2021四川省泸州市高三上学期文科数学第一次教学质量诊断性试题答案

2021届四川省泸州市高三上学期文科数学第一次教学质量诊断性试题答案

一、选择题

二、填空题：

13．3；   14．0；    15.；    16．.

三、解答题：

17．解：（Ⅰ）因为

  ····························································1分

，··························································2分

因为，所以，··················································3分

所以，·······················································4分

即，·························································5分

所以；·······················································6分

（Ⅱ）图象上所有点横坐标变为原来的倍得到函数的图象，

所以函数的解析式为，···········································8分

 因为，所以， ·················································9分

 所以，······················································11分

故在上的值域为．···············································12分

18．解：（Ⅰ）因为，·······················································2分

所以，·························································3分

又因为，·······················································4分

点处的切线方程为．

所以,··························································5分

；····························································6分

（Ⅱ）在上有且只有一个零点，··········································7分

因为，························································8分

当时，，·······················································9分

所以在上为单调递增函数且图象连续不断，·····························10分

因为，，······················································11分

所以在上有且只有一个零点．·······································12分

19．解：（Ⅰ）因为，

由正弦定理得，··················································2分

因为，所以，···················································3分

所以，························································4分

因为，所以，

所以，························································5分

所以，所以．···················································6分

（Ⅱ）解法一：设的边上的高为，的边上的高为，

因为，························································7分

所以，························································8分

所以，是角的内角平分线，所以，·····································9分

因为，可知，···················································10分

所以，························································11分

所以.·························································12分

解法二：设，则，···························································7分

因为，，

所以，································8分

所以，································9分

所以，，

因为，所以，···················································10分

，可知，······················································11分

所以，

所以.·························································12分

解法三：设，，则，

在中，由及余弦定理可得：，

所以，························································7分

因为，可知，···················································8分

在中，

即，··························································9分

在中，，······················································10分

即，·························································11分

所以．························································12分

20.解：（Ⅰ）第一步：在平面ABCD内作GH‖BC交CD于点H；··························2分

 第二步：在平面SCD内作HP‖SC交SD于P；·······························4分

第三步：连接GP，点P、GP即为所求．··································5分

（Ⅱ）因为是的中点，，

所以是的中点，而，

所以是的中点，···················································6分

所以，

连接，交于，连，设在底面的射影为，

因为，

所以，································7分

即为的外心，

所以与重合，···························8分

因为，，

所以，································9分

所以，·······························10分

因为//平面，···························11分

所以．·························································12分

21.解：（Ⅰ）当时，，·······················································1分

所以，·························································2分

因为,

由得，·························································3分

所以，或，

所以在上单减，上单增，············································4分

所以函数在上的最小值为；··········································5分

（Ⅱ）原不等式.······················································6分

因,,所以,

令,····························································7分

即，令，即，

所以在上递增；··················································8分

①当即时,

因为,所以，

当,,即，所以在上递增，

所以，

故，························································9分

②当即时，

因为,,即，

所以在上递减，所以，

故·························································10分

③当即时,

又在上递增，

所以存在唯一实数,使得,即，

则当时，即，当时即，

故在上减，上增，

所以.························································11分

所以，

设（），则，

所以在上递增,所以.

综上所述.·····················································12分

22.解： (Ⅰ) 解法一：设曲线与过极点且垂直于极轴的直线相交于异于极点的点E，且曲线上任意点F，边接OF，EF，则OF⊥EF，              2分

在△OEF中，，··················································4分

解法二：曲线的直角坐标方程为，·····································2分

即， 所以曲线的极坐标方程为；······································4分

（Ⅱ）因曲线的参数方程为与两坐标轴相交，

所以点，·······················································6分

所以线段极坐标方程为，···········································7分

，,

······························8分

，····························································9分

当时取得最大值为．··············································10分

23.解：(Ⅰ)由······························································2分

，

          解得或（舍去），······················································4分

当且仅当时取得“=，

          即的最小值为．························································5分

（Ⅱ）由，，··························································7分

因使不等式成立，

所以

即，····························································9分

即的取值范围是··················································10分

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