2022高三數(shù)學(xué)高考第一輪復(fù)習(xí)第十九章綜合檢測

第十九章綜合檢測（推理與證明）一選擇題:（以下題目從4項(xiàng)答案中選出一項(xiàng)，每題5分，共40分）會(huì)合ABCDABCD0(ABAD)AC0R,則ab0abC,則ab0abR,則復(fù)數(shù)abicdiac,bdab2cd2ac,bdR,則ab0abC,則ab0abn(n3)(x1,x2,x3,,xn)a(a1,a2,a3,,an)b(b1,b2,b3,,bn)naibin1n3n2n4n2cosi1a(1,1,1,1)b(1,1,1,nn1)nnnnn(ai2)(bi2)i1i1cosn2n2，ysinxsinxytanxycos2x3nnn22Cx22xM中，我們把M的最小值叫做x22x的上確界,若a,bR,且ab1，則122a的上確界為bA．9B．9C．1D．224[解析]B12(ab)(12)(5b2a)9，2ab2ab22ab2M912的上確界為9，b222a82022深圳二模如圖，圓周上按順時(shí)針方向標(biāo)有1,2,3,4,5五個(gè)點(diǎn)。一只青蛙按順時(shí)針方向繞圓從一個(gè)點(diǎn)跳到另一點(diǎn)。若它停在奇數(shù)點(diǎn)上，則下一次只能跳一個(gè)點(diǎn)；若停在偶數(shù)點(diǎn)上，則跳兩個(gè)點(diǎn)。該青蛙從這點(diǎn)跳起，經(jīng)2022次跳后它將停在的點(diǎn)是（）1AB．C．D．[解析]A52[每兩次跳3個(gè)點(diǎn)，每跳10次回到5這個(gè)點(diǎn)，故跳2022次后它停在5這個(gè)點(diǎn)，跳2022次后它將停在的點(diǎn)是1]二填空題:（本大題共7小題，每題5分，滿分30分．其中13～15題是選做題，考生只34能選做二題，三題全答的，只計(jì)算前兩題得分）92022中山一模察看下列的圖形中小正方形的個(gè)數(shù)，則第6個(gè)圖中有_______個(gè)小正方形，第n個(gè)圖中有個(gè)小正方形[解析]28，(n1)(n2)設(shè)第n個(gè)圖中有個(gè)小正方形a112，a2123，2a31234，an1234(n1)(n1)(n2)210在數(shù)列中，知足aana(n2),aa,ab,設(shè)saaa,則合情n1n112n12n推理推出=____________,=_______________[解析]a100a；s1002ba112022江蘇模擬已知f(x1)2f(x),f(1)1（xN*）,猜想的表達(dá)式為f(x)2[解析]f(x)2[f(1)1,f(2)2f(3)2,f(x)2]x,,1134x122022韶關(guān)一模在實(shí)數(shù)集上定義運(yùn)算：xyx(1y)，若不等式(xa)(xa)1對隨意實(shí)數(shù)都建立，則實(shí)數(shù)的取值范圍是[解析](1，3)[(xa)(xa)1(xa)(1xa)1x2xa2a10，224a24a301a3]22濟(jì)寧市2022—2022學(xué)年度高三復(fù)習(xí)第一階段質(zhì)量檢測設(shè)等邊ABC的邊長為,是ABC內(nèi)隨意一點(diǎn),且到三邊、、的距離分別為、、,則有d1d2d3為定值3a;由以上平面圖形的特性類比到空間圖形:設(shè)正四面體ABCD的棱2長為,是正四面體ABCD內(nèi)隨意一點(diǎn),且到平面、平面、平面ACD、平面的距離分別為、、，則有h1h2h3為定值[解析]6a314（07韶關(guān)調(diào)研）設(shè)是等比數(shù)列的前項(xiàng)和,關(guān)于等比數(shù)列,有命題若S3,S9,S6成等差數(shù)列,則a,a,a成等差數(shù)列建立；關(guān)于命題：若S,S,S成等差數(shù)列,則285mnl________________成等差數(shù)列請將命題補(bǔ)充完整,使它也是真命題只需一個(gè)切合要求的答案即可[解析]amk,ank,alk(kN)開放題，答案不唯一152022RtABC111222habSABC[]1111h2a2b2c2:16若a0,求證：a212a12a2aa212a12a2aa212a124a2a需證a214a214a21222(a1)26a2a2a2a需證a212(a1)8a22a需證a211(a212)10a22a2需證a212此式顯然建立，故a212a12建立a2a2a1217f(x)cosxsinx,(0,),2f(x)f(x3)[]f(x)2sin(x)44(0,)f(x)f(x3)T228(0,)T2(0,)102(0,),f(x)f(x3)12218ABCA1B1C1PMBB1PNBB11CC1MN2DEFDE2DF2EF22DFEFcosDFE1CC1//BB1CC1PM,CC1PN,CC1平面PMNCC1MN52ABCA1B1C1SABB21A1SBCC21B1SACC21A12SBCCB1SACCAcos111CC1B1BCC1A1A7CC1平面PMN,MNP,PMNPM2PN2MN22PNMNcosMNPPM2CC2PN2CC2MN2CC22(PNCC)(MNCC)cosMNP1011111SPNCC,SACCAMNCC,SABBAPMBB11111111SABB2ASBCC2BSACC2A2SSACCAcos141111[]n=1,;n=4,=;n=5,2n,,2kk222k2k22k2(k1)2k22k1(k1)22(k1)2202k21)22k11)2n1(k(kNyf(x)[a,b][a,b]yf(x)xDyx3[a,b]1bb311y2xkyx3aa3ba

a1a1a0b0b1b1

y2xkyf(x)y2xka2akx2xkb2bkk02k5x25x4k0x2且xk25164k0225164k04524k0k5k4k02k9k[2,9)f(x)aln(1ex)(a1)x,A(x1,f(x1)),B(x2,f(x2)),C(x3,f(x3))24yf(x)2x2x1x3f(x)aln(1ex)(a1)x,f(x)aexx(a1)ex恒建立,(,)1e(a1)1x0eA(x1,f(x1)),B(x2,f(x2)),C(x3,f(x3))f2>f3,2=x1x362BA(x1x2,f(x1)f(x2)),BC(x3x2,f(x3)f(x2)BABC(x1x2)(x3x2)[f(x1)f(x2)][f(x3)f(x2)]8x1x20,x3x20,f(x1)f(x2)0,f(x3)f(x2)0BABC0,B(,)29BABC即：(x1x2)2[f(x1)f(x2)]2(x3x2)2[f(x3)f(x2)]2x2x1x3x2[f(x1)f(x2)]2[f(x3)f(x2)]22f(x2)f(x1)f(x3)2aln(1ex2)2(a1)x2a[ln(1ex1)(1ex3)(a1)(x1x3)2aln(1ex2)2(a1)x2a[ln(1ex1)(1ex3)2(a1