【文档说明】辽宁省鞍山市普通高中2021一2022学年高三第二次质量监测 数学 PDF版答案.pdf，共(5)页，447.501 KB，由envi的店铺上传

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答案第1页共4页参考答案及评分标准1．A2．D3．B4．B5．B6．C7．D8．C9．BD10．AB11．ACD12．ABC13．12i+14．1−15．√32𝜋16．𝑥24+𝑦23=1(𝑥𝑦≠0)17．解(1)𝑓(𝑥)=−√3

2𝑠𝑖𝑛𝑥−12𝑐𝑜𝑠𝑥=−𝑠𝑖𝑛(𝑥+𝜋6)··································2分令−𝜋2+2𝑘𝜋≤𝑥+𝜋6≤𝜋2+2𝑘𝜋，则−2𝜋3+2𝑘𝜋≤𝑥≤𝜋3+2𝑘𝜋

,𝑘∈𝑍所以，单调减区间是[−2𝜋3+2𝑘𝜋,𝜋3+2𝑘𝜋],𝑘∈𝑍.·····································5分(2)由𝑎2−𝑏2=𝑎𝑐⋅

𝑎2+𝑐2−𝑏22𝑎𝑐−12𝑏𝑐得：𝑏2+𝑐2−𝑎2=𝑏𝑐，即𝑐𝑜𝑠𝐴=𝑏2+𝑐2−𝑎22𝑏𝑐=12，由于0<𝐴<𝜋，所以𝐴=𝜋3.················································

················7分在△𝐴𝐵𝐶中，0<𝐵<2π3，𝑓(𝐵)=−𝑠𝑖𝑛(𝐵+𝜋6)，于是𝜋6<𝐵+𝜋6<5𝜋6，则12<𝑠𝑖𝑛(𝐵+𝜋6)≤1，−1≤

−𝑠𝑖𝑛(𝐵+𝜋6)<−12，所以−1≤𝑓(𝐵)<−12.·······································································1

0分18．(1)解:若选①：∵𝑆𝑛=2𝑛−3𝑛−1，当𝑛≥2时，𝑎𝑛=𝑆𝑛−𝑆𝑛−1=2𝑛−3𝑛−1−[2𝑛−1−3(𝑛−1)−1]=2𝑛−1−3，当𝑛=1时，𝑎1=𝑆

1=−2满足上式，故𝑎𝑛=2𝑛−1−3.·····························6分若选②：𝑎𝑛+1=2𝑎𝑛+3,𝑎1=−2,易得𝑎𝑛+1+3=2(𝑎𝑛+3),于是数列{𝑎𝑛+3}是以𝑎1+3=1为首项，2为公比的等比数列

，∴𝑎𝑛+3=2𝑛−1，∴𝑎𝑛=2𝑛−1−3.······················································6分(2)由（1）得𝑏𝑛=𝑛⋅2𝑛−1，从而𝑇𝑛=1×20+2

×21+3×22+⋯+𝑛⋅2𝑛−1，12321222322nnTn=++++，作差得−𝑇𝑛=20+21+22+⋯+2𝑛−1−𝑛⋅2𝑛=1−2𝑛1−2−𝑛⋅2𝑛=(1−𝑛)2𝑛−1,于是𝑇𝑛=(𝑛−1)2𝑛+1.···············

······················································12分19．(1)证明：在梯形𝐴𝐵𝐶𝐷中，𝐴𝐵//𝐶𝐷，1ADCDBC===，故梯形𝐴𝐵𝐶𝐷为等腰梯形，因为23BCD=，则∠𝐴𝐷𝐶=2𝜋3，所以，∠�

�𝐴𝐶=∠𝐴𝐶𝐷=𝜋6，又因为∠𝐴𝐵𝐶=𝜋−∠𝐵𝐶𝐷=𝜋3，则∠𝐴𝐶𝐵=𝜋−∠𝐴𝐵𝐶−∠𝐵𝐴𝐶=𝜋2，∴𝐴𝐶⊥𝐵𝐶，答案第2页共4页因为𝐶𝐹⊥平面𝐴𝐵𝐶𝐷，𝐴𝐶⊂平面𝐴𝐵𝐶𝐷，∴𝐴𝐶⊥�

�𝐹，∵𝐵𝐶∩𝐶𝐹=𝐶，∴𝐴𝐶⊥平面𝐵𝐶𝐹，因为四边形𝐴𝐶𝐹𝐸为矩形，则𝐴𝐶//𝐸𝐹，因此，𝐸𝐹⊥平面𝐵𝐶𝐹.······················6分

(2)解：因为𝐶𝐹⊥平面𝐴𝐵𝐶𝐷，𝐴𝐶⊥𝐵𝐶，以点𝐶为坐标原点，𝐶𝐴、𝐶𝐵、𝐶𝐹所在直线分别为𝑥、𝑦、𝑧轴建立如下图所示的空间直角坐标系，在Rt∆ABC中，AC=BCtanπ6=√3，·····················8分

则𝐴(√3,0,0)、𝐵(0,1,0)、𝐶(0,0,0)、𝐸(0,0,1)、𝐹(√3,0,1)，设点𝑀(𝑡,0,1)，其中0≤𝑡≤√3，设平面𝑀𝐴𝐵的法向量为(),,mxyz=，𝐴𝐵⃗⃗⃗⃗⃗=(−√3,1

,0)，𝐴𝑀⃗⃗⃗⃗⃗⃗=(𝑡−√3,0,1)，由{𝑚→⋅𝐴𝐵→=√3𝑥−𝑦=0𝑚→⋅𝐴𝑀→=(𝑡−√3)𝑥+𝑧=0，取𝑥=1，可得𝑚⃗⃗=(1,√3,√3−𝑡)，··········10分易知平面FCB的一个

法向量为𝑛⃗=(1,0,0)，|𝑐𝑜𝑠<𝑚⃗⃗,𝑛⃗>|=|𝑚⃗⃗⃗⋅𝑛⃗||𝑚⃗⃗⃗|⋅|𝑛⃗|=1√4+(𝑡−√3)2，所以，当𝑡=0,即M与F重合时，|𝑐𝑜𝑠<𝑚⃗⃗,𝑛⃗>|取最小值，此时平面𝑀𝐴𝐵与平面FCB所成锐

二面角最大，此时，平面𝑀𝐴𝐵与平面FCB所成锐二面角的余弦值为√77.·····12分20．(1)解：由题可得()11234535x=++++=，2234435y++++==，··················

···················································2分5151iiixy==，522222211234555iix==++++=．则51522150.65iiiiixyx

ybxx==−==−，···································································4分30.631.2aybx=−=−=所以0.61.2yx=

+··············································································6分（2）解：由题可知X的所有可能取值为0，1，2，3，2123353(1)10CCPXC===，12233563(

2)105CCPXC====，0323351(3)10CCPXC===则X的分布列为答案第3页共4页X123P310351109()5EX=····················································

··································12分21．(1)依题意，𝐹1(−1,0)，𝐹2(1,0)，|𝑃𝐹1|=|𝑃𝐹2|=√2，由椭圆定义知：椭圆长轴长2𝑎=|𝑃𝐹1|+|𝑃𝐹2|=2√2

，即𝑎=√2，而半焦距𝑐=1，即有短半轴长𝑏=1，所以椭圆C的标准方程为：2212xy+=．················································4分(2)依题意，设直线l方程为𝑥=𝑚𝑦+1，由{𝑥=𝑚𝑦+1𝑥

2+2𝑦2=2消去x并整理得22(2)210mymy++−=，设𝑀(𝑥1,𝑦1)，𝑁(𝑥2,𝑦2)，则𝑦1+𝑦2=−2𝑚𝑚2+2，𝑦1𝑦2=−1𝑚2+2，···················6分假设存在点𝑇(𝑡,0

)，直线TM与TN的斜率分别为𝑘𝑇𝑀=𝑦1𝑥1−𝑡=𝑦1𝑚𝑦1+1−𝑡，𝑘𝑇𝑁=𝑦2𝑥2−𝑡=𝑦2𝑚𝑦2+1−𝑡，𝑘𝑇𝑀⋅𝑘𝑇𝑁=𝑦1𝑦2(𝑚𝑦1+1−𝑡)(𝑚𝑦2+1−𝑡)=

𝑦1𝑦2𝑚2𝑦1𝑦2+𝑚(1−𝑡)(𝑦1+𝑦2)+(1−𝑡)2=−1𝑚2+2𝑚2(−1𝑚2+2)+𝑚(1−𝑡)(−2𝑚𝑚2+2)+(1−𝑡)2=−1[−1−2(1−𝑡)+(1−𝑡)2]𝑚2+2(

1−𝑡)2，要使𝑘𝑇𝑀⋅𝑘𝑇𝑁为定值，必有−1−2(1−𝑡)+(1−𝑡)2=0，即𝑡=±√2，·······10分当𝑡=√2时，∀𝑚∈𝑹，𝑘𝑇𝑀⋅𝑘𝑇𝑁=−12(1−√2)2=−32−√2，当2t=−时，∀�

�∈𝑹，𝑘𝑇𝑀⋅𝑘𝑇𝑁=−12(1+√2)2=√2−32，所以存在点𝑇(±√2,0)，使得直线TM与TN的斜率之积为定值．·················12分22．(1)因为𝑔(

𝑥)=𝑥2−(𝑎+2)𝑥+𝑓(𝑥)=𝑥2−(𝑎+2)𝑥+𝑎𝑙𝑛𝑥，所以𝑔′(𝑥)=2𝑥−(𝑎+2)+𝑎𝑥=2𝑥2−(𝑎+2)𝑥+𝑎𝑥=2(𝑥−1)(𝑥−𝑎2)𝑥，令𝑔′(𝑥)=0，得𝑥1=1或𝑥2=𝑎2>1.所以𝑔′(𝑥)>0时

，0<𝑥<1或2ax；𝑔′(𝑥)<0时，1<𝑥<𝑎2.所以𝑔(𝑥)在(0,1)和(𝑎2,+∞)上单调递增；在(1,𝑎2)上单调递减.·······················4分答案第4页共4页(2)因为ℎ(𝑥)

=𝑎𝑙𝑛𝑥−𝑥2，所以ℎ′(𝑥)=𝑎𝑥−2𝑥=𝑎−2𝑥2𝑥.当𝑎≤0时，ℎ′(𝑥)=𝑎−2𝑥2𝑥<0，可得ℎ(𝑥)在(0,+∞)上单调递减，此时ℎ(𝑥)不可能存在两个不同的零点，不符合题意.································

····································6分当𝑎>0时，ℎ′(𝑥)=𝑎−2𝑥2𝑥=−2(𝑥−√𝑎2)(𝑥+√𝑎2)𝑥.令ℎ′(𝑥)=0，得𝑥=√𝑎2.当ℎ′(𝑥)>0时，0<𝑥<√�

�2；当ℎ′(𝑥)<0时，𝑥>√𝑎2.所以ℎ(𝑥)在(0,√𝑎2)上单调递增，在(√𝑎2,+∞)上单调递减.而当𝑥=𝑒1𝑎时，ℎ(𝑒1𝑎)=1−(𝑒1𝑎)2<0，𝑥=𝑒𝑎时，ℎ(𝑒𝑎)=𝑎2−(𝑒𝑎)2<0,(𝑒𝑎>𝑎).所

以要使ℎ(𝑥)存在两个不同的零点𝑥1,𝑥2，则ℎ(𝑥)极大值=ℎ(√𝑎2)=𝑎𝑙𝑛√𝑎2−𝑎2=𝑎2(𝑙𝑛𝑎2−1)>0，即𝑙𝑛𝑎2−1>0，解得𝑎>2𝑒.·······························

································8分因为ℎ(𝑥)存在两个不同的零点𝑥1,𝑥2，则𝑎𝑙𝑛𝑥1−𝑥12=𝑎𝑙𝑛𝑥2−𝑥22，即𝑎(𝑙𝑛𝑥1−𝑙𝑛𝑥2)=𝑥12−𝑥22.

不妨设𝑥1>𝑥2>0，则𝑙𝑛𝑥1−𝑙𝑛𝑥2>0，则𝑎=𝑥12−𝑥22𝑙𝑛𝑥1−𝑙𝑛𝑥2，要证𝑎<𝑥12+𝑥22，即证𝑥12−𝑥22𝑙𝑛𝑥1−𝑙𝑛𝑥2<𝑥12+𝑥22，即证𝑙𝑛𝑥1−𝑙𝑛𝑥2>𝑥12−𝑥22𝑥12+

𝑥22，即𝑙𝑛𝑥1𝑥2>𝑥12𝑥22−1𝑥12𝑥22+1，设𝑡=𝑥1𝑥2>1.即证𝑙𝑛𝑡>𝑡2−1𝑡2+1(𝑡>1)，·······························10分令221()ln(1)1th

tttt−=−+，则ℎ′(𝑡)=1𝑡−4𝑡(𝑡2+1)2=(𝑡2−1)2𝑡(𝑡2+1)2>0，所以ℎ(𝑡)在(1,)+上单调递增，所以ℎ(𝑡)>ℎ(1)=0，即𝑙𝑛𝑡>𝑡2−1𝑡2+1，所以𝑎<𝑥12+𝑥22成立.综上有2𝑒<

𝑎<𝑥12+𝑥22.·····································································12分获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com