2024届福建省泉州市高中毕业班质量检测（一）数学解答题答案_2023年8月_01每日更新_30号_2024届福建省泉州市高中毕业班质量检测（一）

保密★使用 前 泉州市 2024 届高中毕业班质量监测（一） 2023.08 数 学 【解答题部分】 本试卷共22题，满分 150分，共 8页。考试用时120分钟。 四、解答题：本题共6小题，共70分。解答应写出文字说明，证明过程或演算步骤。 17．（10分） a 数列 a 中，a 1，且a  n ． n 1 n1 a 1 n （1）求 a 的通项公式； n 2n （2）令b  ，记数列 b 的前n项和为S ，求S ． n a n n 8 n 【命题意图】本小题主要考查等差数列的定义、通项公式与数列求和等基础知识，考查运算 求解能力，考查函数与方程、化归与转化等思想，体现基础性，导向对发展数 学运算等核心素养的关注． 【试题解析】解法一： a 1 1 （1） 由a  n ，可得  1．····················································1分 n1 a 1 a a n n1 n 1 因为a 1，所以 1.···································································2分 1 a 1 1 所以数列{ }是首项为1，公差为1的等差数列.··································· 3分 a n 1 1 所以 n，即a  .····································································4分 a n n n 2n （2） 因为b  ，所以b n2n.····························································5分 n a n n 又S b b b ， n 1 2 n 所以S 21222 828①.·······················································6分 8 ①式两边同乘以2，得2S 22 2238281②，·························· 7分 8 ②－①，得S (222 2328)829，······································8分 8 高三数学试题 第1页（共24页）2(128) 所以S  829 2729 3586.····································10分 8 12 解法二： a （1） 由a  n ，可得a a a a ． n1 a 1 n1 n n1 n n 1 1 1 1 两边同除以a a ，得1  ，即  1．····························· 1分 n1 n a a a a n n1 n1 n 1 因为a 1，所以 1.·································································· 2分 1 a 1 1 所以数列{ }是首项为1，公差为1的等差数列.··································3分 a n 1 1 所以 n，即a  .···································································4分 a n n n 2n （2） 因为b  ，所以b n2n.····························································5分 n a n n 所以b  2,b 8,b  24,b 64,b 160,b 384,b 896,b  2048，·······9分 1 2 3 4 5 6 7 8 所以S b b b  2824641603848962048 3586.······10分 8 1 2 8 高三数学试题 第2页（共24页）18．（12分） 泉州是历史文化名城、东亚文化之都，是联合国认定的“海上丝绸之路”起点．著名的 “泉州十八景”是游客的争相打卡点，泉州文旅局调查打卡十八景游客，发现 90%的人至少 打卡两个景点．为提升城市形象，泉州文旅局为大家准备了 4 种礼物，分别是世遗泉州金属 书签、闽南古厝徽章、开元寺祈福香包、小关公陶瓷摆件．若打卡十八景游客至少打卡两个 景点，则有两次抽奖机会；若只打卡一个景点，则有一次抽奖机会．每次抽奖可随机获得 4 种礼物中的1种礼物．假设打卡十八景游客打卡景点情况相互独立． （1）从全体打卡十八景游客中随机抽取3人，求3人抽奖总次数不低于4次的概率； （2）任选一位打卡十八景游客，求此游客抽中开元寺祈福香包的概率. 【命题意图】本小题主要考查离散型随机变量的分布列、条件概率、全概率公式等基础知识； 考查运算求解、推理论证能力等；考查化归与转化思想等．体现基础性、应用 性和综合性，导向对发展数学运算、数学抽象等核心素养的关注． 【试题解析】解法一： （1） 设3人抽奖总次数为X ，则X 的可能取值为3,4,5,6．··························· 1分 9 由题意知，每位打卡十八景游客至少打卡两个景点的概率为 ， 10 1 只打卡一个景点的概率为 ．··························································2分 10 9 1 27 依题可得P(X 4)C1 ( )2  ，···········································3分 3 10 10 1000 9 1 243 P(X 5)C2( )2  ，···········································4分 3 10 10 1000 9 729 P(X 6)( )3  ，···················································· 5分 10 1000 27243729 所以P(X 4) P(X 4)P(X 5)P(X 6) 0.999． 1000 ····································································································6分 （2） 记事件A“每位打卡十八景游客至少打卡两个景点”, 则 A“每位打卡十八景游客只打卡一个景点”， 事件B“一位打卡十八景游客抽中开元寺祈福香包”，······················· 7分 9 1 3 7 1 则P(A) ，P(A) ，P(B| A)1( )2  ，P(B| A) ，··········· 9分 10 10 4 16 4 高三数学试题 第3页（共24页）则P(B) P(AB AB) P(AB)P(AB)  P(A)P(B| A)P(A)P(B| A) 9 7 1 1 67      ．·····················································12分 10 16 10 4 160 解法二： （1）设3人抽奖总次数为X ，则X 的可能取值为3,4,5,6．····························1分 9 由题意知，每位打卡十八景游客至少打卡两个景点的概率为 ， 10 1 只打卡一个景点的概率为 ．··························································2分 10 9 1 1 依题可得P(X 3)C0( )0( )3  ，······································5分 3 10 10 1000 1 999 所以P(X≥4)1P(X 3)1( )3  0.999．····························6分 10 1000 （2） 同解法一．··················································································12分 解法三: （1）设3人抽奖总次数为X ，设抽取的 3 位打卡十八景游客中至少打卡两个景点的 人数为Y ，则X 2Y  3Y 3Y ，················································1分 由题可知，Y ~ B  3,0.9 ，······························································· 4分 1 故P(X 4) P(3Y 4) P(Y 1)1P(Y 0)1( )3 0.999．··········6分 10 （2） 同解法一．··················································································12分 高三数学试题 第4页（共24页）19．（12分） △ABC的内角A,B,C所对的边分别为a,b,c，且满足ccosB(b2a)cosC 0． （1）求C ；   （2）若CD平分ACB ，且AD2DB，CD  2，求△ABC的面积． 【命题意图】本小题主要考查解三角形、三角恒等变换等基础知识，考查推理论证、运算求 解等能力，考查数形结合和化归与转化等思想，体现综合性与应用性，导向对 发展直观想象、逻辑推理及数学运算等核心素养的关注． 【试题解析】解法一： （1）因为ccosB(b2a)cosC 0， 所以由正弦定理，可得sinCcosB(sinB2sinA)cosC 0，·················· 1分 即sinCcosBsinBcosC2sinAcosC 0，sin(BC)2sinAcosC 0，···2分 所以sinA2sinAcosC 0，···························································· 3分 1 又sin A0，所以cosC  ，·························································4分 2 2 因为C(0,，所以C  ．··························································5分 3    2 1 （2） 因为AD2DB，所以CD  CB CA，···········································6分 3 3  2 4 2 1 2 4  两边平方，得CD  CB  CA  CBCA，····································7分 9 9 9 即4a2 b2 2ab36①．································································· 8分 b AD 又因为CD平分ACB ，所以   2，即b2a②．························9分 a DB 由①②，解得a 3,b6．····························································· 10分 1 2 9 3 所以S  absinC 9sin  ．··········································· 12分 △ABC 2 3 2 高三数学试题 第5页（共24页）解法二： a2 c2 b2 a2 b2 c2 （1）在△ABC中，由余弦定理，得cosB  ，cosC  ， 2ac 2ab ·································································································1分 又因为ccosB(b2a)cosC 0， a2 c2 b2 a2 b2 c2 a2 b2 c2 所以    0，····································2分 2a 2a b 即a2 b2 c2 ab，······································································ 3分 a2 b2 c2 1 所以cosC   ，··························································4分 2ab 2 2 因为C(0,，所以C  ．··························································5分 3   AD （2） 在△ABC中，AD2DB，所以 2，···········································6分 DB b AD 又因为CD平分ACB ，所以   2，即b2a①．························7分 a DB 在△ACD中，由余弦定理，得CA2 CD2 AD2 2CACDcosACD， 4 即b2 4 c2 2b②．····································································8分 9 在△BCD中，由余弦定理，得CD2 CB2 BD2 2CDCBcosDCB， 1 即a2 4 c2 2a③．····································································9分 9 由①②③解得a 3，b6．···························································10分 1 2 9 3 所以S  absinC 9sin  ．··········································· 12分 △ABC 2 3 2 解法三：（1）同解法一．············································································5分 （2） 过D点作DE∥AC交CB于点E ．······················································6分 因为ACB 120，且CD平分ACB ， 所以ACD CDE DCE 60，··············································· 7分 所以△CDE 为等边三角形，所以CD CE  DE 2．····························8分 DE BE BD 1 又因为    ，所以BC 3，AC  6．···························10分 AC BC BA 3 1 2 9 3 所以S  absinC 9sin  ．··········································· 12分 △ABC 2 3 2 高三数学试题 第6页（共24页）解法四：（1）同解法一．············································································5分   AD （2） 在△ABC中，AD2DB，所以 2．···········································6分 DB b AD 又因为CD平分ACB，所以   2．··········································7分 a DB 1 3 如图，以C 为原点建系：C(0,0)，B(a,0)，A(a, 3a)，D( a, a)．··8分 3 3 1 3 由CD  2，所以( a)2 ( a)2 4，所以a 3，··································9分 3 3 则b6．····················································································10分 1 2 9 3 所以S  absinC 9sin  ．··········································· 12分 △ABC 2 3 2 高三数学试题 第7页（共24页）20．（12分） 已知函数 f(x)(x2)(aexx). （1）当a4时，求曲线y  f(x)在(0, f(0))处的切线方程； （2）讨论 f(x)的单调性. 【命题意图】本小题主要考查运用导数求切线方程，判断函数的单调性等基础知识；考查推 理论证、运算求解等能力；考查化归与转化、数形结合等数学思想；体现综合 性、应用性与创新性，导向对发展逻辑推理、数学运算、直观想象等核心素养 的关注． 【试题解析】 （1） 由已知 f(x)(x2)(aexx)， 则 f(x)aex xaex(x2)(x2) (x1)(aex2)，····························1分 当a4时， f(0)8， f(0)2，·················································· 3分 则曲线y  f(x)在(0, f(0))处的切线方程为y82x，即2x y80．·4分 （2） 由（1）知， f(x)(x1)(aex2)， ①当a≤0时，aex 20， 当x(,1)时， f(x)0， f(x)在(,1)单调递增；···························5分 当x(1,)时， f(x)0， f(x)在(1,)单调递减；···························6分 2 ②当a0时，由 f(x)(x1)(aex2)0，得x 1，x ln ，············ 7分 1 2 a 2 （i）当0a 时，x  x ， 1 2 e 2 2 当x(,1)(ln ,)时， f(x)0， f(x)在(,1),(ln ,)单调递增； a a ···································································································8分 2 2 当x(1,ln )时， f(x)0， f(x)在(1,ln )单调递减；························· 9分 a a 2 （ii）当a 时，x  x 1， f(x)≥0， f(x)在R单调递增；··············10分 1 2 e 2 （iii）当a 时，x  x ， 1 2 e 2 2 当x(,ln )(1,)时， f(x)0， f(x)在(,ln ),(1,)单调递增； a a ·································································································11分 高三数学试题 第8页（共24页）2 2 当x(ln ,1)时， f(x)0， f(x)在(ln ,1)单调递减； a a 综上可得：①当a≤0时， f(x)在(,1)单调递增，在(1,)单调递减； 2 2 2 ②当0a 时， f(x)在(,1),(ln ,)单调递增，在(1,ln )单调递减； e a a 2 ③当a 时， f(x)在R单调递增； e 2 2 2 ④当a 时， f(x)在(,ln ),(1,)单调递增，在(ln ,1)单调递减.··· 12分 e a a 高三数学试题 第9页（共24页）21．（12分） 如图，三棱锥P ABC 中，PAPB，PA PB，AB2BC2，平面PAB 平面ABC. （1）求三棱锥P ABC 的体积的最大值； （2）求二面角PACB的正弦值的最小值. 【命题意图】本小题主要考查线面垂直的判定定理，面面垂直的性质定理，三棱锥体积，二 面角，三角形面积等基础知识；考查空间想象能力、推理论证及运算求解能力； 考查数形结合思想、化归与转化思想等；体现基础性、综合性与应用性，导向 对发展逻辑推理、数学运算、直观想象等核心素养的关注． 【试题解析】解法一：（1）取AB的中点O，连接PO． 因为PA PB，所以PO  AB．························································1分 又因为平面PAB 平面ABC，平面PAB平面ABC  AB，PO平面PAB， 所以PO平面ABC．···································································· 2分 因为PAPB，PA PB，AB 2BC 2， 所以PO 1，BC 1．····································································3分 所以三棱锥 PABC 的体积为 1 1 1  1 V  S PO   ABBCsinABC PO sinABC，··········4分 PABC 3 △ABC 3 2  3 1 因为ABC 0,，所以0sinABC≤1，V „ ， PABC 3  当且仅当sinABC 1，即ABC  时，等号成立. 2 1 故三棱锥PABC 的体积的最大值为 .···············································5分 3 （2）由（1）可知PO平面ABC，所以PO  AC .······························6分 过O作OD  AC于D，连结PD． 高三数学试题 第10页（共24页）因为POOD O ，所以AC 平面POD，·········································7分 又PD平面POD，所以PD  AC ， 所以PDO为二面角P ACB的平面角.··········································8分 PO 1 在Rt△PDO中，tanPDO   ，············································9分 OD OD 1 1 因为 0OD≤ BC  ，当且仅当BC  AC 时等号成立.······················10分 2 2 所以tanPDO的最小值为2.··························································11分 2 5 此时sinPDO取得最小值 . 5 2 5 故二面角P ACB的正弦值的最小值为 .··································· 12分 5 解法二：（1）同解法一．············································································5分 （2）由（1）可知PO平面ABC，   以O为坐标原点，向量OB,OP为x轴，z 轴正方向，建立如图所示的空间直角  坐标系Oxyz．则P(0,0,1)，A(1,0,0)，B(1,0,0)，AP(1,0,1)．  设C(x ,y ,0)(0 x 2,y ≠0)，则 AC (x 1,y ,0) ．···························6分 0 0 0 0 0 0 设平面PAC 的法向量为m(x,y,z)．   mAP 0, xz 0, x 1 则   即  取x 1，则m (1, 0 ,1)．·······7分 mAC 0,   x 0 1  x y 0 y 0, y 0 又平面ABC的法向量为n(0,0,1)．··················································8分 设二面角P ACB的大小为， 0,． |mn| 1 cos  所以 |m||n|  x 1 2 ．················································9分 2 0   y  0 因为BC 1，所以 x 1 2  y 2 1，·················································10分 0 0 高三数学试题 第11页（共24页） x 1 2  x 1 2 令t  0   t 0 ，则t  0 ，  y  1 x 1 2 0 0 整理可得 t1  x 2  22t  x 10， 0 0 所以 22t 2 4  t1 ≥0，解得t≥3.···········································11分 1 3 5 所以当t 3，即x  ，y  时，cos取得最大值 ， 0 2 0 2 5 2 5 此时sin取得最小值 . 5 2 5 故二面角P ACB的正弦值的最小值为 .···································12分 5 解法三：（1）同解法一．············································································5分 （2）由（1）可知 PO平面 ABC ，  以B为坐标原点，向量 BA 为y轴正方向，建立如图所示的空间直角坐标系·····  Oxyz．则P(0,1,1)，A(0,2,0)，B(0,0,0)，AP(0,1,1)．  设C(x ,y ,0)(1 x 1,y ≠0)，则AC (x ,y 2,0)．··························6分 0 0 0 0 0 0 设平面PAC 的法向量为m(x,y,z)．   mAP 0, yz 0, 2 y 则   即  取y 1，则m (1, 0 ,1)．·········7分 mAC 0,  x 0 x(y 0 2)y 0, x 0 又平面ABC的法向量为n(0,0,1)．··················································8分 设二面角P ACB的大小为， 0,． |mn| 1 cos  所以 |m||n| 2 y  2 ．···············································9分 2 0   x  0 因为BC 1，所以x 2  y 2 1，·······················································10分 0 0 高三数学试题 第12页（共24页）y 2 令t  0 ，表示圆x 2  y 2 1上的点与点(0,2)的斜率， x 0 0 0 y 2 所以t≤ 3或t≥ 3，所以( 0 )2≥3， x 0 1 3 5 2 5 即x  ,y  时，cos取得最大值 ，此时sin取得最小值 . 0 2 0 2 5 5 2 5 故二面角P ACB的正弦值的最小值为 .··································· 12分 5 高三数学试题 第13页（共24页）22．（12分） x2 y2 2 已知椭圆E:  1  a b0 的离心率是 ，上、下顶点分别为A,B．圆 a2 b2 2   O:x2  y2 2与x轴正半轴的交点为P，且 PAPB1 ． （1）求E的方程； （2）直线l与圆O相切且与E相交于M,N两点，证明：以MN 为直径的圆恒过定点． 【命题意图】本小题主要考查椭圆的标准方程，直线与圆的位置关系等基础知识；考查运算 求解、逻辑推理和创新能力等；考查数形结合、函数与方程等思想；体现基础 性、综合性与创新性，导向对直观想象、逻辑推理、数学运算等核心素养的关 注． 【试题解析】解法一：（1）由已知得A(0,b)，B(0,b)，P( 2,0).····························1分     则PA( 2,b)，PB( 2,b)， PAPB2b2 1 ，所以b2 3.·······3分 c 2 因为e  ，又b2 c2 a2，所以c2 3，a2 6. a 2 x2 y2 故E的方程为  1.·································································4分 6 3 （2）当直线l的斜率存在时，设l的方程为ykxm，即kx ym0.··· 5分 m 因为直线l与圆O相切，所以  2 ，即m2 2k2 2.···················6分 k2 1 设M(x ,y )，N(x ,y )，则 y kx m，y kx m. 1 1 2 2 1 1 2 2  y kxm,  由x2 y2   1,  6 3 化简，得(2k2 1)x2 4kmx2m2 60，············································7分  4km x x  ,   1 2 2k2 1 由韦达定理，得 ····················································8分  2m2 6 x x  ,  1 2 2k2 1 所以 y y (kx m)(kx m) 1 2 1 2 k2x x km(x x )m2 1 2 1 2 2m2 6 4km k2 km m2 2k2 1 2k2 1 高三数学试题 第14页（共24页）m2 6k2  ，·····································································9分 2k2 1 2m2 6 m2 6k2 3(m2 2k2 2) 所以x x  y y     0，·····················10分 1 2 1 2 2k2 1 2k2 1 2k2 1 故OM ON ，即以MN 为直径的圆过原点O.···································· 11分 当直线l的斜率不存在时，l的方程为 x 2 或 x 2 . 这时M( 2, 2)，N( 2, 2)或M( 2, 2)，N( 2, 2). 显然，以MN 为直径的圆也过原点O. 综上，以MN 为直径的圆恒过原点O.················································12分 解法二：（1）同解法一.·············································································· 4分 （2）设直线l与圆O相切于点Q(x ,y )，M(x ,y )，N(x ,y ).···············5分 0 0 1 1 2 2 y 当x y 0时，则k  0 . 0 0 OQ x 0 x 因为直线l与圆O相切，所以l OQ，所以k  0 .···························· 6分 l y 0 x 则直线l的方程为y y  0 (xx )， 0 y 0 0 因为x 2  y 2 2，故l的方程可化为x x y y2.································ 7分 0 0 0 0 x x y y 2,  0 0 由x2 y2   1,  6 3 化简，得(2x 2  y 2)y2 4y y46x 2 0，(2x 2  y 2)x2 8x x86y 2 0. 0 0 0 0 0 0 0 0 ···································································································8分 46x 2 86y 2 所以 y y  0 ，x x  0 .············································· 9分 1 2 2x 2  y 2 1 2 2x 2  y 2 0 0 0 0 86y 2 46x 2 126y 26x 2 所以x x  y y  0  0  0 0  0，·················10分 1 2 1 2 2x 2  y 2 2x 2  y 2 2x 2  y 2 0 0 0 0 0 0 故OM ON ，即以MN 为直径的圆过原点O.···································· 11分 当x y 0时，则Q( 2,0)或Q(0, 2)， 0 0 这时M( 2, 2)，N( 2, 2)或M( 2, 2)，N( 2, 2). 显然，以MN 为直径的圆也过原点O. 综上，以MN 为直径的圆恒过原点O.················································12分 高三数学试题 第15页（共24页）