2024届四川省雅安市高三下学期第三次诊断性考试数学（理）答案

雅安市高2021级第三次诊断性考试数学（理科）参考答案一、选择题（本大题共12小题，每小题5分，共60分）AACDAABCDDBB二、填空题(本大题共4小题，每小题5分，共20分)413.1514.-115.16.1429三、解答题(本大题共6小题，共70分)（）17.1a12，Snan12时，，，当n1a1a22a24,a22a1当时，n2Sn1an2,两式相减得an12an(n2)，，，数列是以2为首项，2为公比的等比数列，.··································································6分n（2）由（1）可知bn1log2an1n，记cnanbn1n2，∴123n，Tn223242(1n)2234n12Tn223242(1n)2，两式相减得2n1212T422232n(1n)2n14(1n)2n1n2n1.n12∴n1.····························································12分Tnn218．（1）补全的列联表如下：年龄段Ⅰ年龄段Ⅱ合计使用频率高150110260使用频率低250290540合计400400800···································································3分所以，数学（理科）答案1/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}所以有99%的把握认为跑腿服务的使用频率高低与年龄有关.······················6分（2）由数表知，利用分层抽样的方法抽取的8人中，年龄在，内的人数分别为5，3，依题意，的所有可能取值分别为为-2,0,2，所以，，，所以的分布列为：-202P所以的数学期望为.·················12分19.（1）证明：因为，为线段的中点，所以，·························································1分在等腰梯形中，作于，则由得，所以，所以，因为，所以所以∽，所以，所以，所以，······································3分因为，，所以平面，···················································4分因为在平面内，所以，··································5分因为，在平面内，所以平面.·········6分（2）解：因为，，所以，，数学（理科）答案2/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}取的中点，连接，则，因为平面，所以，又所以平面，··································7分所以如图，以为原点，以所在的直线为轴，以所在的直线为轴，建立空间直角坐标系，·························································8分则，C(0,0,0)，令平面PCD法向量31CDmxy0y3x223，33zxCPmxz0322取，m3，3，1由（1）知平面，则平面的法向量，mn313设二面角APCD所成平面角为，则cos，mn13313所以二面角APCD的余弦值为.································12分1320.（1）由已知得，.··········································1分kMFy2c2c设，则203，．····················3分k2cy2cMF10所以椭圆的方程为．·············································4分（2）①当直线的斜率为0时，的方程：，不妨设，，，数学（理科）答案3/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}，，，所以；·························································5分②当直线的斜率不为0时，如图，设的方程：，，．由，得．则，.·······································7分,······························10分又，所以．···············································11分综上，．所以，，成等差数列.····························12分x21.（1）当a=1时，f(x)excosx,x0,，2f(x)excosxxsinxexcosxxsinx，x0x∵x0,，∴ee1，∴ecosx0，∵xsinx0，f(x)0，2∴在上单调递增，∴2，f(x)0,f(x)minf(0)1,f(x)maxf()e22∴f(x)的值域为1,e2.··················································4分（2）法一：令h(x)f(x)g(x)，x①当a0时，h(x)esinxaxcosx1sinxaxcosx0在x0,上恒2成立.····································································6分②当0a1时，数学（理科）答案4/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}h(x)excosxaxsinxacosxexaxsinx(1a)cosx(1a)cosx0，∴h(x)在0,上单调递增，∴h(x)h(0)0成立.·························8分2③当a2时，m(x)h(x)excosxaxsinxacosx，m(x)ex（a1）sinxasinxxcosx0，∴m(x)在x0,上单调递增，即h(x)在x0,上单调递增，22∴h(0)2a0,h()e2a0,22∴存在x0,使得当x0,x时h(x)0,故h(x)在x0,x上单调递减，02000则不合题意·············································分h(x0)h(0)0,.10④当1a2时，令m(x)h(x)excosxaxsinxacosx，则m(x)ex（a1）sinxasinxxcosx0，∴m(x)在x0,上单调递增，即h(x)在x0,上单调递增，22∴h(x)h(0)2a0，即h(x)在0,上单调递增，h(x)h(0)0成立.2综上，a的取值范围是，2.············································12分法二：令，，令011a0得a2.①当时，，令，，在上单调递增，恒成立.································10分数学（理科）答案5/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}②当时，，，，使x00,这与x0恒成立矛盾.综上，.······························································12分22.（1）曲线是以为圆心的半圆，所以半圆的极坐标方程为，曲线以为圆心的圆，转换为极坐标方程为.故半圆，圆的极坐标方程分别为：，5分（2）由（1）得：.点到直线的距离.所以，其中，当时，的面积的最大值为4.···········10分23.（1）当时，不等式为，当时，可以化为，解得；当时，可以化为，得；当时，可以化为，解得，不等式不成立；综上,可得不等式的解集为·································5分（2）当时，当时等号成立，由可得（舍）或，故,由柯西不等式可得，即得当且仅当时，即时取等号．·····························10分数学（理科）答案6/6{#{QQABaYSAogCgQoAAABgCQQlwCgAQkAAACIoOxBAAMAAACAFABCA=}#}