2023年11月诊断性测试数学答案(1)_2023年11月_0211月合集_2024届THUSSAT中学生标准学术能力诊断性测试高三上学期11月月考_THUSSAT中学生标准学术能力诊断性测试2024届高三上学期11月月考数学

中学生标准学术能力诊断性测试 2023年 11月测试 数学参考答案 一、单项选择题：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 2 3 4 5 6 7 8 C A B B A B D D 二、多项选择题：本题共 4小题，每小题 5分，共 20 分．在每小题给出的四个选项中，有多 项符合题目要求．全部选对的得5分，部分选对但不全的得2分，有错选的得0分． 9 10 11 12 BD AC AB ACD 三、填空题：本题共4小题，每小题5分，共20分． 7 14． 13．1 2 10+ 2 15． 16．35 4 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） 2csinB sinC 2sinCsinB sinC （1） = ，所以由正弦定理得 = ， b cosB sinB cosB 1  cosB = ，得B = ············································································ 3分 2 3 3 + 3 tan A+tanB tanC =−tan(A+B)=− =− 4 =−5 3 ················· 5分 1−tan AtanB 3 1−  3 4 （2） ABC内切圆的面积为，所以内切圆半径r=1， 由圆的切线性质得c+a−b=2 3,b=c+a−2 3=3+ 3 ····························· 7分 由余弦定理得b2 =c2+a2−ac，  ( 3+ 3 )2 =c2 +a2 −ac=(a+c)2 −3ac，将a+c=3+3 3代入， 第1页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}1  ac=8+4 3,S = acsin =2 3+3 ·············································· 10分 ABC 2 3 1 （或a+b+c=6+4 3,S = r(a+b+c)=2 3+3 ······················ 10分） ABC 2 18．（12分） （1）过D作AB的垂线交AB于H点，设AC= AB=a， 2 ( ) 则BC = 2a,BD=2a,HD= HB = BD = 2a,AH = BH −BA= 2−1 a， 2 AD= AH2 +DH2 = 5−2 2a， 由题意得，二面角C−SA−D的平面角为CAD， ········································· 2分 ( ) 34 5+2 2 DH 2 cosCAD= = = ············································ 4分 DA 5−2 2 17 （2）分别以 AB，AC，AS 为 x 轴，y 轴，z 轴建立直角坐标系，A(0,0,0),B(a,0,0) ， C(0,a,0),S(0,0,h) ，则E ( a,0,(1−)h ) ,F ( 0,a,(1−)h ) ， SE SF SB=SC且 = =,SE = SF ····················································· 6分 SB SC 又 SAB SAC,BSA=CSA，那么SAE SAF ， 则AE = AF ····························································································· 8分 故SEF 与AEF 都是等腰三角形，取EF的中点G，则SG与AG均垂直于EF， 1 1  1 1  1 1  G  a, a,(1−)h  ,SG =  a, a,−h  ,AG =  a, a,(1−)h  ， 2 2  2 2  2 2  平面AEF ⊥平面SBC等价于SG ⊥ AG ················· 10分 a22 a22 SGAG = + −h(1−h)h=0， 4 4 2 又 SCB =60,a =h,= ························ 12分 3 19．（12分） 1 （1） 2a =a +2,2(a −2)=a −2，即a −2= (a −2) ， n+1 n n+1 n n+1 2 n 第2页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}1 a −2 是公比为 的等比数列 ······························································· 2分 n 2 n−1 n−1 1 1 a −2=3 ,a =2+3 ····················································· 4分     n 2 n 2  1 1 2 1 n−1  1 n （2） S =a +a + +a =2n+31+ +   + +    =2n+61−    n 1 2 n  2 2 2   2  ············································································································· 6分 n n 1 1 1 S −2n−6=−6 , S −2n−6 =6  ，     n 2 n 2 2023 即2n−1 6069,n取最小值14 ·································································· 8分 2 n−1 C 2(n+1) （3） C =n   , n+1 = 1，得n2， n 3 C 3n n 4 即C C ，有C C C ，又C =,C =C = ， n+1 n 3 4 5 1 2 3 3 故 C  中最大项为C ,C ······································································· 10分 n 2 3 7 4 7 又 b  中最小值为2,(b ) −(C )  ，即2 −  ， m m min n max 3 3 3 7 (3−7)(+1)0，又0, ················································· 12分 3 20．（12分） （1）由题意可知：X的所有可能取值为2.3，0.8，0.5， 1 3 4 P(X =2.3)=   =0.3 ······································································ 1分 2 4 5 X =0.8包含的可能为“高低高”“低高高”“低低高”， 1 1 4 1 3 4 1 1 4 P(X =0.8)=   +   +   =0.5 ·········································· 2分 2 4 5 2 4 5 2 4 5 P(X =0.5)=1−0.3−0.5=0.2 ··································································· 3分 X的分布列为： X 2.3 0.8 0.5 P 0.3 0.5 0.2 第3页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}·············································································································· 4分 数学期望E(X)=1.19 ················································································ 5分 （2）设升级后一件产品的利润为Y， Y的所有可能取值为2.3−a,0.8−a,0.5−a ····················································· 6分 1  3 4 6b+3 P(Y =2.3−a)=  +b    = ························································· 7分 2  4 5 10 1  3 4 1  1 4 1  1 4 5−6b P(Y =0.8−a)=  −b    +  +b    +  −b    = ············· 8分 2  4 5 2  4 5 2  4 5 10 6b+3 5−6b 1 P(Y =0.5−a)=1− − = ························································ 9分 10 10 5 6b+3 5−6b 1 E(Y)=(2.3−a) +(0.8−a) +(0.5−a) =1.19+(0.9b−a) 10 10 5 ············································································································ 11分 E(Y) E(X)0.9b−a0， 1 10a 1 即： b ( a0,0.4) （备注： 不写出不扣分） ······························· 12分 2 9 2 21．（12分） （1） AB + AF + BF =4 2，即 AF + BF + AF + BF =4 2， 1 1 2 2 1 1 又 AF + AF = BF + BF =2a,4a=4 2,a= 2 ······························· 2分 1 2 1 2 c 2 又 e= = ,c=1,b=1， a 2 x2 故椭圆E的方程是 + y2 =1 ······································································ 4分 2 （2）依题意知直线BC的斜率存在，设直线BC:y=kx+m， x2 x2 代入 + y2 =1，得 +(kx+m)2 =1， 2 2 1  即 +k2  x2 +2kmx+m2 −1=0 ①， 2  第4页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}=(2km)2 −4   1 +k2   ( m2 −1 ) =−2m2 +2+4k2 0， 2  即2k2 −m2 +10 ②， 设B(x ,y ),C(x ,y ) ，则A(x ,−y ) ， 1 1 2 2 2 2 −2km m2 −1 x +x = x x = 1 2 1 ③， 1 2 1 ④ ····················································· 6分 +k2 +k2 2 2 A,F ,B三点共线，F (1,0) ，直线AB不与坐标轴垂直， 2 2 y −y  1 = 2 ,(x −1)(kx +m)=−(x −1)(kx +m) ， x −1 x −1 2 1 1 2 1 2 2kx x +m(x +x )−k(x +x )−2m=0， 1 2 1 2 1 2 2k ( m2 −1 ) 2km2 2k2m  − + −2m=0 1 1 1 ， +k2 +k2 +k2 2 2 2 2km2−2k−2km2+2k2m−m−2k2m=0， m=−2k,直线BC: y =k(x−2) ， 1 由②得：2k2 −4k2 +10,k2  ， 2 3 k BC = 1+k2 x −x ，点F (−1,0) 到直线BC:kx−y−2k =0的距离d = ， 1 2 1 1+k2 1 3 S = BC d = k x −x ································································· 8分 F 1 BC 2 2 1 2 2    4k2  4 ( 4k2 −1 ) x −x = (x +x )2 −4x x =   − 1 2 1 2 1 2 1 1  +k2  +k2  2  2 第5页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}16k4 −4 ( 4k2 −1 )1 +k2    2  2−4k2 = = ， 2 2 1  1  +k2 +k2     2  2  k2( 2−4k2) S =3 (k 0) ， F 1 BC ( 1+2k2)2 t−1 设2k2+1=t 1，则k2 = ， 2 (t−1)(2−t) −t2 +3t−2 1 2 1 S =3 =3 =3 −2 +3 −1   F 1 BC t2 t2 t t 2 1 3 1 =3 −2 − + ··································································· 10分   t 4 8 1 3 4 4 1 所以当 = 时，即t = ,2k2 +1= ,k2 = （符合题意）， t 4 3 3 6 3 2 6 3 2 S 的最大值为 ，所以当k = 时，FBC的面积取最大值为 FBC 1 1 4 6 4 ············································································································ 12分 22．（12分） （1）定义域为 (−,−1) (1,+) ， 由题意知 ( x2 −x ) ln ( x −1 ) =ax，则 (x−1)ln ( x −1 ) =a有三个不同的实数根， 当x 1时，令g(x)=(x−1)ln(x−1) ， g(x)=ln(x−1)+1,g(x) 在x(1,+) 上单调递增，  1 又g  1+  =0，  e  1  1  g(x) 在x  1,1+ 上单调递减，在x  1+ ,+ 上单调递增，  e  e   1 1 g(x) g  1+  =− ·············································································· 2分  e e 第6页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}当x−1时，令h(x)=(x−1)ln(−x−1) ， x−1 1 2 x+3 h(x)=ln(−x−1)+ ,h(x)= + = ， x+1 x+1 (x+1)2 (x+1)2 又h(−3)=0，h(x) 在 (−,−3) 上单调递减，在 (−3,−1) 上单调递增， h(x)h(−3)=2+ln20， h(x) 在 (−,−1) 上单调递增 ····································································· 5分 又h(−2)=0， 当x→−1−时，h(x)→+，当x→−时，h(x)→−， 当x→1+时，g(x)→0− ，当x→+时，g(x)→+， 1 − a0 ··························································································· 6分 e a （2）由已知可知 (x−1)ln ( x −1 ) = 有3个零点x,x ,x ， x 1 2 3 不妨设x x x ， 1 2 3 1 − a 1 显然−2 x −1，由（1）中函数g(x) 性质且  e − ， 1 1 1 e 1+ 1+ e e 1 1 x 1+  x 2， 2 e 3 2 2 只需证x +x 2+ ，即x 2+ −x ···················································· 8分 2 3 e 3 e 2   1   2   2  又 g  2  1+  −x  −g(x )=  1+ −x  ln  1+ −x  −(x −1)ln(x −1)0，   e 2  2  e 2   e 2  2 2 上面不等式证明如下：  2   2   1 令(x)=  1+ −x  ln  1+ −x  −(x−1)ln(x−1),x  1,1+ ，  e   e   e 第7页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#} 2   1 (x)=−ln  1+ −x  −ln(x−1)−2,x  1,1+   e   e  2   =ln  1+ −x  (x−1)  −20，  e    1 (x) 在x  1,1+ 上单调递增，  e  1 (x)  1+  =0 ··········································································· 10分  e 1 2 1 又−x −1− ,2+ −x 1+ ， 2 e e 2 e   1  g  2  1+  −x   g(x ) ，又显然有g(x ) g(x ) ，   e 2  2 2 3  1 2  1+  −x  x ，  e 2 3 2 x +x 2+ ， 2 3 e 2 x +x +x  ·················································································· 12分 1 2 3 e 第8页 共8页 {#{QQABLYQUggCgQBBAAQhCAwEQCgAQkBACAAoGxAAIMAIBgRFABAA=}#}