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=}#}