数学理科答案_2023年11月_01每日更新_02号_2024届四川省绵阳市高三上学期第一次诊断性考试_2024届四川省绵阳市高三上学期第一次诊断性考试理科数学

绵阳市高中2021级第一次诊断性考试 理科数学参考答案及评分意见 一、选择题：本大题共12小题，每小题5分，共60分． BCDAC ADBBD CC 二、填空题：本大题共4小题，每小题5分，共20分． 13．7 14． 15．9 16．1 三、解答题：本大题共6小题，共70分． 17．解：（1）由a，a，a 成等比数列，则 ，·································2分 1 2 4 ∴ ， 可解得 ，···················································································3分 ∴数列{a}的前 项和 ；································5分 n （2） ①，················································6分 当 时， ，可得 ,························································7分 可得 ②，······································································8分 由②式－①式，得 ，·············································9分 ∴ ·······································································11分 ．·························································································12分 8．解：（1）∵ ，则 ，·····················································1分 1 第 1 页 共 6 页，···························································2分 又 ，························································································4分 ∴ ；········································································5分 ∴ 由题意， ，···················································6分 （2） ··································································7分 ∵ ······································8分 ∴ Z，·······························································9分 ∴ ，·····················································10分 ∴ ∴ 的最小值为 ．··········································································12分 19．解：（1）∵ 为奇函数， ∴ ，解得：m=2．···························································5分 （2）当m>0时，2x2+m>0 ， ∴函数 不可能有两个零点．································6分 第 2 页 共 6 页当m <0时，由 ，解得： 或m2，································7分 要使得f(x)仅有两个零点，则 ，··········································8分 即 ，此方程无解． 故m=0，即 ，·······························································9分 令 ，则 ， ，解得： 或 ， 解得： ， 故 在 ， 上递增，在 上递减，···························10分 又 ， 故函数 仅有一个零点．·························································12分 20．解：（1）∵cos(CB)sinA=cos(CA)sinB (cosCcosB+sinCsinB)sinA=(cosCcosA+sinCsinA)sinB··································2分 ∴ ∴cosCcosBsinA=cosCcosAsinB································································3分 又∵△ABC为斜三角形，则cosC≠0， ∴cosBsinA=cosAsinB，·········································································5分 ∴sin(AB)=0，又A，B为△ABC的内角， ∴A=B；····························································································6分 由△ABC的面积S= ， （2） ∴S= absinC= ，则bsinC=1，即 =sinC，···········································7分 由S= acsinB= ，则csinB=1，即 =sinB，··········································8分 由（1）知A=B则a=b， 第 3 页 共 6 页∴ =sin2Bsin2C，·······································································9分 又sinC=sin(A+B)=sin2B， ∴ =sin2Bsin22B=sin2B4cos2Bsin2B=sin2B4(1sin2B)sin2B···················10分 令sin2B=t，令f(t)=t4(1t)t=4t23t， 2B<1，即0＜t<1， 又因为0＜sin ∴当t= 时，f(t)取最小值，且f(t) = ，············································11分 min 综上所述： 的最小值为 .·······················································12分 ．解：（1）当 时， ， 21 ，····································2分 令 得： ；令 得： 或 ，·······················3分 ∴ 的单调递减区间为： 和 ；单调递增区间为： ．·····5分 （2） 等价于 （*）·········6分 令 ，则 ， ∴ 在 上递减，在 上递增。 ∴ 的最小值为 ，即： ，··················································8分 （*）式化为： ，当t=1时，显然成立． 当 时， ，令 ，则 ，······················9分 ，当 时，易知 ， 第 4 页 共 6 页故易得： 在 上单调递增，在 上单调递减，·······················10分 ∴ ，······································································11分 ∴实数a的取值范围为： ．·······················································12分 22．解：（1）曲线C 的参数方程为C ： （t为参数）， 1 1 由 得C 的普通方程为： ；······································2分 1 曲线C 的参数方程为C ： （ 为参数）， 2 2 所以C 的普通方程为： ；············································4分 2 （2）曲线C 的极坐标方程为： ，···············5分 1 ∴ ，·················································································6分 由 得： ， ∴射线： 与曲线C 交于A ，·····································7分 1 曲线C 的极坐标方程为 ， 2 由 得： ， ∴射线： 与曲线C 交于B ，····································9分 2 则 = = ．······················10分 第 5 页 共 6 页23．解：（1） ···································1分 ∴ ，································2分 解得 ，······························································4分 ∴不等式的解集为 ；··················································5分 （2）证明：由 ，可得 的最小值为 ，················6分 则 ， ， ∴ ·································7分 ··························8分 ，当且仅当 时，等号成立，············9分 ∴ ．·······························································10分 第 6 页 共 6 页