2021年新高考数学全国卷模拟试卷(附参考答案和详解)

绝密★启用前年普通高等学校招生模拟考试（2）数学（适用新高考地区）分★祝考试顺利★IA第I卷85{x||x1|2},B{y|y2,x[0,2]}AB（）Ax(1,3)[1,3)R，iai2i(abi)2（），ba3344AB(4,，AB）34,43,3443,,55555555f(x)ln(x1)2）πππ,f(x)sinx上单调递减，则（）332322332P(3,1)lxy1l）22ππππ63630）49132919a21blog1，c1，3）32312abcacbcabcba45530AC，B5C）0CC5个x2y2x2y20k911k（）9k25k9））x00xa1πy2xπx126l:xy20C:(x1)(y1)2C2222P:xR,xx10xx10"，220P00xR,xx10"2第Ⅱ卷.45a5aRx7ax2．xABCDABCD1，E，FAA，BC111111DEDF．1x2y2a2b2．C6a}1，33．1n）求数列a}n）若数列a}kS35kkn22，c2，A在ABCA，B，Ca，b，ca．4Cb和π2A31mm50005050mm(3,4](1,3]20，BACAB'''90ABAC''BC''的M，AB和ABC，N平面ACC.MN'-MNC6x2y2C:ab0)F(．3a2b2CO为坐标原点，Tx3F作TFP，QOPTQOPTQ(x)ax3bxcx2c16．f，ab28f(x)[在f(x)参考答案与解析|x1|213xy2,x[0,2]x1yAB([1,3)．a2b1，ai2i，(abi)(2i)34i．则22，(3,4)A(1,3)B4,1134,．555f(x)ln(x1)2f(x)ln[(x)1]ln(x1)f(x)f(x)22ππ2π,kZ32xπ343π2π4π,T30.32lx331，lyk(x3)1d|k1|10k3lπD．321k542055255105P202590a1b0c1．，，.A0CABCCCD2x2y20k919kk59k2259k234k5x2y2125k225k9234k325k934k25k与πxay0xay0a1Ax1212|2Cy0BCl22D.2333273182CA18有2323．273ra2TC(x),)rrr525rxr1xr)r7r1Ca10a2.151EAA61112，S112DED1FBC1FDED1h，11111h13216．所以VVS31EDFFDED1DED1EAFC11116．则VVS3DD111321EDF1ADCADC122yx30．2设A(x,y)B(x,y)，，11222yx30,(4ba)y12bybab0，由x2222222bxayab222222b2yy．ba2122Myy2．12b2故2，ba22ba2212．b212，ee21a22．2A(x,y)B(x,y)，．1122Mx2yy2x，．1212x22y21A，Bab2x22y212①1abx222y222②abx21x22y21y22①②得，a2b2yybxxb2．21212xxayy1a221221，2ba221b2212e．a22.a}daa(nd.nn1d123.由a1a3，13d2,a32.n(32n)]32nS2.nna2nnS352kk35,2kkk5.又kN*k7.214ABCcosAsinA．44ac7及a2，c2sinC．sinAsinC4由a2bccos,22得b2b20,b0b1C7,b424A，sinA43cos2A2cos2A1,47AAA.4πππ321cos2Acos2cossin2sin.3338[3,2)58(1,3]0.50.20.7.50002050201980'AC'AB，．ABCAB'''M为AB'N为BC''MNAC．'又，MNACC'AC'ACC'，平面AACC'．MN''ABMP，NP．PM，NAB与BC'''ABCAB''''PNC'MPAA，，，MPACC'PNACC'，又MPNPPMPNAACC，平面''而MNMPNMN平面AACC'．'y-AAB，AC，AA为Oxyzxz设AA'1ABACB((0,'(0,CABC1M,N,.2222'AMNAM'00得，MN设mx,y,zmm1111xz22111yz2211.mMNC设nx,y,z222由，NC0nMN0得nxyz222221yz0.2222n.'-MNC0mn3(1)(1)0,22.6,cca3aa6,bc2b2,22x2y21．62T(3,m)，m0TFk.3(TF0PQk1当m，mPQPQxmy2,0PQx2,当mxmy2将xmy2(my422216m8(m22设P(x,y)Q(x,y)，11224m2yy,yy,3m312m212212xxm(yy)4.31212m2OPTQxy(,)(3,).xmy1122xx.312m24myy312m2m1.OPTQ1S2S2|OF||yy|2OPTQOPQ12222423mm322(x)axbxc，f3()3,故fxax2bf(x)x2c16，ffc即12ab8abcc化简得12ab0,4ab8,a1,b12.f(x)x12xc,3()3123(2)(2).fxx2xx()0令fxx