2025年春季學期高一年級校聯(lián)體第一次聯(lián)考數(shù)學參考答案一、單選題故選：B2．答案：A所以故選：A3．答案：B四邊形ABCD的面積為即B正確．故選B．4．答案：C解析：方法一（割補法由于，而截去的正四棱錐的高為3，所以方法二臺體的體積公式）棱臺的體積為故答案選C.5．答案：BEQ\*jc3\*hps24\o\al(\s\up8(–),e)EQ\*jc3\*hps24\o\al(\s\up8(––),e2)5．答案：BλEQ\*jc3\*hps23\o\al(\s\up2(–),e)3，可得λEQ\*jc3\*hps23\o\al(\s\up2(–),e)=0，EQ\*jc3\*hps24\o\al(\s\up7(–),e)故選：B．6．答案：DEQ\*jc3\*hps20\o\al(\s\up6(+),4)EQ\*jc3\*hps23\o\al(\s\up14(C),2)故選：D．7．答案：A332故選：A．8．答案：B解析：的中點,EQ\*jc3\*hps23\o\al(\s\up5(-),A)EQ\*jc3\*hps23\o\al(\s\up5(-),A)EQ\*jc3\*hps23\o\al(\s\up5(-),A)EQ\*jc3\*hps23\o\al(\s\up5(-),A)EQ\*jc3\*hps23\o\al(\s\up5(-),A)故選：B.二、多選題9．答案：ACD解析：由題意，復數(shù)對于B項：z=12i，所以復數(shù)z的虛部等于－2，所以B項錯誤；對于D項：z=12i，對應(yīng)的點(1,2)在復平面的第三象限，所以D項正確。故選：ACD．10．答案：CD解析：因為與EQ\*jc3\*hps24\o\al(\s\up8(→),b)的夾角為鈍角，所以且所以+4≥2，所以||的最小值為2，所以B不正確；EQ\*jc3\*hps23\o\al(\s\up9(→),b)故選：CD.11．答案：AC解析：對于A，根據(jù)余弦定理得所以b2+c2—bc=12，所以b22所以A正確．對于B，因為所以△ABC外接圓的半徑為2，所以B錯誤.23對于D，因為2sinA=3≠1，所以sinB≠2sinAsinB，所以b≠2asinB，所以D故選：AC．三、填空題12．答案4EQ\*jc3\*hps23\o\al(\s\up8(→),b)EQ\*jc3\*hps21\o\al(\s\up6(→),b) 44解析：」BD=3DC，:AD—AB=3(AC—AD 44解析：如圖，因為△ABC是邊長為6的正三角形，則其外接圓的半徑所以外接球的表面積為15．答案：(1)π(2)223:a-.2=0，············································································2分2:22cosa-,2=0，·········································4分 ···········································································5分2266解析：(1)由正弦定理得3sinAsinB=sinBcosA．·······················2分因為B∈(0,π)，所以sinB≠0，············································6所以S△分17．答案1）表面積9π,體積3π（29π+解析1）設(shè)圓錐的高為h，由題意得：:圓錐側(cè)面積S1=πrl=23.3π=6π,···········································4分:又圓柱的底面半徑為r=3，高（母線）為h=3·······························9分:圓柱的體積為分:剩下幾何體的體積為分S圓柱側(cè)分S剩圓錐18．答案(1)（2）1解析1）以A為坐標原點，AC所在直線為x軸，建立如圖所示平面直角坐標系，則A(0,0),B(0,2),C(3,0)···································································1分EQ\*jc3\*hps20\o\al(\s\up6(-),A)································································2分EQ\*jc3\*hps20\o\al(\s\up8(-),A)EQ\*jc3\*hps21\o\al(\s\up7(-),A)EQ\*jc3\*hps18\o\al(\s\up6(-),A)(64)聯(lián)立①②解得點G的坐標為|(5,5,(64)所以··········································································9分所以············································································10分（2）因為BE上的點滿足x+y=2，所以點H的坐標為(|(,,)，·······························································14分所以分解析：(1):bcosC+ccosB=2acosA，整理得：sin(B+C)=sinA=2sinAcosA，············:sinA≠0，·················································································4分即··································