高三数学参考答案_2025年1月_250120湖南益阳市2024-2025学年高三上学期期末考试_数学

益阳市 2024 年下学期普通高中期末质量检测 高三数学（参考答案） 一、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． 题号 1 2 3 4 5 6 7 8 答案 C C B A A B D B 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，部分选对的得部分分，有选错的得0分． 题号 9 10 11 答案 BCD AC BD 三、填空题：本题共3小题，每小题5分，共15分． 5 1 1 12． 3i 13． 14．[- , ] 4 e4 e4 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．(本小题满分13分) x2 y2 2 1 解：(Ⅰ)点P( 2,1)代入方程C:  1(a 0,b 0)得  1① a2 b2 a2 b2 且2b2c②，a2 b2 c2③···································································3分 由①②③可解得：a2 4,b2 2， x2 y2 所以椭圆C:  1··········································································5分 4 2 2 (Ⅱ)设直线l的方程： y  xm，点A(x ,y )、B(x ,y ). 2 1 1 2 2 x2 y2 直线方程代入椭圆C:  1得2x2 2 2mx2m2 40， 4 2 由 8m2 8(2m2 4)32(4m2)0，得2m2 x x  2m，x x m22·································································8分 1 2 1 2 则弦长 AB  1k2 (x x )2 4x x  3(4m2) ， 1 2 1 2 2 点P( 2,1)到直线l： y  xm的距离 2 2  21m 2 m 2 d    m2 ······················································· 10分 2 3 3 1( )2 2 2 1 1 2 1 ABP面积S  AB d  3(4m2) m2  m2(4m2)  2········12分 2 2 3 2 当且仅当m2 4m2即m 2 时等号成立， 2 故ABP面积取得最大值 2 时直线l的方程为y  x 2························13分 2 高三数学参考答案 第 1 页 共 5 页 {#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}16．(本小题满分15分) b2cosA 2cosC 【详解】(Ⅰ)法一：由  可得bsinC 2cosAsinC2sinAcosC , sinA sinC bsinC2sin(AC)2sinB····································································3分 由正弦定理可得：bc2b·········································································5分 所以c2·······························································································7分 b2cosA 2cosC b2cosA 2cosC 法二：由  可得  ·····································2分 sinA sinC a c b2 c2 a2 a2 b2 c2 b bc ab 由余弦定理得：  ···········································4分 a c b2 c2 a2 a2 b2 c2 化简得：bc  ，即bc 2b，所以c2··················7分 b b (Ⅱ)法一：tanC 2tanB，则sinCcosB2sinBcosC， 所以3sinCcosB2sinBcosC2sinCcosB2sinBC2sinA·····················9分 由正弦定理可得：3ccosB2a，又因为c2，所以3cosBa························ 11分 1 3 3 所以VABC面积为：S  acsinB3cosBsinB sin2B ···························13分 2 2 2 π 3 当sin2B1即B 时取等. 所以VABC面积的最大值为 ·····························15分 4 2 1 2sin Asin B 2sin Asin B 法二：S  absinC   2 sinC sin(AB) 2sin AsinB 2tan AtanB   ················································ 9分 sin AcosBcosAsinB tan AtanB tanBtanC 3tanB 又tan Atan(BC)  ································.12分 tanBtanC1 2tan2B1 3tan2 B 2 2tan2 B1 6tanB 6tanB 3  则S     ,当B  时取等··········15分 3tanB 2tan2 B2 4tanB 2 4 tanB 2tan2 B1 法三：tanC 2tanB，易知B,C都为锐角 如图，作边BC上的高AD， 则有BD2 AD2  AB2  4, tanC 2tanB AD 2AD   ,即BD2CD·····································································11分 CD BD 1 1 3 3 3 BD2AD2 3 S  BCAD  BDAD BDAD   ABC 2 2 2 4 4 2 2  当且仅当BD  AD ,即B  时取等·························································· 15分 4 法四：tanC 2tanB，则sinCcosB2sinBcosC，ccosB 2bcosC , a2 c2 b2 a2 b2 c2 余弦定理可得  ，即a23b2 3c2 12······················ 10分 2a a 由余弦定理可得：a2 b2c22bccosAb244bcosA, 高三数学参考答案 第 2 页 共 5 页 {#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}则有12a23b2 4b244bcosA, b22 化简可得b2bcosA2，即cosA ····················································12分 b 1 b2 2 所以S  bcsinAbsinAb 1( )2  b45b24 2 b 5 3 当b2  时，S  ············································································15分 2 max 2 法五：同法四可得a23b2 3c2 12··························································· 10分 如图过点C作底边AB的高CD， 不妨设AD 1d,BD 1d,CD h, 则有b2 (1d)2 h2,a2 (1d)2 h2, 则a23b2 3(1d)23h2(1d)2h2 12， 1 9 整理可得h2 (d  )2  ,······································································13分 2 4 9 3 1 所以h2  ，即h ，当且仅当d  时取等， 4 2 2 1 3 则有S  ch ·················································································15分 2 2 17．(本小题满分15分) 解：(Ⅰ)设ACBD O ，连接PO，过点A作PO的垂线，垂足即为点A在平面PBD内 的射影H. ·····························································································2分 下面证明: AB  AD,CD  BC ，点A,C在线段BD的中垂线上，即有AC BD PA平面ABCD,BD 平面ABCD ，PA BD 又PAAC=A，PA,AC 平面PAC ，BD平面PAC ·························4分 又BD平面PBD，平面PBD 平面PAC 又平面PBD平面PAC  AO，AH  PO，AH 平面PAC AH  平面PBD ，故点H为点A在平面PBD内的射影·······························6分 (Ⅱ)由(Ⅰ)可知，可建立如图空间直角坐标系O-xyz， AB AD2,，又BAD120， BCD60，PB PD且PB PD，易知OA1，OB  3，PB  2OB  6, 3 在RtPAB中，PA PB2AB2  2 ，在RtPAO中，PO 3，OH  3 1 2 则O(0,0,0),A(0,1,0),B( 3,0,0), H(0, , )······································9分 3 3  设平面HAB的法向量为n x,y,z ，   2 2 AB ( 3,1,0)，AH (0, , ) 3 3    3x y 0 H  nAB 0  则  ， 2 2 nAH 0  y z  0 3 3  不妨取x1,则n(1, 3, 6)  平面DAB的法向量m 0,0,1  ································································12分 高三数学参考答案 第 3 页 共 5 页 {#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}    mn 6 二面角H ABD的平面角为，|cos||cos m,n|    ··········14分 m n 10 10 又(0,)，sin 1cos2 5 10 二面角H ABD的正弦值为 ···························································15分 5 18．(本小题满分17分) 解：(Ⅰ)记事件A为抽到一件合格品，事件B为抽到另一件为不合格品， C1 ×C1 160 P(AB)= 80 20 = ········································································2分 C2 495 100 C2 -C2 476 P(A)= 100 20 = ········································································4分 C2 495 100 P(AB) 160 40 P(B| A)= = = ······························································5分 P(A) 476 119 1 (Ⅱ)(i)由题：若 X ~B(100, ) ，则EX50,DX25·······································7分 2 1 又 PX kCk ( )100 PX 100k, 100 2 所以 P0 X 25 1 P(0 X 25 或 75 X 100) 1 P  X 50 25 ··········· .9分 2 2 由切比雪夫不等式可知， P  X 50 25  25  1 252 25 1 所以 P0 X 25 ·········································································11分 50 (ii)设随机抽取100件产品中合格品的件数为X ，假设厂家关于产品合格率为95%的说法 成立，则X :B100,0.95，所以EX95,DX4.75·····························13分 由切比雪夫不等式知， PX 80P  X 95 15  950.05 0.021·············15分 152 即在假设下100个元件中合格品为80个的概率不超过0.021，此概率极小，由小概率 原理可知，一般来说在一次试验中是不会发生的，据此我们有理由推断工厂的合格率 不可信.·······························································································17分 19．(本小题满分17分) 1 解：(Ⅰ) f(x)的定义域为(1,)， f(x) ··············································2分 x1  f(0)1， f(0)0，  f(x)在原点处的切线方程为： y01(x0)，即 y  x·························4分 (Ⅱ)将x看成变量，n看成常数， 1 1 1 1 x 设g(x) ( x)ln(x1)   ln(x1) ，x≥0， x1 2n 2n x1 x1 高三数学参考答案 第 4 页 共 5 页 {#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}1 1 1 x 所以g(x) 2n  1  2n ·····················································6分 (x1)2 x1 (x1)2 1 1 所以x[0, )时，g(x)＜0，x( ,)时，g(x)＞0， 2n 2n 1 1 所以g(x)在[0, )上单调递减，在( ,)上单调递增·······························8分 2n 2n 1 1 所以g(x)≥g( ) ln( 1) 2n 2n 1 1 1 1 1 又 f( )=ln( 1) ， f( )≤ ( x)ln(x1) ····························10分 2n 2n 2n x1 2n  1 1 1 x f( )≤ ( x) ln(x1)  21 x1 21 x1  (Ⅲ)由(2)可知：  1 1 1 x f( )≤ ( x) ln(x1)  2n x1 2n x1 对n累加，有 1 1 1 1 1 1 nx f( ) f( )＜ (   ) nln(x1) ···················12分 21 2n x1 21 22 2n x1 1 1 (1 ) 1 2 2n nx     nln(x1) x1 1 x1 1 2 1 1 nx 1nx  (1 )  nln(x1)＜  nln(x1) ································15分 x1 2n x1 x1 1 1 1n(e1) n1 令xe1，则 f( ) f( )＜ n  ···························17分 21 2n e e 高三数学参考答案 第 5 页 共 5 页 {#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}