2025江苏省盐城市高三上学期11月期中考试数学试卷（含答案）_2024-2025高三（6-6月题库）_2024年11月试卷_1115江苏省盐城市2024-2025学年高三上学期11月期中考试（全科）

{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}盐城市 2025 届高三年级第一学期期中考试
数学参考答案
1.C 2.C 3.B 4.A 5.D 6.D
7.B 8.B 9.ACD 10.AB 11.ABD
   
12. 0,2 13. 12 14.8 tan 2n
15.解：⑴因为函数 f(x)为偶函数，所以xR，都有 f(x) f(x)，················2分
即  ex kex  sin  x    ex kex  sinx，
即xR，都有  k1  ex ex  sinx0，···············································4分
所以k 1.··························································································6分
⑵当k 0时， f(x)exsinx， f(x) f(x)exsinxexsin(x)(ex ex)sinx ，
所以(ex ex)sinx0.············································································9分
当x0时，显然不合题意；
当x  0,2  时，ex ex 0，则sinx0，此时，x  0,  ；
当x  2，0  时，ex ex 0，则sinx0，此时，x  ,0  ，·············12分
综上，不等式 f(x) f(x)0的解集为  ,0    0,  .·····························13分
 
16.解：⑴因为 f(x)sinxcosx 2sin(x )，所以 f(A) 2sin(A ) 2，
4 4

所以sin(A )1，············································································· 4分
4

 
又因为A 0, ，所以A .································································6分
4
1 2 1 2
⑵因为 ABBC  BABC   BC ，所以BABC  BC ，
2 2
1
法一：所以BA在BC方向上的投影向量为 BC ，
2
过点A作边BC的垂线，垂足为H ，则H 是边BC的中点，
所以AB  AC ，即bc，······································································9分
在ABC中，因为a2 b2 c2 2bccosA，即12b2  2b2，
第 1 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}2 2
所以b2  ，·············································································· 12分
2
1 2 21
所以S  bcsin A b2  .················································15分
ABC
2 4 4
1 a2 c2 b2 a2 c2 b2 1
法二：所以cacosB  即ca   ，
2 2ac 2 2
即a2 c2 b2 1即bc，···································································· 9分
在ABC中，因为a2 b2 c2 2bccosA，即12b2  2b2，
2 2
所以b2  ··················································································12分
2
1 2 21
所以S  bcsin A b2  .················································15分
ABC
2 4 4
17．解：⑴设AD  x，
1 1  9
则S   AB ACsinBAC  63sin  3，
BAC 2 2 3 2
1 1  3x
S   AB ADsinBAD  6xsin  ，
BAD 2 2 6 2
1 1  3x
S   AC ADsinCAD  3xsin  ，
CAD 2 2 6 4
由S  S S ············································································3分
BAC BAD CAD
9 3 3
得 3  x x，
2 2 4
得x 2 3 ，即AD 的长为2 3 .································································5分

⑵在VABC 中，由AB 6，AC 3，BAC  及余弦定理可得
3

BC2  AB2  AC2 2AB ACcosBAC 62 32 263cos 27，
3
得BC 3 3，·······················································································7分

又AB 6，AC 3，得AB2  AC2  BC2，得ACB  ，
2
1  
而CAD  BAC  ，∴ADC ACDCAD  ，
2 6 3
又VPCD为等腰三角形，∴VPCD为等边三角形，·······································9分
第 2 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}
在VACD 中，AD 2 3，AC 3，CAD  ，由余弦定理可得
6

CD2  AD2  AC2 2AD ACcosDAC (2 3)2 32 22 33cos 3，
6
得CD  3 ，······················································································ 11分
 
则PD  3，BD 2 3 ，设ABP ，则DBP  ，BPD  ，
6 6
PD BD
在△DBP 中，由正弦定理得  ，
sinDBP sinBPD
3 2 3

则   ，···································································13分
sin( ) sin( )
6 6
  1 3 1 3
得2sin( )sin( )，则2( cos sin) cos sin，
6 6 2 2 2 2
sin 3
得cos3 3sin，得tan  ，
cos 9
3
即tanABP的值为 ．·······································································15分
9
18.⑴证：由(a 1)2 4S ，得(a 1)2 4S (n2)，
n n n1 n1
所以(a 1)2 (a 1)2 4a ，即(a 1)2 (a 1)2，
n n1 n n n1
得(a a 2)(a a )0，··································································2分
n n1 n n1
因数列 a  各项为正，所以a a 0，于是a a 2(n2)，
n n n1 n n1
所以数列
a 
是公差为2的等差数列，·······················································3分
n
又(a 1)2 4S 4a ，所以a 1，
1 1 1 1
所以a 12(n1)2n1.····································································· 4分
n
⑵解：由⑴知，a 2n1，
n
1
所以T  (12n1)nnnn2，
n 2
2n2 1
由T 2n2 (n1)2，即nn2 2n2 (n1)2，所以 2，
n n
2n2 1
令c  2，················································································6分
n n
2n11 2n2 1 n(2n11)(n1)(2n2 1) 2n2(n1)1
所以c c     ，
n1 n n1 n n(n1) n(n1)
当n1时，c c ，即c c ；
n1 n 2 1
第 3 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}当n2时，c c ，即c c ；
n1 n 2 3
当n3时，c c ；············································································ 8分
n1 n
20 1
所以c  21，故的最大值为1.··········································9分
2 2
⑶解：由⑴得b b b 22n1，所以b b b 22n1，
n n1 n2 n1 n2 n3
于是b 4b ，又b b 1，bb b 21，所以b 22b ，·······················10分
n3 n 1 2 1 2 3 3 1
由P (b b Lb )(b b Lb )(b b Lb )
3k 1 4 3k2 2 5 3k1 3 6 3k
1(14k) 4
所以P 4(b b Lb )4[ ] (4k 1)，
3k 1 4 3k2 14 3
4 n
所以当n3k时，P  (43 1)；···························································12分
n 3
由P (b b Lb )(b b Lb )(b b Lb )b ，
3k1 4 7 3k1 2 5 3k1 3 6 3k 1
1(14k) 7 4
因b 4b 4，所以P 7(b b Lb )17 1 4k  ，
4 1 3k1 2 5 3k1 14 3 3
P 1也适合上式，
1
7 n1 4
所以当n3k 1时，P  4 3  ；·····················································14分
n 3 3
由P (b b Lb )(b b Lb )(b b Lb )b b ，
3k2 4 7 3k1 5 8 3k2 3 6 3k 1 2
1(14k) 10 4
又b 4b 4，所以P 10(b b Lb )210[ ]2 4k  ，
5 2 3k2 1 4 3k2 14 3 3
P 2也适合上式，
2
10 n2 4
所以当n3k 2时，P  4 3  ；···················································16分
n 3 3
4 n
 (43 1),n3k,kN*
3

7 n1 4
综上所述，P  4 3  ,n3k 1,kN .···········································17分
n 3 3
10 n2 4
 4 3  ,n3k 2,kN
 3 3
19.解：⑴ f(x)(x1)ex，··········································································1分
当x1时， f(x)0， f(x)在  1，  单调递增;
第 4 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}当x1时， f(x)0， f(x)在  ，1  单调递减;
1
所以[f(x)]  f(1) ，无极大值.····················································4分
极小值
e
注：未说明无极大值的扣1分.
f(lnx)
⑵由a3axln3 0，a 0得ax3axln3xlnx0，
x
得3axln3ax  xlnx，
则 f  ln3ax   f  lnx  ，且lnx0，即x1，············································6分
lnx
由⑴中 f(x)的单调性及ln3ax 0，lnx0知ln3ax lnx，变量分离得：a ，
xln3
lnx
设(x) ，x(1,)，································································8分
xln3
1lnx
则(x) ，x(1,)，
x2ln3
当x(1,e)时，(x)0，(x)在(1,e)上单调递增;
当x(e,)时，(x)0，(x)在(e,)上单调递减;
1 1
所以(x) (e) ，即 0a ，
max eln3 eln3
1
故实数a的取值范围为 (0, ]．··························································10分
eln3
⑶由 f(m) f(n)k(mn)2及mn0，设mnt 0,n x0，
则上式可化为 f(xt) f(x)kt2，
即对任意的x,t 0，(xt)ext xex kt2恒成立，
设h(x)(xt)ext xex kt2，x 0，·················································12分
则h(x)(xt1)ext (x1)ex，x 0，
则h(x)(x1)(ext ex)text 0，
则h(x)在(0,)上单调递增，所以h(x)h(0)tet kt2，
由题意知对任意的t 0都有tet kt2 0，··············································· 14分
et et
变量分离得k  ，t 0，设g(t) ，t 0，·····································15分
t t
第 5 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}ettet et t1 
则g(t)  ，t 0，
t2 t2
当t 1时， g(t)0，g(t)在  1，  上单调递增;
当0t 1时， g(t)0，g(t)在  0，1  上单调递减;
所以g(x)  g(1)e，
min
得k的取值范围为(,e]．····································································17分
第 6 页 共 6 页
{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}