甘肃省兰州市城关区兰州第一中学2021-2022学年高考考前提分数学仿真卷含解析

2021-2022高考数学模拟试卷05Mx|x1,Nx|x21．设全集U（）2．．1）377．．991320“3xxxx1m40n）．8．9）?．i213（a0b0C）22f(x)使xa30且x1mn2）．[2,4]UxZx1x30＝)U．．．．A0,023）．．2,0)(0,2]在[)f(x)x．．．．）．．．．30ll与2y）xxx2222．132222234物1到这5372），AB．53axdx20xyx1log(aa)log(aa)a的各项都是正数，且a,a,4a．n2231234245y2aCPC:12412（C（l:x与Fk0,ABEFME22BNNAMlC1xC点O3．2C1,BAOB12331A22212）BC231D13B（，，D.XX23sinxcosx2cosx1.fx2，，，，ABCabc，，200元200400元4001.0元/（xy1月份100100126080%a，b100Y1月份的用电费用，求Y.⊥△⊥（；参考答案5AM.NMUUA13coscoscoscossinsin444442222111127112.22222222223.C.nn1m40n112m1m，，m；2；nn213m1，mm25；314m1，mm2n415，m1mm35116；8n516m1，mm；28n617，m1mm4；242n718，m1mm2；2n819，m1mm1；2n9110，m1的值为10.n.B由30020010203040由30020010203040，i401050i40？．BAcca,ebb2.22a0l:bxay0F,0bc2到lbacF222ba22c22a2b2ccc2239a又2aa92c3222cc293cAa,b,c,eb.Df(1)0m10n2f(x)由0,24ax12a.xa302,x1m1|1n10n2ex2f(x)x10,2ax3(x1)2(x1)4422a30x12x2x1x1x10,240,2[2,3]2.2a3yx1x1．Ax1x301x3UCA由UT5226,0,0,3626T52T，26622T2，gAsin20，又66kkZ02，32或，33g03，又A0A23，，，A2，23gx2sin2x即，3yfxygx32yfx2sin2x2sin2x.33A.Asin(x)(x)cos(x)xxxfx轴22yf(x)；f(x)xx22而f()0Bf)0；5A..D111tis3,t1,ist1,is3361111,it121231211111122311t12(1)12123121).A33300lyxcxyyc3333lyxcxccbacbbbb222222333项.AA.C被5372n.被5372，57ana23n1n则令n2a35n12202058n.n35585758233559189.24513、{xR2x{xR|xB{，AB．A2a551axx|，3233015即a2．axyxzy2xy2xz为，yBB1z121，z1．：（.2.1aa,a,4an2231aa4a则，321aq4aaq，2111q40q2，q4q1或log(aa)log(aa)234245aalog34aa2245aqaq123411aqaqq32111log224，.2.x2y2）12（54PF（(x,y)PFaP12yk(x1),k0A(x,y),B(x,y)（l111222y10k5k22x,xx12yN(5,y)0,M,Nx11，x315k45k04222yy002P(x,y)PF(c,y),PF(c,y)，12a42xycx4c，22222a2122,x[a,a]a．x0PFPF124c3c12，a25x2y21C，54（l1yk(x1),k0，(1),ykx5k)x10kx5k200．(4x2y22222541,10k5k22A(x,y),B(x,y)2,xx12设设xx1，45k45k112222N(5,y)0,M,NM(3,0)yy2yy0101．3x2x31110k5k22kk5k2y2k(xk(xx)xk5k45k而4yyyk(x220111212x3x3x3x302221111lBN//xBN．A(x,y),B(x,y)2，112xx,xx1212）34cos）（C.1（C和CAOB.12x22cosC（y2sin1Cx4xy0．221C4cos1（，34sincos3．3sin，2（kZkk则当当或63k（kZ23，62k（kZ．3BkCC1A2k,.和6321OAOBAOB3S．ABC2故AOB3．.16）281（B77~B(10,)DP(kP(k1)P(k1)k)1、1、1P(kPk)P(k)P(k)DX.111))（B.C12233331117（D，C())C()2323333337设~(10,)，B7k则k)C()()，kk10k7C()()P(kk1k19k7k10.Pk)k7C()()kk10k10707kk207k50由由由1得k1得kk271P(kP(k)，，PPPkk7kk2(1)()1k得PkPkkP(k2k)2.168118A)B，3311)11)()]1C，C121233333327DX，XP81681787E(X).题.17k,k,kZ））63fx2sin2x（.2k,kZ2k2x6262）fB2sin2B2Bb7.（63fx23sinxcosx2cos2x13sin2xcos2x2sin2x（）.6取262k,k,kZ2k2x2k,kZx.63fB2sin2B2,（）611B0,2B,B，故2B，.6663627bac2accosB49，b.222bca5781222222A.2577.0.5x,0x200{0.8x60,200x400x140,x140）a0.0015，b0.0020.）yy260x400:（1Px4000.80a,ba,b;3（）x值YYx200,y,,,.y0.5x0；x400y0.52000.8x2000.8x60；200x400y0.52000.82001.0x400x140与xy0.5x,0x200y{0.8x60,200x400.x140,x140260x400（yPx4000.800.12100b0.30.8a，0.0015b0.0020{100a0.050.2（，，，，，，X50y0.55025Py250.1，当x当x当x当x当x当xPy750.2，150y0.515075Py1400.3，250y0.52000.850140350y0.52000.8150220Py22