人教A版高中數(shù)學(xué)選修2測(cè)試題(全套)和答案及解析

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁眉頁腳。本資料屬于網(wǎng)絡(luò)整理，如有侵權(quán)，請(qǐng)聯(lián)系刪除，謝謝！pqpq．xxB．xxB0000．設(shè)UCAU<B<<使>β有2BB1/>使b>1>a>b22x13xA是Ba>．20題qa．pp2/x．2x2200nnaan1參考答案則xy22xy則或22xy22005．3/222222(2)(xaxa)02xxBAB且BAa4a4a2(2)(xaax2AB22當(dāng)0當(dāng)xxxxa212124/否命題,真假性相同,故④正確．.22min2p的否命題為:若ax2bxc0p20bxc.2212Ummm．由,,c在Rx即且25/p或p且,pax,0022所q0或n11nnn*1nn1n1第二章圓錐曲線與方程測(cè)試題1,x40yxyxyx2222FFx1P||2121253C73或76/121是114414,y4x22!2錯(cuò)誤!錯(cuò)誤!8FFx1P且|21212F129y21,C的2錯(cuò)誤!63C28F2,0,F40,2錯(cuò)誤錯(cuò)誤錯(cuò)誤!22121222x3x3654x2PC1點(diǎn)QC5y1點(diǎn)R2212線C5y1則||223Py2x點(diǎn)P,P在y2知Fy11CB,21|F||F11y3在y27/3則2212若1212y2,221Ce2圓C于,BO1CDC4B5A6D7CACC1223||F|4．2112241且故!5得c2,．7,e228P|||||2,|||1212129y2y4,22設(shè)xy錯(cuò)誤xy〔x1,y錯(cuò)誤x00000000200008/2224224222212||,||－12212122212121ab1212ccb222bb,b2解22212caabb33aa18322×2cbac92221212121y2x則2xyxy,y3x4xx易2112212111pp3yxF,0,－設(shè)x22211py23x,yxy2221122p1y=－p,x=,〔x〔y－21231221223,p而故．y=221612124x3y22222211．1P1,①2129/22222S|FF|×4．122,AB．|＋|＝AB|且,222pp22221122xx即xx,12122212又xxxx1212121212y8．222222222a且a,2e；212122②21由e1,①2C②,22Cykx22222設(shè)xyx,y則xxxx,11221212＝S1212222121212221221/．5932＋－－ABAD0ABAC．．．．AD3c＝＝cbc>4＋a－b2a,bab1237．．．33335PAn,<PA或6D且OBOCxOD＝0則x4B26D67ABCD,3,111111則11BC．52ab－＝〔c則x14B4C．629ABCD1111<＋AAAD＋－>3<>；AB2AB11211111②；AC1AB11AA1；AB11ABCD·AA·111111個(gè)2個(gè)3個(gè)4個(gè)/10·BCBC·CDCD··AB－則bc||＝|＝a,b則a與bnn<nn1212BzAA2a與b夾角為則＋3AA12AA23AAn1n1nb量ABCD,1111O是BDB．1111〔第17題ABC,CA＝棱11112,N、AA1AB11〔第18題1·；BNCM12BA,．CB11ABCD,點(diǎn)E11111〔第19題1DA；112E為,E13DD．1420P,CD＝＝AB/2F1；〔第20題2,C求k參考答案1D4B得a1,b則ab2a,b＝．3|a||b|5BD·BC0AB．>．．．14．或．46．．＝＝＋2OC＋．CD1BC1AD1AC11CD11BC與CD．1111．C．1/30>．10以CCB,,z1>提示：以D,zDA·．DE11112