2024届高三定时训练3(−2,0]2369ea=4A={x|5x7}B={x|x3x5}AB={x|5x7}a22aa−1+1A=ABa22aa−1+1A2a−13a+15a2a2a=2a4{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}aaa|2a4fx(x)(x3a=)(−)−a3fx()0(,3)aa=3fx()0a3fx()0(3,)af(x)=(x−3)(x−a)9x[4,+)fx()−9ax+x−3999x+=x−3++32(x−3)+3=9x−3x−3x−39x−=3x=6x−3a9a{aa|9}f(x)=3sin2x+cos2x+131=2(sin2xx+cos2)+122π=2sin(2x+)+16{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}ππ3π2kxkπ+++22πkZ262π2πkπxk++π63π2π[kkπ++,π]kZ63π1fA()=2，sin(2)A+=62π5ππA(0,π)2A+=A=663a=23a2=b2+c2−2bccosAb22+c−bc=12bc=b22+c−122bc−12bc12bc==2313S=bcsinA=bc3324△ABC33fx()Rx0f(x)=x2exx0时−x0f(x)=−f(−x)=−[(−x)22e−−xx]=−xe−xx2e−x,0fx()=2xxxe,0f(ax22−31)x−+ax−31x−−f(5−ax)+ax−5{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}xRfaxxaxxfaxax(31)3122−−+−−−+−(5)5xRh()()x=+fxxhx()Raxx2a−x−315−xRaxa2x−+(3+)40a0a0(36+aa)2−1019aa(1,9)22n=122aa11=+a1=22n222anSn=+ann，2可得2Sn(Sn−Sn−−11)=(Sn−Sn)+2n22Snn−=S−12n222222222Sn=S1+()()()S2−S1+S3−S2++Sn−Sn+1=n+n2Sn=n+n=n(n+1)an=Sn−Sn−1=n(n+1)−(n−1)n............................................5分a1=2{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}annnnn=+−−(1)(1)annnnn+1=++−+(1)(2)(1)212nn+故aa−=−nn+1nnnn++++−21121(11)2(2)nnnnnn+++−−++=(2)(11)nnnn++++−2(1)21222nnnnn++−−+−22=(2)(11)nnnn++++−2(1+n22−1)−2n+2n=...............................................8分(n+2+n)(n+1+n−1)(1+n2−1)2=n2+2n2−1n2+2n=(n2+2n)21+n22−1n+2naann+1−0anamn−22am1−22m21fx()(0，+)f()x=+mxm0fx()01m0x−(0,)fx()0m{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}1x−(,+)fx()0fx()m11x=−ln()−=0mmm=−1xfx(g)(x)ln1x+m+1e−x(0，)+xln1x+xx2ex+ln()xe=−x()x=xx21q(x)=+x2exlnxq(x)=(x2+2x)ex+0x1qx()q()0q(1)022x0x0xx00e+=ln0xx00lnx0−lnxx00e=−=−lnex0xh(x)=xeh(x00)=−h(lnx)xx00=−ln()x(0,x0)(,)x0+x0lnxx00+−111(xx)(0)=e−=+=1x0x0x0m(−,0]{#{QQABaYCAggCAAAIAAAhCUwXSCkGQkBCCAKoGBEAIoAABQANABAA=}#}