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河南省实验中学2022——2023学年下期期中试卷答案（高一）一、单选题（共8小题）1-4DACB5-8ACBD二、多选题（共4小题）9.ABC10.AD11.ACD12.ABD三、填空题（共4小题）13.√514.315.[2−√2

，2+√2]16.（1，2）四．解答题（共6小题）17．解：（1）已知向量𝑎→=(−4，3)，𝑏→=(1，−2)，则𝑐𝑜𝑠𝜃=𝑎→⋅𝑏→|𝑎→|⋅|𝑏→|=−4+3×(−2)√(−4)2+32×√1+(−2)2

=−105×√5=−2√5，····················（3分）所以𝑠𝑖𝑛𝜃=√1−𝑐𝑜𝑠2𝜃=√1−(−2√5)2=√55，即sinθ的值为√55；············

··········································（5分）（2）已知向量𝑚𝑎→+𝑏→与向量𝑎→−𝑏→垂直，则(𝑚𝑎→+𝑏→)⋅(𝑎→−𝑏→)=0，····························

····················（7分）即𝑚𝑎→2+(1−𝑚)𝑎→⋅𝑏→−𝑏→2=0，又𝑎→2=(−4)2+32=25，𝑎→⋅𝑏→=−4×1+3×(−2)=−10，𝑏→2=12+(−2)2=5，所以25m﹣10（1

﹣m）﹣5＝0，即35m＝15，解得𝑚=37．····························································（10分）18．解：（1）在△ABC中，由正弦定理得：√3sinAcosC﹣sinCsinA＝0，···········（2分

）因为0＜A＜π，所以sinA＞0，从而√3cosC＝sinC，······················································（3分）又C∈（0，π），可得cosC≠0，所以tanC=√3，·························

·······························（5分）所以C=𝜋3．·····························································（6分）（2）在△AB

C中，b＝6，S△ABC=12absinC=12×6×a×sin𝜋3=6√3，··········（8分）得a＝4，·····························································（9分）由余弦定理得：c2＝a2+

b2﹣2abcosC，即c2＝62+42﹣2×6×4×cos𝜋3=28，······（11分）所以c＝2√7．························································（12分）19.解：（1）设�

�𝑃→=𝑡𝐸𝐶→，由题意𝐵𝐸→=23𝐵𝐴→=23𝑏→，·所以𝐸𝐶→=𝐸𝐵→+𝐵𝐶→=𝑎→−23𝑏→，𝐵𝑃→=𝐵𝐸→+𝐸𝑃→=𝐵𝐸→+𝑡𝐸𝐶→=23𝑏→+𝑡(𝑎→−23𝑏→)=𝑡𝑎→+23(1−𝑡)𝑏→①，··········

···············································（2分）设𝐷𝑃→=𝑘𝐷𝐴→，由𝐵𝐷→=13𝐵𝐶→=13𝑎→，𝐷𝐴→=𝐷𝐵→+𝐵𝐴→=𝑏→−13𝑎→，

𝐵𝑃→=𝐵𝐷→+𝐷𝑃→=13(1−𝑘)𝑎→+𝑘𝑏→②，·························································（4分）由①、②得，𝑡

𝑎→+23(1−𝑡)𝑏→=13(1−𝑘)𝑎→+𝑘𝑏→，所以{𝑡=13(1−𝑘)23(1−𝑡)=𝑘，解得{𝑡=17𝑘=47，·······································（6分）所以𝐵𝑃→=17𝑎→+47𝑏→；····

···············································（7分）（2）证明：由𝐴𝐶→=𝑎→−𝑏→，得𝐴𝐹→=15𝐴𝐶→=15(𝑎→−𝑏→)，所以𝐵𝐹→=𝐵𝐴→+𝐴𝐹→=15𝑎→+

45𝑏→，·········································（9分）所以𝐵𝐹→=75𝐵𝑃→，······················································（10分）因为𝐵𝐹→与𝐵𝑃→有公共点B，所以B

，P，F三点共线．···············································（12分）20．证明：（1）根据题意，结合勾股定理易知在直角梯形ABCD中，AC⊥BC，且AC=BC=√2，又

PA⊥AB，PA⊥AC,PA＝1∴PC=√3,PB=√5,∴△PBC为直角三角形，················································（2分）在Rt△PAB中，M为PB的中点，则AM=12𝑃𝐵．在

Rt△PBC中，M为PB的中点，则CM=12𝑃𝐵，∴AM＝CM．···························································（4分）（2）连接

DB交AC于F，由相似三角形的性质易知∵DC=∥12𝐴𝐵，∴DF=12𝐹𝐵．·························（5分）取PM中点G，连接DG，FM，则DG∥FM，··················

············（6分）又DG⊄平面MAC，FM⊂平面AMC，∴DG∥平面AMC，·····················································（7分）连DN，GN，则GN∥MC，········

·······································（8分）又GN⊄平面MAC，MC⊂平面AMC，∴GN∥平面AMC，······················································（9分）又GN∩DG＝G，GN、DG

⊂平面DNG，∴平面DNG∥平面ACM，··············································（11分）又DN⊂平面DNG，∴DN∥平面ACM．······················································（1

2分）21．解：（Ⅰ）f（x）=𝑎→•𝑏→=（cos（𝜋3−x），﹣sinx）•（sin（x+𝜋6），sinx）＝（cos（𝜋3−x）（sin（x+𝜋6）﹣sin2x＝（12cosx+√32sinx）2﹣sin2x=14（cos2x﹣sin2x）+√32sinxcosx＝14c

os2x+√34sin2x=12sin（2x+𝜋6），········································（3分）∴函数的周期T=2𝜋2=𝜋，··············································（4分）由2kπ+𝜋

2≤2x+𝜋6≤2kπ+3𝜋2，k∈Z，即kπ+𝜋6≤x≤kπ+2𝜋3，k∈Z，即函数的单调递减区间为[kπ+𝜋6，kπ+2𝜋3]，k∈Z．··························（6分）（Ⅱ）∵f（C）=−12，∴12sin（

2C+𝜋6）=−12，∴sin（2C+𝜋6）＝﹣1，即2C+𝜋6=2kπ−𝜋2，得C＝kπ−𝜋3，k∈Z，∵0＜C＜π，∴当k＝1时，C=2𝜋3，··························

············（8分）由余弦定理得c2＝a2+b2﹣2abcosC，∵c＝2√3，∴12＝a2+b2﹣2abcos2𝜋3=a2+b2+ab≥2ab+ab＝3ab，即ab≤4，······························

······························（10分）则三角形的面积S=12absinC≤12×4×√32=√3，当且仅当a＝b时取等号，即三角形的面积的最大值为√3．············

·····························（12分）22．解：（1）证明：连接A1C1，BC1，∵E，F，G分别为所在棱的中点，∴A1C1∥GF，EF∥BC1，∵AD1∥BC1，∴AD1∥EF，又AD1⊂平面ACQ，EF⊄平面ACQ，

∴EF∥平面ACQ，·····················（2分）同理可证GF∥平面ACQ，又GF∩EF＝F，∴平面EFG∥平面ACQ；·········（4分）（2）线段CD上存在点P，当DP=13DC时，满足DQ∥平面D1PH，······

····（5分）证明如下；如右图，取CD上靠近D点的三等分点为P，连接PD1，连接PH并延长交AB于点M，连接D1M，则平面D1PH与平面D1PM为同一平面，取线段D1M的中点为N，连接QN，NP，··································（7分）由平行

关系及H为AC的中点，得△AMH≌△CPH，则AM=23AB=23CD，因为Q，N分别为AD1，MD1的中点，所以QN=12AM=13AB=13CD，且QN∥AM，又DP∥AM且DP=13DC，即QN∥DP且

QN＝DP，·························（9分）所以四边形QDPN为平行四边形，故QD∥NP，···························（10分）又QD⊄平面D1PH，NP⊂平

面D1PH，故QD∥平面D1PH．·················（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