2025—2026学年度第一学期期中学业水平诊断数学答案_251113山东省烟台市2025—2026学年度第一学期期中学业水平诊断（全科）

2025~2026 学年度第一学期期中学业水平诊断 高三数学参考答案 一、选择题： 1.C 2.B 3.D 4.C 5.A 6.D 7.B 8.B 二、选择题 9.BC 10.ABD 11.AD 三、填空题 π 12.1 13.(0, ) 14.5,21 6 四、解答题 π π 15.解：（1）由题知， f(x)=sin(x+ )+2cos(x+ ) ， 6 3 3 1 1 3 所以 f(x)= sinx+ cosx+2( cosx− sinx)， 2 2 2 2 3 3 即 f(x)= cosx− sinx， 2 2 π 所以 f(x)= 3cos(x+ ). ···························································· 2分 6 π 因为 f(x)图象关于点( ,0)对称， 3 π π π 所以 + = +kπ,kZ， ·························································· 4分 3 6 2 所以=1+3k， ············································································ 5分 又因为 02，所以=1. ························································ 6分 π （2）由（1）知， f(x)= 3cos(x+ ). 6 1 将函数 f(x)图象上各点横坐标缩短为原来的 倍（纵坐标不变）， 2 π 得到y= 3cos(2x+ )， ··································································· 8分 6 π π 再将得到的图象向左平移 个单位，故得到函数g(x)= 3cos(2x+ ). ······ 10分 12 3 π π π 4π 当x[0, ]时，2x+ [ , ]， 2 3 3 3 π π π π 4π 故当2x+ [ ,π]，即x[0, ]时，函数g(x)单调递减，当2x+ [π, ]，即 3 3 3 3 3 π π x[ , ]时，函数g(x)单调递增. ·················································· 11分 3 2 3 所以g(x)[− 3, ]. ································································· 13分 2 高三数学试题答案（第 1 页，共 7 页）16.解（1）当0 x8时，W =R(x)−(200+54x) =297x−ax3−200−54x， 整理得W =243x−ax3−200； ····························································· 2分 因为x=5时，年利润为890万元，所以2435−125a−200=890， ··········· 3分 所以a=1， ······················································································· 4分 所以W =243x−x3−200， ·································································· 5分 5400 当x8时，W =R(x)−(200+54x) =2539− −200−54x x 5400 =2339− −54x. ················································· 6分 x 243x−x3−200,0 x8  综上：W = 5400 . ··············································· 7分 2339− −54x,x8   x （2）当0 x8时，W =243x−x3−200，W=243−3x2. 所以当0 x8时，W0恒成立， 所以W 在 (0,8) 上单调递增， ······························································· 9分 所以当x=8时，W =1232. ························································ 10分 max 5400 当x8时，W =2339− −54x. x 5400 5400 5400 因为 +54x2 54x =1080，当且仅当 =54x， x x x 即x=10时取“=”. ········································································· 12分 所以当x=10时，W =1259. ··························································· 13分 max 因为12321259， 所以当年产量为10万件时，该企业生产该电子产品所获年利润最大. ············ 15分 17.解：（1）证明：因为a−b=2bcosC， 由正弦定理得：sinA−sinB=2sinBcosC， ·········································· 1分 因为A+B+C=π，所以sin(π−B−C)−sinB=2sinBcosC， ················ 2分 即sin(B+C)−sinB=sinBcosC+cosBsinC−sinB=2sinBcosC， 所以cosBsinC−sinBcosC=sinB， ··················································· 3分 即sin(C−B)=sinB， ········································································ 4分 因为B,C(0,π)，所以C−B(−π,π)， 所以C−B=B或C−B+B=π（舍）， ·················································· 6分 综上，C =2B. ··················································································· 7分 b c b c （2）法一：由正弦定理得， = ，即 = ， sinB sinC sinB sin2B 高三数学试题答案（第 2 页，共 7 页）c 所以cosB= . ················································································· 8分 2b 在 ABC中，由余弦定理得： a2 +c2 −b2 a2 +(c−b)(c+b) a2 +2a(c−b) cosB= = = ， ······················· 9分 2ac 2ac 2ac 1 (c+b)+2(c−b) 整理得： a+2(c−b) 2 5 3b c ， ··········· 11分 cosB= = = − = 2c 2c 4 4c 2b c 3 c 解得 = 或 =1， ········································································· 13分 b 2 b c 3 因为C =2B，所以cb，所以 = . ················································· 14分 b 2 c 3 综上，cosB= = . ······································································ 15分 2b 4 法二：因为b+c=2a，由正弦定理得， sinB+sinC=2sinA， ···················· 8分 由（1）知，C =2B， 所以sinB+sin2B=2sin(B+2B)， ······················································· 9分 即sinB+sin2B=2sinBcos2B+2cosBsin2B， 故sinB+2sinBcosB=2sinB(2cos2B−1)+4cos2BsinB， ·················· 11分 因为sinB0，所以1+2cosB=2(2cos2 B−1)+4cos2 B， 即8cos2 B−2cosB−3=0， ································································ 13分 3 1 解得cosB= 或cosB=− ， ····························································· 14分 4 2 3 因为C B，所以cosB= . ································································ 15分 4 18.解：（1）在数列{a }中，由S =n2 +n知， n n 当n2时，S =(n−1)2 +n−1， ························································· 1分 n−1 所以S −S =2n，即a =2n，且当n=1时，S =a =2， ······················ 2分 n n−1 n 1 1 所以a =2n. ······················································································· 3分 n 由bb =2n得，b b =2n+1， n n+1 n+1 n+2 b 所以 n+2 =2，故数列{b }的奇数项成以b =1，2为公比的等比数列，偶数项成以b =2， b n 1 2 n 2为公比的等比数列， ············································································ 4分 n−1 n−1 所以当n为奇数时， b =12 2 =2 2 ， ···················································· 5分 n n n 当n为偶数时， b =222 −1 =22 . ···························································· 6分 n 高三数学试题答案（第 3 页，共 7 页） n−1 2 2 ,n为奇数 所以b = . ······································································· 7分 n n  22,n为偶数  n−1 2n2 2 ,n为奇数  （2）由（1）知，c = n ， n  (3−n)22 ,n为偶数  2(n−1)2(n+1)  n+1 n2 2 ,n为奇数  即c = n . ······················································ 8分 n −2  (3−n)22 ,n为偶数  (n−1)(n+1) 令A =c +c + +c +c ， n 1 3 2n−3 2n−1 则A =121+322+ +(2n−3)2n−1+(2n−1)2n，① n 所以2A =122 +623+ +(2n−3)2n +(2n−1)2n+1，② ························ 9分 n 所以①-②得， −A =2+2(22 +22 + +2n)−(2n−1)2n+1 n 22(1−2n−1) =2+2 −(2n−1)2n+1 1−2 =−6−(2n−3)2n+1， ··································································· 11分 所以A =6+(2n−3)2n+1. ·································································· 12分 n n n n+2 −2 (3−n)22 1 22 2 2 又因为 = ( − ) ， ··············································· 14分 (n−1)(n+1) 4 n−1 n+1 令B =c +c + +c +c ， n 2 4 2n−2 2n 1 21 22 22 23 2n−1 2n 2n 2n+1 所以B = ( − + − + + − + − ) n 4 1 3 3 5 2n−3 2n−1 2n−1 2n+1 1 2n−1 = − . ·········································································· 16分 2 2n+1 1 2n−1 13 2n−1 所以T = A +B =6+(2n−3)2n+1+ − = +(2n−3)2n+1− ， 2n n n 2 2n+1 2 2n+1 13 (16n2 −16n−13)2n−1 所以T = + . ···················································· 17分 2n 2 2n+1 高三数学试题答案（第 4 页，共 7 页）1 mx2 +2x−1 19.解： f(x)的定义域为(0,+)， f(x)=mx+2− = ， ············ 1分 x x m （1）因为曲线 f(x)= x2 +2x−lnx在(1, f(1))处的切线与直线x+2y−1=0垂直， 2 所以 f(1)=m+2−1=2，所以m=1. ·············································· 3分 （2）令g(x)=mx2 +2x−1， 当m0时，g(x)=mx2 +2x−1的图象是开口向上的抛物线，且=4+4m0， −1− 1+m −1+ 1+m 设mx2 +2x−1=0的两根分别为x = ,x = ，则x  x ， 1 m 2 m 1 2 因为g(x)=mx2 +2x−1的图象过点(0,−1)，所以x 0 x ， 1 2 −1+ 1+m −1+ 1+m 所以当0 x 时， f(x)0，当x 时， f(x)0， m m −1+ 1+m −1+ 1+m 所以 f(x)的单调递减区间为(0, ) ，单调递增区间为( ,+). m m ············································· 5分 1 1 1 当m=0时，f(x)=2− ，所以当0 x 时，f(x)0，当x 时，f(x)0， x 2 2 1 1 所以 f(x)的单调递减区间为(0, )，单调递增区间为( ,+)， ···················· 6分 2 2 当m0时，g(x)=mx2 +2x−1的图象是开口向下的抛物线，且=4+4m， 若−1m0，=4+4m0， −1− 1+m −1+ 1+m 设mx2 +2x−1=0的两根分别为x = ,x = ，则x  x ， 1 m 2 m 1 2 因为g(x)=mx2 +2x−1的图象过点(0,−1)，所以0 x  x ， 2 1 −1+ 1+m −1− 1+m 所以当0 x 或x 时， f(x)0， m m −1+ 1+m −1− 1+m 当  x 时， f(x)0， m m −1+ 1+m −1− 1+m 所以 f(x)的单调递减区间为(0, )， ( ,+)，单调递增区间为 m m 高三数学试题答案（第 5 页，共 7 页）−1+ 1+m −1− 1+m ( , )， ································································· 8分 m m 若m−1，=4+4m0， f(x)0恒成立， f(x)的单调递减区间为(0,+)， ···························································· 9分 −1+ 1+m 综上，当 m0时， f(x) 的单调递减区间为 (0, ) ，单调递增区间为 m −1+ 1+m ( ,+)， m 1 1 当m=0时， f(x)的单调递减区间为(0, )，单调递增区间为( ,+)， 2 2 −1+ 1+m −1− 1+m 当−1m0时， f(x)的单调递减区间为(0, )， ( ,+)，单 m m −1+ 1+m −1− 1+m 调递增区间为( , )， m m 当m−1时， f(x)的单调递减区间为(0,+). ······································· 10分 （3）m=0时， f(x)=2x−lnx， 1 因为0a b且lna−lnb=2(a−b)， 2 所以2a−lna=2b−lnb，即 f(a)= f(b). ············································· 11分 1 1 由（2）知， f(x)的单调递减区间为(0, )，单调递增区间为( ,+)， 2 2 1 所以当a=b= 时，a+b=1. ····························································· 12分 2 1 当ab时，则0a b，下面证明a+b1，即b1−a. ·················· 13分 2 1 1 1 因为0a b，所以1−a( ,+),b( ,+)， 2 2 2 所以只需证 f(b) f(1−a)，又因为 f(a)= f(b)， 1 所以只需证 f(a) f(1−a)，0a ， ··············································· 14分 2 1 令F(x)= f(x)− f(1−x)，x(0, )， 2 1 1 1 所以F(x)= f(x)− f(1−x)=2− +2− =4− ， x 1−x x−x2 1 1 因为x(0, )，所以(x−x2)（0, ），即(x−x2)4 2 4 高三数学试题答案（第 6 页，共 7 页）1 1 所以F(x)0，所以F(x)在(0, )上单调递减，且F( )=0， ·················· 15分 2 2 所以F(x)0，即F(a)0，即 f(a) f(1−a)，即a+b1. ·················· 16分 综上，a+b1，因为a+b−t 0，所以t 1. ········································ 17分 高三数学试题答案（第 7 页，共 7 页）