1、g20744g1212g10806g7072g16901g20168 g12636 1 g20133 g1953 4 g20133 g20744g1212g10806g7072g16901g20168 g12636 2 g20133 g1953 4 g20133陈州高中2018-2019（下）第一次月考高二理数试题g1072g451g17977g6425g20168g726g7516g20168g195312g4671g20168g712g8703g4671g201685g2102g712g195360g2102.g3416g8703g4671g20168g13577g2090g11444g3
2、339g1114g17977g20137g1117g712g2586g7481g1072g20137g7263g12630g2616g20168g11550.1g714g5154g11797ag729(g7133g7122g7125)g712bg729(1g7125g712g7131)g712g2121a(ag7113b)g729Ag714(0g71234g71210) Bg714(g7133g71219g7127) Cg71444 D.232g714g13567g17911g9961P(4g712g7132)g11444g6347g10393g13551g11444g7735g2038g71
3、45g12347g1130Ag714y2g729xg6214x2g729g7138y Bg714y2g729xg6214y2g7298x Cg714y2g729g7138x Dg714x2g729g7138y3g714g3416g19375g7145g1411ABCDA1B1C1D1g1117g712ABg711BCg711CC1g713D1C1 g12665g1214A.AD1 BAC1 C.AD D.AB4g714g5154g11797P(8g712a)g3416g6347g10393g13551y2g7294px(p0)g1082g712g1092Pg2144g10070g9961g11
4、444g17421g12267g113010g712g2121g10070g9961g2144g2038g13551g11444g17421g12267g1130Ag7142 Bg7144 Cg7148 Dg714165.g4018g3374g6256g12138g712g3416g2064g1413g1411ABCDg1117g712ABg309g5283g19858BCDg712BCg309CDg712g1092ABg729BCg7291g712CDg7292g712g9961Eg1130CDg1117g9961g712g2121AEg11444g19375g1130A. 2B. 3Cg7
5、142D. 56g714g5154g11797Fg7263g6347g10393g13551y2g729xg11444g10070g9961g712Ag712Bg7263g16917g6347g10393g13551g1082g11444g1108g9961g712|AF|g711|BF|g7293g712g2121g13551g8677ABg11444g1117g9961g2144yg17828g11444g17421g12267g1130Ag71434 Bg7141 Cg71454 Dg714747.g4018g3374g6256g12138g712g9961Pg3416g8595g714
6、5g5522ABCDg6256g3416g5283g19858g3910g712PAg309g5283g19858ABCDg712PAg729ABg712g2121PBg1086ACg6256g6208g11444g16386g7263Ag71490Bg71460Cg71445D.308g714g17911g6347g10393g13551y2g7292px(p0)g11444g10070g9961g1420g1072g7569g11556g13551g1236g6347g10393g13551g1214g9961A(x1g712y1)g712B(x2g712y2)g712g2121y1y2x
7、1x2g1130Ag7144 Bg714g7134 Cg714p2 Dg714g713p29g714g7481g1072g1114g8595g1081g16386g5522g11444g1108g1114g20134g9961g3416g6347g10393g13551y2g7292px(p0)g1082g712g2582g1072g1114g20134g9961g3416g2511g9961g712g2121g16917g1081g16386g5522g11444g17897g19375g7263Ag7142 3p Bg7144 3p Cg7146 3p Dg7148 3p10g714g40
8、18g3374g712g3416g11556g1081g7969g7713ABCA1B1C1g1117g712g287ACBg72990g7122ACg729AA1g729BCg7292g712Dg1130AA1g1082g1072g9961g714g14613g1212g19858g16386B1DCC1g11444g3927g4671g113060g712g2121ADg11444g19375g1130A. 2B. 3Cg7142D. 2211.g1086g11556g135512xg713yg7114g7290g5283g15996g11444g6347g10393g13551yg729
9、x2g11444g2103g13551g7145g12347g1130Ag7142xg713yg7113g7290 Bg7142xg713yg7133g7290 Cg7142xg713yg7111g7290 Dg7142xg713yg7131g729012g714g4018g3374g712g5154g11797g6347g10393g13551 g11444g10070g9961g1130Fg712g17911g9961Fg11444g11556g13551ABg1236g6347g10393g13551g1214g9961Ag712Bg712g1236g6347g10393g13551g1
10、1444g2038g13551g1214g9961Cg712g14613 g712g2121Ag7144Bg7145Cg7146Dg7147g1212g451g3739g12458g20168g726g708g7516g4671g20168g19534g4671g20168g712g8703g4671g201685g2102g712g195320g2102).13.g5154g11797g2556g7458g13551x2mg713y2g7291g11444g2595g10070g9961g5792g4013g7263g6347g10393g13551y2g7298xg11444g10070g
11、9961g712g2121mg729_g71414.g5154g11797g11556g13551l1g11444g7145g2625g2625g18431g1130ag729(2g7124g712x)g712g11556g13551l2g11444g7145g2625g2625g18431g1130bg729(2g712yg7122)g712g14613|a|g7296g712g1092ag309bg712g2121xg711yg11444g1644g7263_15.g11556g16386g1081g16386g5522ABCg11444g1108g7569g11556g16386g178
12、97BCg7293g712ACg7294g712PCg309g5283g19858ABCg712PCg72995g712g2121g9961Pg2144g7116g17897ABg11444g17421g12267g7263_g71416g714g5154g11797g6347g10393g13551Cg726y2g7294xg11444g10070g9961g1130Fg712g2038g13551g1130lg712g17911g6347g10393g13551Cg1082g11444g9961Ag1420g2038g13551lg11444g3506g13551g712g3506g173
13、79g1130Mg712g14613g440AMFg1086g440AOF(g1958g1117Og1130g3456g7735g2511g9961)g11444g19858g12319g1147g8708g11303g2971g712g2121g9961Ag11444g3456g7735g1130_g714g20744g1212g10806g7072g16901g20168 g12636 3 g20133 g1953 4 g20133 g20744g1212g10806g7072g16901g20168 g12636 4 g20133 g1953 4 g20133g1081g451g1640
14、3g12676g20168g726g195370g2102.g16403g12676g5316g1993g2090g7095g4487g16932g7230g451g16881g7230g17911g12347g6214g9540g12743g8597g20692g71417g714(10g2102)g7785g6558g1083g2119g7569g1318g8818g6347g10393g13551g11444g7735g2038g7145g12347g714(1)g6347g10393g13551g11444g10070g9961g7263g2556g7458g1355116x2g713
15、9y2g729144g11444g5142g20134g9961g727(2)g6347g10393g13551g11444g10070g9961Fg3416xg17828g1082g712g11556g13551yg729g7133g1086g6347g10393g13551g1236g1214g9961Ag712|AF|g7295g71418g714g70812g2102g709g4018g3374g712g5154g11797g9961Pg3416g8595g7145g1411ABCDABCDg11444g4649g16386g13551BDg1082g712g287PDAg72960.
16、(1)g8818g5426g19858g11556g13551DPg1086CCg6256g6208g16386g11444g3927g4671g727(2)g8818DPg1086g5283g19858AADDg6256g6208g16386g11444g3927g4671g71419.g70812g2102g709g5154g11797g6347g10393g13551y2g729g713xg1086g11556g13551yg729k(xg7111)g11560g1236g1214Ag712Bg1108g9961g714(1)g8818g16881g726OAg309OBg727(2)g
17、5507g440AOBg11444g19858g12319g12665g1214 10g7206g712g8818kg11444g1644g71420.g70812g2102g709g4018g3374g712g3416g3339g7969g19285PABCDg1117g712AB/CDg712g1092 90BAP CDP = = g714g7081g709g16881g7230g726g5283g19858PABg309g5283g19858PADg727g7082g709g14613PA=PD=AB=DCg712 90APD = g712g8818g1212g19858g16386AP
18、BCg11444g1417g5462g1644g71421.g70812g2102g709g5154g11797g6347g10393g13551Cg726y2g7294xg712Fg7263g6347g10393g13551Cg11444g10070g9961g712g17911g9961Fg11444g11556g13551lg1086g6347g10393g13551Cg1236g1214Ag712Bg1108g9961g712Og1130g3456g7735g2511g9961g714(1)g4018g7628lg11444g7116g10679g11301g712g8818g1301
19、ABg1130g11556g5556g11444g3382g11444g7145g12347g727(2)g14613|FA|g7292|BF|g712g8818g11556g13551lg11444g7145g12347g71422g714g70812g2102g709g4018g3374g712g2064g1413g1411g7263g3382g7713g11444g1072g18200g2102g712g4531g7263g11105g11801g5522ABCD(g2554g1958g1973g18200)g1301ABg17897g6256g3416g11556g13551g1130
20、g7163g17820g17828g7163g17820120g5575g2144g11444g712Gg7263DFg909g11444g1117g9961g714(1)g16878Pg7263CEg909g1082g11444g1072g9961g712g1092APg309BEg712g8818g287CBPg11444g3927g4671g727(2)g5507ABg7293g712ADg7292g7206g712g8818g1212g19858g16386EAGCg11444g3927g4671g7141陈州高中2018-2019（下）第一次月考理数答案一选择题：1-5CAABB 6
21、-10CBBBA11-12DB二填空题：13. 3 14. 1或-3 15. 3 16. (2，2 2)三解答题：17.解：(1)由双曲线方程得x29y2161，其左顶点为(3，0)因此抛物线的焦点为(3，0)设其标准方程为y22px(p0)，则p23所以p6因此抛物线的标准方程为y212x(2)当抛物线开口向右时，设抛物线的标准方程为y22px(p0)，A(x0，3)，依题意得92px0，x0p25.解得p1，或p9当抛物线开口向左时，设抛物线的标准方程为y22px(p0)，A(x0，3)，依题意得92px0，p2x05.解得p1或p9综上所述，所求抛物线的标准方程为y22x或y218x18
22、.解：如图，以D为坐标原点，DA为单位长度建立空间直角坐标系Dxyz.则DA(1，0，0)，CC(0，0，1)连接BD，BD，在平面BBDD中，延长DP交BD于点H.设DH(m，m，1)(m0)，由DH，DA60及DHDA|DH |DA |cosDH，DA，可得 2m 2m21，解得 m 22 ，所以DH22，22，1 .(1)因为cosDH，CC 11 2 22，所以DH，CC45，即异面直线DP与CC所成的角为45.(2)平面AADD的一个法向量是DC(0，1，0)因为cosDH，DC22 022 1101 212，所以DH，DC60，即DP与平面AADD所成的角为30.219.(1)证明
23、：如图，由方程组 y2x，yk（x1），消去x并整理，得ky2yk0设点A(x1，y1)，B(x2，y2)，由根与系数的关系知y1y21k，y1y21因为kOAkOBy1x1y2x2 y1y21 y2y22 1y1y21，所以OAOB(2)设直线与x轴交于点N，显然k0令y0，则x1，即点N(1，0)所以Sg440OABSg440OANSg440OBN12|ON|y1|12|ON|y2|12|ON|y1y2|121 （y1y2）24y1y212 1k24 10，所以k16（2）在平面PAD内作PF AD ，垂足为F，由（1）可知，AB 平面PAD，故AB PF ，可得PF 平面ABCD以F为坐
24、标原点，FA的方向为x轴正方向，| |AB 为单位长，建立如图所示的空间直角坐标系F xyz 由（1）及已知可得 2( ,0,0)2A ， 2(0,0, )2P ， 2( ,1,0)2B ， 2( ,1,0)2C 所以 2 2( ,1, )2 2PC = ， ( 2,0,0)CB = ， 2 2( ,0, )2 2PA= ， (0,1,0)AB = 设 ( , , )x y z=n 是平面PCB的法向量，则 0,0,PCCB = =nn 即2 2 0,2 22 0,x y zx + = =可取 (0, 1, 2)= n 设 ( , , )x y z=m 是平面PAB的法向量，则 0,0,PAA
25、B = =mm 即2 2 0,2 20.x zy = =可取 (1,0,1)=m 则 3cos , | | | 3= = n mn m n m ，所以二面角A PB C 的余弦值为 33 321.解：设A(x1，y1)，B(x2，y2)(1)因为y24x，所以F(1，0)，准线为x1又直线l的斜率为1，所以直线l的方程为yx1代入y24x，得x26x10由根与系数的关系，得 x1x26x1x21，易得AB的中点，即所求圆的圆心的坐标为(3，2)又由抛物线的定义，知|AB|x1x228，所以所求圆的半径r4，所以以AB为直线的圆的方程为(x3)2(y2)216(2)因为|FA|2|BF|，所以F
26、A2BF又FA(x11，y1)，BF(1x2，y2)，所以x112（1x2）y12y2易知直线l的斜率存在且不为0，设直线l的斜率为k，则直线l的方程为yk(x1)，代入y24x，得k2x2(2k24)xk20，由根与系数的关系，得x1x22k24k2x1x21又x112(1x2)所以x11x21或x12x212，所以k2 2，此时0，所以直线l的方程为y2 2(x1)或y2 2(x1)22.解：(1)因为APBE，ABBE，AB，AP平面ABP，ABAPA，所以BE平面ABP，又BP平面ABP，所以BEBP，又EBC120，因此CBP30.(2)法一：4取EC的中点H，连接EH，GH，CH.
27、因为EBC120，所以四边形BEHC为菱形，所以AEGEACGC 3222 13.取AG中点M，连接EM，CM，EC，则EMAG，CMAG，所以EMC为所求二面角的平面角又AM1，所以EMCM 1312 3.在BEC中，由于EBC120，由余弦定理得EC22222222cos12012，所以EC2 3，因此EMC为等边三角形，故所求的角为60.法二：以B为坐标原点，分别以BE，BP，BA所在的直线为x，y，z轴，建立如图所示的空间直角坐标系由题意得A(0，0，3)，E(2，0，0)，G(1，3，3)，C(1，3，0)，故AE(2，0，3)，AG(1，3，0)，CG(2，0，3)，设m(x1，y1，z1)是平面AEG的一个法向量由mAE0，mAG0，可得 2x13z10，x1 3y10.取z12，可得平面AEG的一个法向量m(3， 3，2)设n(x2，y2，z2)是平面ACG的一个法向量由nAG0，nCG0，可得x2 3y20，2x23z20.取z22，可得平面ACG的一个法向量n(3， 3，2)所以cosm，n mn|m|n|12.因此所求的角为60.
