陕西省高三教学质量检测卷（三）文科数学试题及参考答案

这是一份陕西省高三教学质量检测卷（三）文科数学试题及参考答案，文件包含陕西省高三教学质量检测卷三数文答案pdf、陕西省高三教学质量检测卷三文数docx等2份试卷配套教学资源，其中试卷共12页， 欢迎下载使用。
 数学（文科）二模参考答案一、选择题答案题号123456789101112答案BDCCACCCDDAB 二、填空题答案 13．14．​​​​​​​15．16．2三、解答题17. 解:(1 )由题知周期,即,则. ··············································2分           又,,,,   .·····························································6分        (2)由(1)知.,,则. ..............................................................8分          由及余弦定理得,,,          即,.          故面积的最大值为. ··········································12分18. 解: (1) , ,...........................................................4分        ,,回归方程为.·······················································8分       (2)当时,,则,故预测数据符合回归模型. ································12分 19.(1)证明:如图,取的中点,连结,.       为的中点,为的中点, 且,...........................2分       四边形为平行四边形,即.       又分别为的中点, ,平面平面.又在平面内, 平面. ...............................................6分      法2:如图,取的中点,连结,,          分别为的中点, , ...................................2分          又为的中点,底面为平行四边形,          ,则,即四边形为平行四边形.  ..............................................4分          ,而平面,平面.  ...............................................6分               (2)如图,作,,,,则.又平面平面,且,平面,即四棱锥高为,       .····································································12分20. 解: (1)由抛物线的定义可知,曲线的方程为. .............................4分      (2)设点,,,由题设直线的方程为.         联立方程得, .........................................6分则.         由得,即,则切线的方程为,即为,同理切线的方程为.把点代入切线方程得,解得,则,即 ,..................................................9分点到直线的距离, 线段, ,故当时,面积有最小值. ........................................12分21.解: (1),当时,由知,即在上单调递减; ..................................3分当时,令,则,令,则.综上可知,当时,在上单调递减; 当时,在上单调递增,在上单调递减.........................................6分     (2)由得,即在上有实数解.         设,则,当时, , ................................................................9分则在上单调递增. ,,由方程在上有实数解, 知,即.................................12分22.解: (1)由得,则有.       即曲线的普通方程为 ..........................................5分   (2)曲线方程可化为,其极坐标方程为.     又由得,,则，亦即有...........................................................7分     ,即,,，满足.又,则,即.  故直线的斜率为..............................................10分法2：设直线的直线方程为,其参数方程为(为参数,且)，代入得，.................................................7分,即，由根与系数的关系知,即与同正,即,又,    ，满足，又,,即. 故直线的斜率为. .......................................10分23.解: (1)当时,.令,       当时,不等式为,得;当时,不等式为,不等式无解;当时,不等式为,得.综上可知,函数的定义域为. .............................5分     (2)由恒成立,得恒立,又由,得,即.解得或.故实数的取值范围是. ........................................ 10分