河南省南陽(yáng)市20182019學(xué)年高二數(shù)學(xué)上學(xué)期期末考試試題理(掃描版)

河南省南陽(yáng)市2018-2019學(xué)年高二數(shù)學(xué)上學(xué)期期末考試一試題理（掃描版）高二理科期末參照答案一．選擇題CBADBABABCCD二．填空題13.401814.215.6516.1513三．解答題17.解：若命題p：方程表示圓為真命題，則，解得．3分若命題q：雙曲線(xiàn)的離心率，為真命題，則，解得．6分命題“”為假命題，“”為真命題，與q必然一真一假．，或，解得或空集Φ綜上可得：實(shí)數(shù)m的取值范圍是．10分解（1）延長(zhǎng)AN,交CD于點(diǎn)G,由相似知ANBN,又PABD2BD,PMDN,NGND因此ANBNAM,即MN//PGNGNDMP又MN平面PCD,PG平面PCD,因此直線(xiàn)MN//平面PCD6分（2）因?yàn)镈ADCDP,以DA,DC,DP為x,y,z軸建立空間直角坐標(biāo)系,設(shè)A1,0,0,則B1,1,0,C0,1,0,P0,0,1,M1,0,1,N1,1,02222則PB1,1,1,平面AMN的法向量為m1,1,1,設(shè)向量PB與m的夾角為,則cos1,3設(shè)PB與平面AMN夾角為,則cos22sin3PBAMN22.12319.1b(12cosC)2acosCccosAsinB(12cosC)2sinAcosCcosAsinCsin(AC)2sinBcosC2sinAcosCcosAsinC2sinBcosCsinAcosC0C2sinBsinAa2b.6212bb2SsinC4sinCb2,a4.2abc10,c4.1222222216.12cosCAD4420.1AEA1B1,A1B1//AB,AEAB,AA1AB,AA1AEA,ABA1ACC1,ACA1ACC1,ABAC,2AAxyz,A(0,0,0),E(0,1,1),F(1,1,0),A(0,0,1),B(1,0,1)322211D(x,y,z),A1DA1B1(0,1),(x,y,z1)(1,0,0),D(,0,1),DF(1,1,1),22AE(0,1,1),DFAE110,DFAE6222zDA1B1C1xBAEFCy2D,DEFABC14714ABCn(0,0,1)DEFn(x,y,z),nEF0nDF0FE(1111,11),,,),DF(,222221x1y1z0x3z2(1222,),1112)xyz0yz(22(12)z2(1),n(3,12,2(1))10DEFABC14,14cosm,nmn142(1)14mn,9(12)24(1)2,1414172,DA1B1124n2ann1an121.n2anSnSn133ann1(n2.)ana1a2a3a4anan1n1a1a2a3an1134567n112345n1n(n1)(n2)n12n(n1)anan2.bnq.b1a342,b4a51163q2bnq82nbnbn2n.5cnnbnn2n12n22n1an1(n1)(n2)n2n1Tn2322(242325242n22n1(3)4)()(2n)2354n1Tn2n228n2TnTnCn1(n1)2n201(n2)(n3)Tn1TnTn(Tn)min210T13220193m40381346m1345.12322.:C2:y28x(2,0),8b24,b2.C2:x2y21.84x12y1281,(1)M(x1,y1),N(x2,y2),4x22y221,84(x1x2)(x1x2)(y1y2)(y1y2)0，兩式相減得84又MN的中點(diǎn)為(1,1),因此x1x22,y1y22.y2y11因此x1.x22明顯,點(diǎn)(1,1)在橢圓內(nèi)部,因此直線(xiàn)MN的斜率為1.5分2橢圓右焦點(diǎn)F2(2,0).當(dāng)直線(xiàn)AB的斜率不存在也許為0時(shí),1111327分mn42228.當(dāng)直線(xiàn)AB的斜率存在且不為0時(shí),設(shè)直線(xiàn)AB的方程為yk(x2)，設(shè)A(x1,y1),B(x2,y2),聯(lián)立方程得yk(x2),x22y28,消去y并化簡(jiǎn)得(12k2)x28k2x8k280，因?yàn)?8k2)24(12k2)(8k28)32(k21)0，因此x1x218k2，x1x28(k21