河南省南阳市第一中学2017届高三上学期第二次周考数学（理）答案含答案

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1、www.ks5u.com周考2理数参考答案1．A2．B3．B4．B5．C6．D7．C8．C9．C10．D11．B12．D13．1：214．15．16．17．······················6分·························6分18（1）∵为的中点，，，∴，，∴四边形是平行四边形，∴，∵，∴．·················································3分又平面底面，∴平面···································5分∴平面平面。·············

2、··········································6分（2）∵，平面底面，平面底面，∴底面，以为原点，为轴，为轴，为轴，建立空间直角坐标系，则，，，，···············8分，，设平面的法向量，则，取，得，平面的法向量．························10分设二面角的平面角为，则，∴，∴二面角的大小为．………………··································12分19．（1），．·············································

3、··········4分，所以对任意正整数，都有成立．·····················8分（3），则，①，②①-②可得．故······12分20．解：（1）∵当时，，在上是增函数，当时，，在上也是增函数，∴当或时，总有在上是增函数，又，故函数的单调增区间为，单调减区间为．·······4分（2）∵存在，使得成立，即··················································6分由（1）的函数在上是减函数，在上是增函数，所以当时，的最小值，的最大值为和中的最大者．，令，∵，∴在上是增函数．·······

4、························8分而，故当时，，即；即，函数在上是增函数，解得；···········································10分当时，，即．即函数在上是减函数，解得．·······················11分综上所述，所求的取值范围为．·····························12分21.（1）∵的定义域为，，∵在处取得极小值，∴，即，经验证是的极小值点，故·················································4分（2）

5、①当时，，∴在上单调递减，∴当时，矛盾．②当时，，令，得；，得．（i）当，即时，时，，即递减，∴矛盾．（ii）当，即时，时，，即递增，∴满足题意．综上:．························································8分（3）证明：由（2）知令，当时，即．当，有．···················12分选做题22.23.解：（1）由＝＋≥，得ab≥2，当且仅当a＝b＝时等号成立．故a3＋b3≥2≥4，当且仅当a＝b＝时等号成立．所以a3＋b3的最小值为4.（2）由（1）知，2a＋3b≥2≥4.由于

6、4>6，从而不存在a，b，使2a＋3b＝6.