2 点A(x ,y ),B(x ,y )在椭圆上，满足 i  i 1(i1,2)．·················································7分 1 1 2 2 4 2 x2 y2 x2 y2 x2x2 y2y2 x2y2 x2y2 xx y y (x y x y )2 (x y x y )2 于是1( 1  1 )( 2  2) 1 2  1 2  2 1  1 2 ( 1 2  1 2)2 2 1 1 2 ≥ 2 1 1 2 ， 4 2 4 2 16 4 8 8 4 2 8 8 ·····························································································································9分 1 因此 |x y x y |≤ 2 ，当且仅当xx 2y y 0时等号成立． 2 2 1 1 2 1 2 1 2 所以△AOB的面积为 2．························································································10分 解法三： 设A(2cos, 2sin),B(2cos, 2sin) ，0,≤2π，O为原点．··································4分 1 由解法二可知 S  |2 2cossin2 2sincos|（利用该结论必须要证明，否则扣3分） △AOB 2 ·····························································································································7分  2|sin()|···········································································8分 ≤ 2．························································································9分 π 3π 当且仅当|| 或 时，所以△AOB的面积为 2．···············································10分 2 2 （ⅱ）解法一： 根据(2)中解法一的取等条件，斜率为k的好直线有且仅有两条． 设l :yk x 2k2 1，l :yk x 2k2 1，··················································11分 MN,PQ 1 1 MQ,NP 2 2  yk x 2k21 2k2 1 2k2 1 联立l ,l 直线方程： 1 1 ，消y，解得x  1 2 ， MN MQ  yk 2 x 2k 2 21 M k 2 k 1  yk x 2k21 2k2 1 2k2 1 联立l ,l 直线方程： 1 1 ，消y，解得x  1 2 ，···············12分 MN NP  yk 2 x 2k 2 21 N k 2 k 1 高三数学试题 第 11 页（共 13 页）2 2k2 1 因此|MN| 1k2 |x x | 1k2 2 ，··························································13分 1 M N 1 |k k | 2 1 2 2k2 1 直线l ,l 的距离d  1 ，·············································································14分 MN PQ 1k2 1 2 2k2 1 2 2k2 1 因此S |MN |d  1k2 2  1 四边形MNPQ 1 |k k | 1k2 2 1 1 4 (2k2 1)(2k2 1) 4 4k2k2 12(k2 k2)  1 2  1 2 1 2 ········································15分 |k k | |k k | 1 2 1 2 4 4k k 2(k2k2) 4 2 (k k )2 ≥ 1 2 1 2  1 2  4 2 ，·······································16分 |k k | |k k | 1 2 1 2 4 4k2k2 12(k2 k2) 4 4k2k2 14k k 2(k k )2 （或者 1 2 1 2  1 2 1 2 1 2 |k k | |k k | 1 2 1 2 4k2k2 14k k （2k k 1)2 4 1 2 1 2 2 4 1 2 2≥4 2 ）··································16分 (k k )2 (k k )2 1 2 1 2 1 当且仅当k k  时，四边形面积的最小值为4 2．（取等条件不写扣1分）··················17分 1 2 2 解法二： 同解法一的l :yk x 2k2 1，l :yk x 2k2 1，······································11分 MN,PQ 1 1 MQ,NP 2 2 可知平行四边形MNPQ的对角线交点为坐标原点， 1 则所求四边形MNPQ的面积S 4S OMN 4 |x y x y |2|x y x y | △ M N N M M N N M 2  yk x 2k21 联立l ,l 直线方程： 1 1 ， MN MQ  yk x 2k21 2 2 2k2 1 2k2 1 k 2k2 1k 2k2 1 解得x  1 2 ，y  2 1 1 2 ············································12分 M k k M k k 2 1 2 1  yk x 2k21 联立l ,l 直线方程： 1 1 ， MN NP  yk x 2k21 2 2 2k2 1 2k2 1 k 2k2 1k 2k2 1 解得x  1 2 ，y  2 1 1 2 ，··········································13分 N k k N k k 2 1 2 1 高三数学试题 第 12 页（共 13 页）( 2k2 1 2k2 1)(k 2k2 1k 2k2 1)( 2k2 1 2k2 1)(k 2k2 1k 2k2 1) S 2 1 2 2 1 1 2 1 2 2 1 1 2 (k k )2 2 1 · 2(k k ) (2k21)(2k21) (2k21)(2k21) (2k2 1)(2k2 1) 2 1 2 1 2 4 1 2 4 1 2 (k k )2 (k k ) (k k )2 2 1 2 1 2 1 ····························································································································15分 以下同解法一. 解法三： 同解法一的l :yk x 2k2 1，l :yk x 2k2 1，······································11分 MN,PQ 1 1 MQ,NP 2 2 2 2k2 1 2 2k2 1 直线l ,l 的距离d  1 ，直线l ,l 的距离d  2 ，·····························12分 MN PQ 1 NP QM 2 1k2 1k2 1 2 设直线l ,l 的倾斜角分别为, ，l ,l 的夹角 MN NP 1 2 MN NP d 则所求的四边形MNPQ的面积S d  2 ，································································13分 1 sin tantan k k 因为l ,l 的夹角正切值为tan|tan()| 1 2  1 2 MN NP 1 2 1tantan 1kk 1 2 1 2 k k k k 所以sin 1 2  1 2 ，·················································14分 (1kk )2 (k k )2 (1k2)(1k 2) 1 2 1 2 1 2 d 2 2k2 1 2 2k2 1 (1k2)(1k 2) (2k2 1)(2k2 1) 所以S d  2  1  2  1 2 4 1 2 ·················15分 1 sin 1k 1 2 1k 2 2 |k 1 k 2 | (k 2 k 1 )2 以下同解法一. 高三数学试题 第 13 页（共 13 页）