2022-2023青岛高三期初测试-参考答案_2.2025数学总复习_数学高考模拟题_2023年模拟题_新高考_山东省青岛市23届高三上学期期初调研检测数学含答案

2022年 (cid:3044) 三 年 (cid:2484) 期 初 (cid:2738) 研 检 测 数学参(cid:2544)答案及(cid:2715)分标准 一、单(cid:20143)(cid:17983)择(cid:20174)：本(cid:20174)共8小(cid:20174)，每小(cid:20174)5分，共40分。 B C A D C B C D 二、多(cid:20143)(cid:17983)择(cid:20174)：本(cid:20174)共4小(cid:20174)，每小(cid:20174)5分，共20分。 9(cid:8008)ACD 10(cid:8008)AC 11(cid:8008)BCD 12(cid:8008)AD 三、填(cid:12464)(cid:20174)：本(cid:20174)共4个小(cid:20174)，每小(cid:20174)5分，共20分。 15 1 5 85 13(cid:8008) (cid:8021) 14(cid:8008)300(cid:8021) 15(cid:8008)(cid:16) (cid:31)a(cid:31)0(cid:8021) 16(cid:8008)[ ， ](cid:8008) 2 e2 2 7 四、(cid:16409)(cid:12682)(cid:20174)：本(cid:20174)共6小(cid:20174)，共70分。(cid:16409)(cid:12682)应写出文字(cid:16938)明，(cid:16887)明(cid:17917)(cid:12353)或演(cid:12749)步(cid:20698)。 17. (cid:8002)10分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)因为a(cid:152)cosB+b(cid:152)cosA=2c(cid:152)cosC 所以由正弦定理得(cid:8020)sinAcosB+sinBcosA=2sinCcosC ······························· 1分 即sin(A+B)=2sinCcosC ·········································································· 2分 因为A+B=π(cid:16)C(cid:8006)所以sinC =2sinCcosC 1 因为0(cid:31)C (cid:31)π(cid:8006)所以cosC = 2 π 所以 C = ································································································ 5分 3 π 2π (cid:8002)2(cid:8003)由(cid:8002)1(cid:8003)知 C = (cid:8006)A= (cid:16)B 3 3 π 2π π 因为(cid:39)ABC为(cid:7144)(cid:6316)三(cid:6316)形(cid:8006)所以0(cid:31) B(cid:31) 且0(cid:31) (cid:16)B(cid:31) 2 3 2 π π 所以 (cid:31) B(cid:31) ··························································································· 7分 6 2 2π 3 1 sin( (cid:16)B) cosB+ sinB a sinA 3 2 2 3 1 由正弦定理得(cid:8020) = = = = + ·········· 9分 b sinB sinB sinB 2tanB 2 π π 3 因为 (cid:31) B(cid:31) (cid:8006)所以tanB(cid:33) 6 2 3 a 1 所以 (cid:143)( ,2). ························································································ 10分 b 2 数学(cid:12682)案 (cid:12642)1(cid:20139)（共6(cid:20139)）18(cid:8008)(cid:8002)12分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)(cid:6371)明(cid:8020)(cid:6787)接AB(cid:8006) 1 由(cid:7483)BB (cid:65)平(cid:7414)ABC(cid:8006)所以BB (cid:65) BC 1 1 因为BC (cid:65) AB(cid:8006)AB BB = B 1 所以BC (cid:65)平(cid:7414)AABB ··························· 2分 1 1 因为AB (cid:141)平(cid:7414)AABB 1 1 1 所以BC (cid:65) AB ······································· 3分 1 因为AA = AB=2(cid:8006)所以四(cid:6767)形AABB是正方形(cid:8006) 1 1 1 所以AB (cid:65) AB(cid:8006) ··································· 4分 1 1 又因为BC AB= B(cid:8006) 1 所以AB (cid:65) 平(cid:7414)ABC ···························· 5分 1 1 因为AC (cid:141)平(cid:7414)ABC 1 1 所以AC (cid:65) AB ······································ 6分 1 1 (cid:8002)2(cid:8003)由(cid:8002)1(cid:8003)得BC (cid:65)平(cid:7414)AABB 1 1 所以点C到平(cid:7414)AAB的(cid:6623)(cid:4912)为BC ································································ 7分 1 1 1 2 2 2 所以V =V = (cid:117) AB (cid:152)AA (cid:152)BC = BC = B 1 (cid:16)A 1 AC C(cid:16)AA 1 B 1 3 2 1 1 1 3 3 (cid:6321)得BC = 2 ···························································································· 8分 因为BA,BC,BB 两两垂直(cid:8006)以 1 数学(cid:12682)案 (cid:12642)2(cid:20139)（共6(cid:20139)） B 为原点(cid:8006)分别以BC,BA,BB 所在直(cid:5223)为 1 x (cid:6720)(cid:8006) y (cid:6720)(cid:8006) z (cid:6720) 建(cid:5002)如图所示(cid:4971)(cid:7256)直(cid:6316)坐标(cid:5177)(cid:8006)则A(0,2,0),A(0,2,2),B (0,0,2),C( 2,0,0) ········ 9分 1 1 (cid:6368)平(cid:7414)ABC的法向(cid:6998)为m=(x ,y ,z ) 1 1 1 1 因为AC =( 2,(cid:16)2，0)(cid:8006)AB =(0，(cid:16)2，2) 1 则m(cid:152)AC = 2x (cid:16)2y =0(cid:8006)m(cid:152)AB =(cid:16)2y +2z =0 1 1 1 1 1 令x = 2(cid:8006)则m=( 2,1,1) ······································································· 10分 1 (cid:6368)平(cid:7414)ABC的法向(cid:6998)为n=(x ,y ,z ) 1 1 2 2 2 因为AC =( 2,(cid:16)2，(cid:16)2)(cid:8006)AB =(0，(cid:16)2，0) 1 1 1 则n(cid:152)AC = 2x (cid:16)2y (cid:16)2z =0(cid:8006)n(cid:152)AB =(cid:16)2y =0 1 2 2 2 1 1 2 令x = 2(cid:8006)则n=( 2,0,1) ······································································· 11分 2 (cid:6368)二(cid:7414)(cid:6316)A (cid:16)BC(cid:16) A的平(cid:7414)(cid:6316)为(cid:84) 1 1 m(cid:152)n 3 3 所以cos(cid:84)= = = |m|(cid:152)|n| 2 3 2 π 所以二(cid:7414)(cid:6316)A (cid:16)BC(cid:16) A的大小为 ····························································· 12分 1 1 6 B z A 1 C A B 1 C 1 x y19(cid:8008)(cid:8002)12分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)由不(cid:5048)式x2 (cid:16)4nx+3n2 (cid:100)0可得(cid:8020) 数学(cid:12682)案 (cid:12642)3(cid:20139)（共6(cid:20139)） n (cid:100) x (cid:100) 3 n (cid:8006) (cid:63)a =2n+1······························································································ 2分 n 1 1 3 T = (cid:117)3n+1(cid:16) n(cid:16) n 4 2 4 当n=1时(cid:8006)b =1(cid:8006) ···················································································· 3分 1 1 1 当n(cid:116)2时(cid:8006)b =T (cid:16)T = (cid:117)3n (cid:16) (cid:8006) n n n(cid:16)1 2 2 因为b =1(cid:6807)合上式(cid:8006) 1 1 1 (cid:63)b = (cid:117)3n (cid:16) ························································································ 6分 n 2 2 3 (cid:8002)2(cid:8003)由(cid:8002)1(cid:8003)可得(cid:8020)c =3n (cid:16)1+((cid:16)1)n(cid:16)1(cid:79)( )n (cid:8006) n 2 3 (cid:63)c =3n+1(cid:16)1+((cid:16)1)n(cid:79)( )n+1 n+1 2 5 3 c (cid:31)c (cid:8006)(cid:63)c (cid:16)c =2(cid:117)3n + ((cid:16)1)n(cid:79)( )n (cid:33)0 n n+1 n+1 n 2 2 4 (cid:63)((cid:16)1)n(cid:79)(cid:33)(cid:16) (cid:117)2n ····················································································· 8分 5 4 当n为奇数时(cid:8006)(cid:79)(cid:31) (cid:117)2n 5 4 由于 (cid:117)2n (cid:7346)着 5 n 的增大(cid:5415)增大(cid:8006)当 n = 1 4 8 时(cid:8006) (cid:117)2n 的最小值为 (cid:8006) 5 5 8 (cid:63)(cid:79)(cid:31) ·································································································· 10分 5 4 当n为偶数时(cid:8006)(cid:79)(cid:33)(cid:16) (cid:117)2n 5 4 由于(cid:16) (cid:117)2n (cid:7346)着 5 n 4 16 的增大(cid:5415)减小(cid:8006)当n=2时(cid:8006)(cid:16) (cid:117)2n 的最大值为(cid:16) (cid:8006) 5 5 16 (cid:63)(cid:79)(cid:33)(cid:16) 5 16 8 (cid:5281)上可知(cid:8020)(cid:16) (cid:31)(cid:79)(cid:31) ············································································ 12分 5 5 20(cid:8008)(cid:8002)12分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)(cid:7380)假(cid:6368)为H ：性别因(cid:5179)与学生体(cid:5489)(cid:7185)炼的(cid:5239)常性无关(cid:5452) 0 根据列(cid:5452)(cid:6220)中的数据(cid:8006)(cid:5239)(cid:6341)(cid:5079)得到200(cid:117)(40(cid:117)80(cid:16)20(cid:117)60)2 (cid:70)2 = (cid:124)9.524(cid:33)7.879= x ····································· 3分 60(cid:117)140(cid:117)100(cid:117)100 0.005 根据小概率值(cid:68)=0.005的独(cid:5002)性检(cid:7598)(cid:8006)我们推断 数学(cid:12682)案 (cid:12642)4(cid:20139)（共6(cid:20139)） H 0 不成(cid:5002)(cid:8006)即(cid:6344)为性别因(cid:5179)与学生体(cid:5489)(cid:7185) 炼的(cid:5239)常性有关(cid:5452)(cid:8006)此推断犯(cid:7153)(cid:6415)的概率不大于 0 . 0 0 5 . ······································· 4分 (cid:8002)2(cid:8003)用A(cid:6220)示事件“(cid:6812)到(cid:5239)常参加体(cid:5489)(cid:7185)炼的学生”(cid:8006)B(cid:6220)示事件“(cid:6812)到男生”(cid:8006)则 n(AB) 80 4 P(B| A)= = = . ······································································· 7分 n(A) 140 7 (cid:8002)3(cid:8003)由(cid:7483)知Χ 的所有可(cid:5535)取值为0，1，2(cid:8006) 1 3 1 1 2 1 1 2 1 3 1 3 2 1 1 11 P(Χ =0)= (cid:117) (cid:117) = (cid:8021)P(Χ =1)= (cid:117) (cid:117) + (cid:117) (cid:117) + (cid:117) (cid:117) + (cid:117) = (cid:8021) 3 4 3 12 3 2 3 3 2 4 3 4 3 3 4 18 2 1 2 2 1 1 11 P(Χ =2)= (cid:117) (cid:117) + (cid:117) (cid:117) = (cid:8021) 3 2 3 3 2 4 36 所以Χ 的分布列为 Χ 0 1 2 1 11 11 P 12 18 36 ····································································· 10分 1 11 11 11 E(Χ)=0(cid:117) +1(cid:117) +2(cid:117) = . ···························································· 12分 12 18 36 9 21(cid:8008)(cid:8002)12分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)(cid:6368)动圆P的半径为R(cid:8006)圆心P的坐标为(x,y) 7 由(cid:7483)意可知(cid:8020)圆C 的圆心为C ((cid:16)1,0)(cid:8006)半径为 (cid:8021)圆 1 1 2 C 2 1 的圆心为C (1,0)(cid:8006)半径为 (cid:8008) 2 2 动圆P与圆C 内切(cid:8006)且与圆C 外切(cid:8006) 1 2 (cid:173) 7 |PC |= (cid:16)R (cid:176) (cid:176) 1 2 (cid:63)(cid:174) (cid:159)|PC |+|PC |=4(cid:33)|CC |=2 ··········································· 2分 1 1 2 1 2 (cid:176) |PC |= +R (cid:176)(cid:175) 2 2 x2 y2 (cid:63)动圆P的圆心的(cid:6709)(cid:6803)E是以C ,C 为焦点的椭圆(cid:8006)(cid:6368)其方(cid:4944)为(cid:8020) + =1(a(cid:33)b(cid:33)0)(cid:8006) 1 2 a2 b2 其中2a=4,2c=2(cid:8006)(cid:63)a=2,b2 =3 x2 y2 从(cid:5415)(cid:6709)(cid:6803)E的方(cid:4944)为(cid:8020) + =1 ································································· 4分 4 3 (cid:8002)2(cid:8003)(cid:8002)і(cid:8003)(cid:6368)直(cid:5223)AB的方(cid:4944)为y =k(x(cid:16)1)(k (cid:122)0)(cid:8006)A(x ,y ),B(x ,y ),则M(x ,(cid:16)y ) 1 1 2 2 1 1(cid:173)y =k(x(cid:16)1) (cid:176) 由(cid:174)x2 y2 可得(cid:8020) (4k2 +3)x2 (cid:16)8k2x+4k2 (cid:16)12=0 + =1 (cid:176) (cid:175) 4 3 8k2 4k2 (cid:16)12 (cid:63)x +x = (cid:8006) x x = ····························································· 5分 1 2 4k2 +3 1 2 4k2 +3 y + y 直(cid:5223)BM 的方(cid:4944)为y+ y = 2 1 (x(cid:16)x )(cid:8006) ···················································· 6分 1 x (cid:16)x 1 2 1 令y =0可得N 点的横坐标为(cid:8020) x (cid:16)x k(x (cid:16)x )(x (cid:16)1) 2x x (cid:16)(x +x ) x = 2 1 y +x = 2 1 1 +x = 1 2 1 2 N y + y 1 1 k(x +x (cid:16)2) 1 x +x (cid:16)2 2 1 1 2 1 2 4k2 (cid:16)12 8k2 2(cid:117) (cid:16) 4k2 +3 4k2 +3 = =4 8k2 (cid:16)2 4k2 +3 (cid:63)N为一个定点(cid:8006)其坐标为(4,0) ··································································· 8分 (cid:8002)іі(cid:8003)根据(cid:8002)і(cid:8003)可(cid:6784)一步求得(cid:8020) | AB|= 1+k2 |x (cid:16)x |= 1+k2 (cid:117) (x +x )2 (cid:16)4x x 2 1 2 1 1 2 8k2 4k2 (cid:16)12 12(k2 +1) = 1+k2 (cid:117) ( )2 (cid:16)4(cid:117) = ·········································· 9分 4k2 +3 4k2 +3 4k2 +3 1 AB(cid:65) DG(cid:8006)(cid:63)k =(cid:16) (cid:8006) DG k 12(k2 +1) 则|DG|= ·················································································· 10分 3k2 +4 AB(cid:65) DG(cid:8006) 1 1 12(k2 +1) 12(k2 +1) 72(k2 +1)2 (cid:63) 四(cid:6767)形ADBG(cid:7414)(cid:4934)S = | AB|(cid:117)|DG|= (cid:117) (cid:117) = 2 2 4k2 +3 3k2 +4 (4k2 +3)(3k2 +4) 72(k2 +1)2 72(k2 +1)2 288 (cid:8002)法一(cid:8003)S = (cid:116) = (4k2 +3)(3k2 +4) 4k2 +3+3k2 +4 49 ( )2 2 288 (cid:5048)号当且仅当4k2 +3=3k2 +4时取(cid:8006)即k =(cid:114)1时(cid:8006)S = ························ 12分 min 49 (cid:8002)法二(cid:8003)令k2 +1=t(cid:8006)(cid:39)k (cid:122)0(cid:8006)(cid:63)t (cid:33)1(cid:8006) 72t2 72 72 则S = = = 12t2 +t(cid:16)1 1 1 1 1 49 (cid:16) + +12 (cid:16)( (cid:16) )2 + t2 t t 2 4 1 1 288 当 = (cid:8006)即k =(cid:114)1时(cid:8006)S = ····························································· 12分 t 2 min 49 数学(cid:12682)案 (cid:12642)5(cid:20139)（共6(cid:20139)）22(cid:8008)(cid:8002)12分(cid:8003) (cid:6321)(cid:8020)(cid:8002)1(cid:8003)当x(cid:143)(0,+(cid:102))时(cid:8006)由(cid:7483)知(cid:8020) f (cid:99)(x) =lnx ··········································· 1分 当0(cid:31) x(cid:31)1时(cid:8006) f (cid:99)(x) (cid:31) 0(cid:8006) f(x)在(0,1)上单(cid:6435)(cid:6819)减 当x (cid:33)1时(cid:8006) f (cid:99)(x) (cid:33) 0(cid:8006) f(x)在(1,+(cid:102))上单(cid:6435)(cid:6819)增 ·········································· 2分 所以当x(cid:143)(0,+(cid:102))(cid:8006) f(x)(cid:116) f(1) = (cid:16)1(cid:8006)又因为 数学(cid:12682)案 (cid:12642)6(cid:20139)（共6(cid:20139)） f ( 0 ) = 0 所以 f(x)最小值为 f(1)=(cid:16)1 ········································································ 3分 (cid:8002)2(cid:8003)因为 f(e) =0, f(0) =0(cid:8006)由(cid:8002)1(cid:8003)知(cid:8020)当x(cid:143)[0,e]时(cid:8006)(cid:16)1(cid:100) f(x)(cid:100)0 ············ 4分 因为 f (cid:99)(e) =1(cid:8006)所以 f(x)在点(e,0)处的切(cid:5223)方(cid:4944)为y = x(cid:16)e ···························· 5分 令g(x) = xlnx(cid:16)2x+e(0(cid:31) x (cid:100)e)(cid:8006)则g(cid:99)(x) =lnx(cid:16)1(cid:100)0 所以g(x)在(0,e]上单(cid:6435)(cid:6819)减(cid:8006)g(x)(cid:116) g(e) =0 所以 f(x)(cid:116) x(cid:16)e ························································································ 6分 所以曲(cid:5223)y = f(x)(0(cid:100) x (cid:100)e)在 x (cid:6720)、 y (cid:6720)、 y = (cid:16) 1 和y = x(cid:16)e之(cid:7256) ··················· 7分 (cid:6368)原点为O(cid:8006)y(cid:6720)与y = (cid:16)1交点为A(cid:8006)y = (cid:16)1和 y = x (cid:16) e 的交点为B(e(cid:16)1,(cid:16)1)(cid:8006) 点(e,0)为C(cid:8006) 所以曲(cid:5223)y = f(x)(0(cid:100) x (cid:100)e)在梯形OABC内(cid:6915) 1 1 所以S (cid:31) S =(e(cid:16)1)(cid:117)1+ (cid:117)1(cid:117)1=e(cid:16) ··················································· 8分 OABC 2 2 m m +n n m m +n n (cid:8002)3(cid:8003)因为(x+ )ex (cid:100) xn+1lnx(cid:8006)所以 (x+ )ex (cid:100) xn+1lnx n x n x 所以(n+ m )e m x +n (cid:100) nxn lnx = xn lnxn =lnxnelnxn ············································ 9分 x m ①当n+ (cid:31)0时(cid:8006) x 因为x(cid:116)1(cid:8006)所以m(cid:31)(cid:16)nx(cid:100)(cid:16)n(cid:8006)所以m+n(cid:31)0 ·············································· 10分 m ②当n+ (cid:116)0时(cid:8006) x 令g(x)= xex (cid:8006)x(cid:143)[0,+(cid:102)) 则g(cid:99)(x)=(x+1)ex (cid:33)0在x(cid:143)[0,+(cid:102))时恒成(cid:5002) 所以g(x)= xex 在x(cid:143)[0,+(cid:102))时单(cid:6435)(cid:6819)增 由(cid:7483)知(cid:8020)lnxn (cid:116)0 m 所以n+ (cid:100)lnxn ··················································································· 11分 x m 所以 (cid:100) xlnx(cid:16)x n m 由(cid:8002)1(cid:8003)知(cid:8020) (cid:100) (cid:16)1 n 所以m+n(cid:100)0 ·························································································· 12分