2018-2019學(xué)年寧夏銀川一中高一上學(xué)期期末考試數(shù)學(xué)試題

+m37．D，nm，，mm，∥m⊥，∥mnm∥n，，BCBC的23234422．．．(m2)xmy7(m2)x(m2)y504m()11．C或2D．22(xy2xy22CC(xy2xy22222．．．x2(y22x2(y22．13π33π5π．．．．5π31()22．1122．14221．8．15ABC2ACABBA1ABC—111130456090ABCDOABO3．．．kx3x2y342322MNyMNk33,3323,0,0334．．．D．))a()2222A.(C.(,0)(0,)B.(22,2)(2,22)223222232322,)(,)D.(,)(2,)222C:(x1(y1Cx(y92222CC，A、B122y|PB||PA|P是24．5242．．D．2OOy6x8y0ABCD．x22AC和BDBxyCC．x2y28y120l:axy2a0：.alC.22llCB=.VABC中，平面VABABC，VABACBC2OVA，、MAB、且VMABOC∥）求證：VBMOC．MOCVABV．31l．2llM(2,1)，N(4,1)ABCD=2E為11111C;11BC.1114(1,0)B(1,0)Cxy6x8y0，22PyCPBOAxCA2|||2P求ABC，111BD1AA=3=2AC，.1111DC2A；1CACBADCABAB73ππ2055y8y120x(y4)4C(0,4)：x2，222|42a|342，再求圓心到直線l:axy2a0d2ard,.a12d22:axy2a02224ldd，|42a|則2a1或a7，a12y207xy140或x.lOAV解：（1M、、AB∥MOCVBMOVBMO面，MOC，∥VBMOC．ACBCO，是AB（2）ABOC，又因為平面VABABCOCABC，OCVABMOCVAB2．2OC1，，（3ABCABVABS3OCVABVAB，133CVABOCS．3VAB63又因為三棱錐VABCCVAB=.3xy11，(0,0)，abab1(1,2)，P2121ab1ab∴+．2112ab(a,a1)2（(2,1)，(4,1)，MN∴((1((1，a2a2a2a2a22∴，(2,1)r(x(4．2y22B作=Rt3Rt69在=在=.在BEC,EC222BEBC故C.得又∩=,BEBBBBB1111113VAAS2在Rt32.ADC1AC1AD1DC12121E-ABC1VABC11111111132,23.則S35.ECEC2CC21AE1AA12AD2DE21ACE111BBC5d,EAC1VdS111311ACE1110552,.dd105BC.111k(1)，ykxA|k4k|1k232，(3,4)2rk4340；xy當(dāng)kx13430或x1.xy7（2(,)Pxy||||2(xy)22||2，22222要|||2|取得最大值只要使|2|2又(|OP|)|OC|r3427，P22max∴||2|)272，22max92152854xyxy4yx5125:y3，x3y6x8y21022x2128,P．55，AA，1Rt△AB，ACBC6AABC1，6BDAB3AB:2BD，33BC△，BAC90ADB．A，AAD1AAD.1又AAAD1AD1A.1CCCCEBE，AE交于11ABACCA11,1,∴┴,11AEBACCB18AC過作CFC交于F111CFAA3則，，11CCF60．13Rt△Rt△AEACsin6023．在在235．=2,=,=AEBE35155cosBEA155CCBA．19（2(,)Pxy||||2(xy)22||2，22222要|||2|取得最大值只要使|2|2又(|OP|)|OC|r3427，P22max∴||2|)272，22max92152854xyxy4yx5125:y3，x3y6x8y21022x2128,P．55，AA，1Rt△AB，ACBC6AABC1，6BDAB3AB:2BD，33BC△，BAC90ADB．A，AAD1AAD.1又AAAD1AD1A.1CCCCEBE，AE交于11ABACCA11,1,∴┴,11AE