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时间:2018-12-18
《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》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
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.
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