【文档说明】湖北省武汉市常青联合体2023-2024学年高二下学期期中考试数学试卷答案.pdf，共(4)页，2.010 MB，由管理员店铺上传

转载请保留链接：https://www.doc5u.com/view-e0b55397872d426753806186839a0915.html

以下为本文档部分文字说明：

12024学年度第二学期期中考试数学试卷参考答案一、单选题1.B2.D3.C4.C5.B6.C7.D8.B二、选择题9.ACD10.ABD11.ABD三、填空题12.913.14.四、解答题15.(1)由,得,·································

·······································································································2分于是,整理得,解得,··

··············································································································5分所以.······························

··································································································································

······································································6分(2)原方程变形为,即,显然,·····················

···9分因此,化简整理,得,··········································································································

·································································12分而,解得,所以.···············································

···········································································································13分16.(1)二项式展开式的通项公式为(

).························2分因为第项和第项的系数比为,所以,化简得,解得,·······························································

·················································································4分所以().令,得,·········

····················································································································································

················6分{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}2所以常数项为第项.···································

·················································································································································

··········7分(2)设展开式中系数最大的项是第项,则,·····························································11分解得.···············································

·················································································································································

·······················13分因为,所以或,所以展开式中系数最大的项是第项和第项.·············15分17.(1)由,得,令,则,解得;················································

············································································································

···········1分当时,,所以,所以,··········3分所以当时,,有,········································6分又满足上式,所以,得,所以数列是

首项为0，公差为1的等差数列.···························································································

····························8分(2)由(1)知,所以,················································································

····················9分所以,故,两式相减,得{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}3,所以.······························

····································································································································

······················································15分18.(1)由函数,可得,····································

····················································································2分令,即,解得;令,即,解得,所以函数在区间单调递减,单调递增,························

······················································4分当时,取得极小值,极小值为,无极大值.·····························································

············6分(2)由不等式恒成立,即恒成立,即对于任意,不等式恒成立,································································

····················································8分设,可得,令,即,解得;令,即,解得,所以在上单调递减;在单调递增,························································

·······················12分所以,当时,函数取得极小值,同时也时最小值,,·············································14分所以,即,所以实数的取值范围为.·······························

··························17分19.(1)函数的定义域为,,当时,,所以在上单调递增;当时,由得,所以在上单调递增;由得,所以在上单调递减;故时,所以在上单调递增;············

············································································································2分当时,在上单调递增,

在上单调递减;·············································4分(2)由f(x)的图象恒在x轴上方，可得f(x)=x−mlnx>0{#{QQABBQQAggg

oQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}4因为x>0且m>0,不等式两边同时除以mx，可得1���>������������···········6分设h(x)=������������可得h’(x)=1−�����������

�2令h’(x)>0，解得0<x<e令h’(x)<0，解得x>e所以h(x)在(0,e)上单调递增，在(e,+∞)上单调递减所以当x=e时h(x)取得最大值，h(e)=1���···················

···························································································8分所以1���>

h(x)max即1���>1���所以m的范围是(0,e)······························································9分(3)证明:,,由(1)可知,当时,在上是增函

数,故不存在不相等的实数,使得,所以.由得,即,···············11分不妨设,则,则,要证,只需证,即证,只需证,·······························································

·····························································14分令,则只需证,即证,令,则,所以在上是增函数,所以,从而,故.·············································

·····························································································································

···········17分{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}