高三数学答案及评分标准_2024-2025高三（6-6月题库）_2024年12月试卷_1223山东省中学联盟百校大联考2024-2025学年高三上学期12月学情诊断（全科）

绝密 ★ 启用前
a2 a2  2(n1)1 2n1,
n n1
a2 a2  2(n2)1 2n3,
n1 n2
2025 届高三上学期学情诊断

a2 a2  2215,
数学答案及评分标准
3 2
2024.12
a2 a2  2113,
2 1
(n1)(32n1)
将以上各式相加，得a2 a2 35(2n3)(2n1) n2 1 ，
n 1 2
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，只有一个选项是正确
的．请把正确的选项填涂在答题卡相应的位置上． 将a 1代入上式即得a2 n2，且当n1时也成立，所以a2 n2，
1 n n
二、
又因为数列  a  为正项数列，所以a n (nN)．·····································································6分
n n
1．A 2．C 3．B 4．B 5．D 6．C 7．A 8．C
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
（2）由（1）可得b
n
(1)nn3n，令c
n
(1)nn，其前2n项和为T
2n
，
部选对得6分，部分选对的得部分分，选对但不全的得部分分，有选错的得0分．
则T 1234(2n1)2nn，··············································································9分
2n
9．AC 10．BCD 11．ACD
又因为3132 32n 
3(132n)

3(32n 1)

32n13
，························································12分
13 2 2
三、填空题：本题共3小题，每小题5分，共15分． 32n13
所以S n ．··························································································· 13分
2n 2
1 π 16．【解析】正弦定理+最值
12． 13． 14．4 3，[2, 6]
2 9 bcosC 3bsinC sinBcosC  3sinBsinC
（1）根据正弦定理， 1可化为 1，1分
ac sin AsinC
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．
sinBcosC  3sinBsinC sin AsinC
15．（1）【法一】构造常数列 sinBcosC  3sinBsinC sin(BC)sinC
sinBcosC  3sinBsinC sinBcosC sinCcosBsinC
由a2 a2  2n1 =(n+1)2 n2 (nN)，a 1，可得a2 (n+1)2 a2 n2 a2 12 0 ，
n1 n 1 n1 n 1
sinC( 3sinBcosB1)0.
 
故数列 a
n
2 n2 是恒为0的常数列，所以a
n
2 n2，······································································· 5分 ·············································································································································5分
π 1
又因为数列  a  为正项数列，所以a n (nN).······································································· 6分 因为C(0, π)，所以sinC 0，故有 3sinBcosB10，进而有sin(B ) ，因为B(0, π)，
n n 6 2
π π 5π π π π
【法二】累加法 所以B ( , )，故有B  ，所以B  ．································································7分
6 6 6 6 6 3
由题意得：n2且nN，有
 2π π
0 A C 

π  3 2 π π
（2）因为B  ，所以 ，进而有 C  ．·················································8分
3  π 6 2
0C 

 2
数学试卷答案 第1页（共9页）
{#{QQABZQYAogAoAAIAARhCEwFSCkGQkgCACagOBEAEoAABiRNABAA=}#}1 1
a b 1 cos|cosn ,n | 
由正弦定理可得sin A

3

sinC ，
1 2 tan2tan21
tan2tan2(
π
)1
2
2
1 1 3
  
所以有 a  sin A  sin( 2 3 π C)  2 3 cosC 1 2 sinC  1  3cosC ， tan2 tan 1 2 1 2 tan2 tan 1 2 1 3
sinC sinC sinC 2 2sinC
1 π
当且仅当tan2 ，即 时等号成立，························································· 14分
1 3 3 1 3cosC 3 1 3 tan2 4
所以S  acsinB  a  (  ) (  )，·······································11分
ABC
2 4 4 2 2sinC 4 2 2tanC
3
即平面AEFG 与平面ABCD所成角的余弦值的最大值为 ．·········································15分
π π 3 3 3 1 3 3 3
因为 C  ，所以tanC  ，所以  S  (  ) ，··························14分
6 2 3 8 ABC 4 2 2tanC 2 18．【解析】（1）显然 y  f(x)的图象经过(2,0)，当x0时， y 2，所以 f(x)的图象经过
3 3
的所有定点的坐标为(2,0)和(0,2)················································································1分
所以ABC面积的取值范围是( , )．·················································································15分
8 2 由题知 f(x)ex ax(x2)(ex a)(x1)  ex 2a  ，····························································2分
17．【解析】直线与平面所成的角，平面与平面所成的角，线面平行的判定定理和性质定理，
若以(2,0)为切点， f(2)e2 2a，切线为y (e2 2a)(x2)；
最值问题
若以(0,2)为切点， f(0)2a1，切线为y (2a1)x2；············································ 4分
（1）由BD//l ，BD平面AEFG ，l 平面AEFG可得，BD//平面AEFG ，····················2分
（注：上述两条切线写出一条即可）
再由BD平面BB D D，平面BB D D平面AEFG  EG，所以BD//EG，·······················4分
1 1 1 1
又因为BE//DG，所以四边形BDGE 为平行四边形，所以BE  DG．·······························5分 （2）①当a0时，ex 2a 0恒成立，
在正四棱柱ABCD A
1
B
1
C
1
D
1
中，BB
1
,DD
1
均垂直于平面ABCD，所以直线AE、AG与平面 所以当x1时, f(x)0, f(x)在(,1)单调递减，
ABCD所成的角分别为EAB,GAD，即EAB,GAD，···································· 6分
当x1时， f(x)0， f(x)在(1,)单调递增； ··················································5分
又因为BE  DG，AB  AD，所以tan tan，从而；········································7分
   ②当a0时，由 f(x)0，得x 1或x ln  2a ． ············································6分
（2）以A为坐标原点，分别以AB,AD,AA 的方向为x轴、y轴、z轴的正方向建立空间直角坐 1 2
1
e
标系． 当ln(2a)1，即a  时， f(x)0恒成立，则 f(x)在R上单调递增，········································ 7分
2
则E(1,0,tan)，G(0,1,tan)，所以AE (1,0,tan)，AG (0,1,tan)，····························9分 当ln(2a)1时，即a  e 时,当x1时， f(x)0, f(x)在(,1)单调递增;
2
  AEn 1  xztan0 当1 xln(2a)时， f(x)0， f(x)在  1,ln(2a)  单调递减；
设平面AEFG的法向量n (x,y,z)，则有 ，令z 1，则有
1
 AGn
1
 yztan0
当xln(2a)时， f(x)0， f(x)在  ln(2a), 单调递增； ··················································9分
x tan,y  tan，所以n (tan,tan,1)，································································· 11分
e
1 当ln(2a)1时，即0a 时，
2
平面ABCD的法向量n (0,0,1)，平面AEFG与平面ABCD所成角为，则
2 当xln(2a)时， f(x)0， f(x)在 ，ln(2a)  单调递增；
数学试卷答案 第2页（共9页）
{#{QQABZQYAogAoAAIAARhCEwFSCkGQkgCACagOBEAEoAABiRNABAA=}#}1 1
当ln(2a) x1时， f(x)0， f(x)在(ln(2a),1)单调递减； 令g(x) x(lnxx1) ，g(x) lnx2x，
ex ex
当x1时， f(x)0， f(x)在  1, 单调递增； ·················································11分 易知g(x)在(0,)上单调递增，x时，g(x)；x0时，g(x)，所以存在
1
综上所述：当a0时， f(x)在(,1)单调递减，在(1,)单调递增； t(0,)，使得g(t) lnt2t 0，即et lnt2t 即et (t)lntt ，
et
e
当0a 时， f(x)在 ，ln(2a)  单调递增，在(ln(2a),1)单调递减，在  1, 单调递增； 1
2
令G(x)ex x ，则G(t)G(lnt)，因为G(x)在R 上单调递增，所以lnt t，即et  ．
t
e
当a  时， f(x)在R上单调递增；
2
当x(0,t)时，g(x)0，g(x)单调递减；当x(t,)时，g(x)0，g(x)单调递增．
e
当a  时， f(x)在(,1)单调递增，在  1,ln(2a)  单调递减，在  ln(2a), 单调递增．···············12分 1
2 所以g(x) g(t) tlntt2t 0 ，所以 f(x2)(lnxx1)e2 0 ．·························· 17分
et
（3）【证法一】当a 0时， f(x2)(lnxx1)e2 0 ex(xlnxx2x)10．
   
19．【解析】（１）【法一】由a 2a 1可得a 12 a 1 ，所以数列 a 1 为公比是
令g(x)ex(xlnxx2x)1，则g(x)ex(xlnxx2xlnx12x1)ex(x1)(lnxx) n n1 n n1 n
1 1 1 2的等比数列，所以a 12n a 1  2n，即得a 2n 1，所以数列的母函数为
令h(x)lnxx,h(x) 1，则h(x)在(0,)上单调递增.又因为h(1)10，h( )1 0， n 0 n
x e e

1 1 f  x  2n 1  xn ．
所以存在t( ,1)，使得h(t)lntt 0，即lnt t即et  .当x(0,t)时，h(x)0，g(x)0，
e t
n0
·························································································································· 4分
g(x)单调递减；当x(t,)时，h(x)0，g(x)0，g(x)单调递增.所以
g(x) g(t)et(tlntt2t)10 ．
1 
【法二】在 Ck1 xn 1Ck1xCk1x2 Ck1 xn 中
所以ex(xlnxx2 x)10也即 f(x2)(lnxx1)e2 0 ．·········································17分  1x k n0 nk-1 k k1 nk-1
【证法二】当a 0时， f(x2)(lnxx1)e2 0  1 lnx x10． 令k 1得 1   C0xn   xn 1xx2 xn ，所以
xex 1x n
n0 n0
1 (x1) 1 (x1)(xex1) 1
令g(x) lnxx1，g(x)  1 ， 12x22x2 2nxn ，
xex x2ex x x2ex 12x
令h(x) xex 1，h(x)(x1)ex 0 ，则h(x)在(0,)上单调递增．又因为h(0)10， f  x a a xa x2a xn①
0 1 2 n
1
h(1)e10，所以存在t(0,1)，使得h(t)tet 10，即et  即lnt t.当x(0,t)时， 2xf  x 2a x2a x22a x32a xn1②
t 0 1 2 n
①②得
h(x)0，g(x)0，g(x)单调递减；当x(t,)时，h(x)0，g(x)0，g(x)单调递增．所
 12x  f  x a  a 2a  x  a 2a  x2  a 2a  x3  a 2a  x n
1 0 1 0 2 1 3 2 n n1
以g(x) g(t) lntt1lntt 0
tet
1
0xx2x3xn 1 ，
1
1x
所以 lnxx10也即 f(x2)(lnxx1)e2 0 ．··············································17分
xex 1 1 1 2 1 1 1
f  x       
12x  1x  12x  12x 12x 1x 12x 1x
1
【证法三】当a 0时， f(x2)(lnxx1)e2 0   x(lnx x1)0．
ex
数学试卷答案 第3页（共9页）
{#{QQABZQYAogAoAAIAARhCEwFSCkGQkgCACagOBEAEoAABiRNABAA=}#}(20 21x22x2 2nxn ) (1xx2 xn )
(20 1)(211)x(22 1)x2 (2n 1)xn 


2n 1

xn ．································································································4分
n0
（２）由题意得G  x S S xS x2S xn ①
0 1 2 n
那么xG  x S xS x2S x3S xn1 ②
0 1 2 n
①②得
 1x  G  x  S  S S  x S S  x2  S S  x3  S S  x n
0 1 0 2 1 3 2 n n1
a a xa x2a xn  g  x ，
0 1 2 n
 
g x
所以G  x  ．·······························································································10分
1x
1 
（3）由公式 Ck1 xn 1Ck1xCk1x2 Ck1 xn 可得
 1x k nk-1 k k1 nk-1
n0
1 
令k 3得 C2 xn
 1 x 3 n2
n0
2   
所以 2C2 (2x)n (n2)(n1)2nxn a xn
 12x 3 n2 n
n0 n0 n0
2
数列{a }的母函数为 f

x

，
n  12x 3
由（２）结论知数列{S }的母函数为
n
 
f x 2 2 4 4 4
G  x      
1x  1x  12x 3 1x 12x (12x)2 (12x)3
   
2C0xn 4C02nxn 4C1 2nxn 4C2 2nxn
n n n1 n2
n0 n0 n0 n0
 
 242n4C1 2n4C2 2n  xn [2n1  n2n2  2]xn ，
n1 n2
n0 n0
所以S 2n1  n2 n2  2．·················································································· 17分
n
数学试卷答案 第4页（共9页）
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