新高考數(shù)學(xué)模擬試卷(含答案)

新高考數(shù)學(xué)模擬試卷(含答案)）．）．(x)x2x3xx13f532xv3）為ABCE1．．41若,,,pqm()n）與331fxx0fxx22x1gtt2t1與．③x0④fx在(x)(x3)ea(2lnxx1))()f在x(1,2)a．(,)）．．22221112321.xAyxy0.7x0.35）34546t4.5,3.5）．t．A1噸fx0f2f3f3f2x22,x0．．.i211,f(x).在a32332．S．nnnn6．33PQP+Qa+bP+Q.合泳泳計男生女生合計1．Pkbc)2K2a，b，c求A；cC.ABCb3，c.，ABCDAB//DC，ABADP2．PBD；PBCPCABMPBD32【參考答案】***試卷處理標(biāo)記，請不要刪除CC.2．BB613,2442張，總的方法數(shù)是C241．B33．DDfxx3xx1?x0x2x3x1x1,2x532ν10ν30311ν332112ν1133363D4．AA.,;12,;B01100110C26440110144由分類計數(shù)原理與信息011066．AA11223131.444411111122242411131，244443144.pq--mn,∴mn,mn∴.,0fx,02，fxx2xx3fx2xx2x與fxx2x3fxxfxx與gxxgx2x2x與11fx|x00gx與xfxx01gx1對，x0x0fxx22x1gtt2t1與.C.9．CCa在)fxx()(()fxxxgxxe)xea0在2xxag1eag2e且(1,2)()2fx在0(1,2)(1,2)gxxex在fx在axe在x)(2)2ega2f(x)(x3)ea(2lnxx1)，x2aax()((()(()，fxexxeaxexxxxxxfx)在a0x)fx((x)0在則x)即xea02xgxxe()(1)0,(1)x令xgxx)ex，gxxex在ag1eag2e2且，fx0(1,2)在f(x)在(1,2)xexa()0(1,2)在(1,2)即(xxea0在xx即axe在(1,2)xgxxex在)2ag(2)2e，aa2ea(2e,)22().10．DD是E是F1112EDDAABBFABADABADABAD∴EF。2323。D11．DD3456x=，4y∵，∴y，，．D12．BB0f3f2f32f,f23,f3.和fx2在xx3fx由0f3f2，f3f2f3f2，f3,3f3f2,2f和323322ffff，0f3f3f2f2..B.二、填空題13．2【解析】【詳解】當(dāng)x≤0時由x）=x2﹣2=0解得x=有1個零點(diǎn)；當(dāng)x＞0函數(shù)（x）=2x﹣6+lnx單調(diào)遞增則（1）＜（3）＞0此時函數(shù)（x）只有一個零點(diǎn)所以共有2個零點(diǎn)故答案為：2當(dāng)21當(dāng)則2．(fx[，ab(fafxfx((fxfxhxgxfxhxx(((和((0(＝(＝(和＝(fxyhxygx點(diǎn)：復(fù)數(shù)的運(yùn)算解析：2，..1(,)9212xf(x)的fxx342211220aaa．fa991366y，2,kkZ，662k,kZ}.664形．2a1Snna2aana,Sn11nS.S2a1Snnnnaa1111a2663.1261.=33199.9%3）.35K2322名學(xué)，11(A,B)(A,B)，(A,B)1(A,B)，1112n21(A,B)…2n3123n133．264⊥．N12MN13線與與所DM2613．2633sinCDM．44，⊥=．與221MNcosDMN22613．26M⊥，=3在△2=ACAD2．3在△．43．4bcA3由c3asinCccosA3sinAsinCcosAsinCsinC10CsinA，62A又0)A.312ABCbcbcsinA3故=，S=,c而abcbccosA故cbb22222（⊥1⊥2λ和1λ2ABCD2,且AB//DC,,2BD22，242BC平面PBC平面PBDABCD,PBD設(shè)E為BDPEPBPD6,BD,PE2，PBDABCDPBDBD，x,y,zABCD，D(2,0