2022屆甘肅省蘭州市五十五中高考適應(yīng)性考試數(shù)學(xué)試卷含解析

2021-2022高考數(shù)學(xué)模擬試卷053c（）ca，，ABCA，，a14cAC，ABCbBC2．．4．5．3222{}＝，+＝λaaa）aan147212．．．．ABCDABCDP,Q2O，ABAD111111）2．．21．．122xA(x,y)xy4,B(x,y)y2AB(．4()22．1．2．3alog0.5blog0.3，，c20.3a,b,c)30.2bccabcba．．a(chǎn)．a(chǎn)cb．A,A1B,B,BA，B21232件A3件BBA）123523．．．．5f(x2)f(x)在，2f(0)0f(23x)022．，))．，((33222323．，．)()()(3311123111a}nlim()lim()（）aa1aanaan32n1242n23．．．x15x16x18x20，，，，1x22x24x25，，）SS6．i6，S7S．i6，S7S．i，S．i6S77251130723，，如16）1．12d0d=15a）0)nd45CC,且abab或alog0.2,blog0.222fxfx,fx,fx13123kaba，ba．1ab02E:2M,N.PABC120，ABAEED2EF，ABCDEFEFAB.EB；ACFB.2ab2bc2aabcabc，，C，，，，SAB4S2223P1,0（、MNCC:y2FCPPAPFa5OMOAMAOAOFOPA（OPM1(a1)xx,aRf(x)2．20(2,f(2))（af(x)（f(x)參考答案5D354512324aCCabsinC4cAsinC,cosCSABCb5c.C4csinAacosC4cA4sinAsinC3sinAcosC4sinC3cosC.34sinC,cosC.sinCcosC122551133SabsinC1bb5.ABC22524cab2abcosC1521518c3222225C.D18dλ∈λ19daan14，15＝1λλ．．C于RCDR112＝2CD11，2∴∥＝＝，222＝，22∴＝2B.2yABx點ABAB為，.Ax，0在log0.5log10,在0log1log0.3log0.21，y在,xR0..DAB.A,A1B,B,B，A，B21233；B521424A；31352A；B.Df(x2)f(x)2f(x)xx和，23x223x2.f(x2)f(x)2f得(0)0f(4)0；，2在f(x)f(x)又在0即x23x2f(23x)0得f3x)f，23x42x；30f(23x)0得f當(dāng)即x23x23x)f，23x023x；223x)0f(2)().(33.Aa1a21{}1。2a}nan111123a23lim(aa)2a42a1由，1a1n1132n14111114lim()。14aaa3n1242nA17xxxS222.12717xxx7S222，127S6i，S，7.A20.20，3，5，，，13，17，819271182C28203,17，7,1322120P．A..Dda,ad.25a2d,a24d,a,a,a(2d)22(24d).d425125.A.aaaa2d1aad14a,a,aa2d0.由2d13631611A..45x2y50,且1131(x3)x2y5Cy31.14．＞11a1,b0ab與1.11a1,b00,ablog0.3log2log0.61ab，abb.a＞..1,42fxfxfxx,x,x1,fx將12323k1fxfxfxk.,k123x,x,x,fx,fx,fx,123123fxfxfxx,x,x1故,12323k1xfxx21x1k1111令tx13,xx121x2x1xxk11t3則y,tk210當(dāng)k即k13,則y；3,y1,1k1當(dāng)k即k210當(dāng)k即k13,則y,,34k22k,1fx當(dāng)k1,時2fxfx33123k22解得1k4；3當(dāng)k1時,fxfx1,fx1232k4k2fx11時,fxfx2當(dāng)k且,331232k411k1,,312k4,21,4:2.1212abm．b(1,1)2(m1(0aay133x22,,2.2243QBPQQE2ykx2MONOMON0lABPa2，B0，PQ:QB3:2設(shè)Qx,y由QQ3QB得PQ2654xQ則yQ5b21x2Ey14ykx2lMx,yNx,y設(shè)由11222ykx2得14kx0x22y124lE16k41214k0即223得k2416kxx14k122又xx1214k2MONcosMON0OMON0xxyy01212xxyyxxkx2kx21kxxkxx4021212121212121216kk2k40即1214k14k22k4②233k22k或22332,,2l225AC,OEM,OM,OFBDOFC.Mx,y,z,OB,OF為n(BCFACFm.AC,MEM,OM,OF.OBDABCDAC.EDEMAD.AEEADABCD，EBEADABCD,EMBD11,AB,OMEF,OMEF,OM,EF,EF22EMOFOMEF.OF＝ACOFOBDOFC，BD又FCOFCBDFC.,OB,OF,OB,OF.為軸3zy0n0，BCFn(.取zACFm..b21222222bc222．將abc422bc又2212SA22故．b222222bc則tanA1A，A451BABC222.則AS212c2b4S4Abca2221bca222222．sinAcosAtanA10A180A45S2bcx2y42）1）．323（x2y2（1t32PMPNttPMPN.據(jù)12x（（C2yx2yx2y2211．C223232x1tcos（（tytsinx2y21，321t3tt40，36得222224tt設(shè)，t，tMN，3212124．當(dāng)0PMPN3.9133）a或）4294DPA1D,,（aAFCxP9a.AFC4Pt,6,3,//xMOMAMP設(shè)Mt（MAMP.9F1,01AFCa1D，4xPDPA,,則a15a4;9AFCa4,P,F則2a3PFa1352a3或4.2MOAMAOAOF，//xMOMA，MAPt,6,3t29,C設(shè)Mt313，MAMPMO219132MAyS2OPAp.x2y22ln20(0,1))）20,()a時fx在在a時fx01,()11在(0,1)和,在1,(0,)(x)在在1時,f1,f(x)時aaaa11)在,1.和aa.(axx1))且f(x)(0,,0a1和a1與a1a0x.1112xx,f(x)10（a時,f(x)f1.x212f(2)2ln2f(x)(2,f(2))y(2(x,又x2y22ln20.即1ax(axlnx(0,).（）f(x)2212(a1)x1(1)(x(x)(a.fxxxx,f(x)0時x(1,),f(x)0,a0時,10時,)在(0,1)f(x)在111a1時,1x和,,f(x)00時x,fx0,時aaa11,1,f(x)在a1時,(0,1)和,在aa1