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

这是一份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，共6页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
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、选择题题号123456789101112答案BAAC DABA CC DD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 欢迎访问“高中试卷网”——http://sj.fjjy.org