綜合測(cè)評(píng)一、單項(xiàng)選擇題(本題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.數(shù)列{an}的通項(xiàng)公式為an=2n+1,則第9項(xiàng)a9=()A.9 B.13 C.17 D.192.已知數(shù)列{an}是首項(xiàng)為3,公差為d(d∈N*)的等差數(shù)列,若2019是該數(shù)列中的項(xiàng),則公差d不可能是()A.2 B.3 C.4 D.53.已知{an}是等差數(shù)列,Sn是其前n項(xiàng)和,若公差d<0,且S2=S7,則下列結(jié)論不正確的是()A.S4=S5 B.S9=0 C.a5=0 D.S2+S7=S4+S54.曲線y=xlnx在點(diǎn)(e,e)處的切線方程為()A.y=2xe B.y=2xe C.y=2x+e D.y=x15.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a1+a4=94,S6=9S3.若bn=log2an,則數(shù)列{bn}的前10項(xiàng)和是(A.35 B.25 C.25 D.356.已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且滿足f(x)=2xf'(1)+lnx,則f'(1)=()A.e B.1 C.1 D.e7.函數(shù)y=x3x的導(dǎo)函數(shù)的單調(diào)遞增區(qū)間為()A.(0,+∞) B.(∞,1)C.33,33 D.(1,+8.已知可導(dǎo)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),若對(duì)任意的x∈R,都有f(x)>f'(x)+1,且函數(shù)y=f(x)2021為奇函數(shù),則不等式f(x)2020ex<1的解集為()A.(0,+∞) B.(∞,0) C.∞,1e D.1e,+∞二、多項(xiàng)選擇題(本題共4小題,每小題5分,共20分.在每小題給出的選項(xiàng)中,有多項(xiàng)符合題目要求.全部選對(duì)的得5分,部分選對(duì)的得2分,有選錯(cuò)的得0分)9.已知數(shù)列{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,且2a3+3a7=6A.2 B.4 C.85 D.10.下列函數(shù)存在極值點(diǎn)的是()A.y=x1x B.y=2|x|C.y=2x3x D.y=xlnx11.若數(shù)列{an}滿足:對(duì)任意正整數(shù)n,{an+1an}為遞減數(shù)列,則稱數(shù)列{an}為“差遞減數(shù)列”.則下列數(shù)列是“差遞減數(shù)列”的有()A.an=3n B.an=n2+1 C.an=n D.an=lnn12.下列四個(gè)說法正確的是()A.當(dāng)x>0且x≠1時(shí),有l(wèi)nx+1lnxB.函數(shù)f(x)=lg(ax+1)的定義域是xx>1aC.函數(shù)f(x)=exx2在x=2處取得極大值D.圓x2+y210x+4y5=0上任意一點(diǎn)M關(guān)于直線axy5a2=0的對(duì)稱點(diǎn)M'也在該圓上三、填空題(本題共4小題,每小題5分,共20分)13.已知函數(shù)f(x)=x3+ax2+bx+a2在x=1處取得極值10,則f(2)的值為.

14.函數(shù)f(x)=x2cosx在區(qū)間[0,π]上的最大值為.

15.已知等比數(shù)列{an}的各項(xiàng)都為正數(shù),且a3,12a5,a4成等差數(shù)列,則a4+a16.在數(shù)列{an}中,an=nsinnπ2+cosnπ2,前n項(xiàng)和為Sn,則a4=,S100=四、解答題(本題共6小題,共70分.解答應(yīng)寫出文字說明、證明過程或演算步驟)17.(本小題滿分10分)(2021河北石家莊模擬)在①a5=6,a1+S3=50,②S12>S9,a2+a21<0,③S9>0,S10<0這三個(gè)條件中任選一個(gè),補(bǔ)充在下面問題中并解決問題.問題:設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若,判斷Sn是否存在最大值,若存在,求出Sn取最大值時(shí)n的值;若不存在,說明理由.

18.(本小題滿分12分)(2021江蘇南京檢測(cè))已知數(shù)列{an}滿足a1=3,a2=5,且2an+2=3an+1an,n∈N*.(1)設(shè)bn=an+1an,求證:數(shù)列{bn}是等比數(shù)列;(2)若數(shù)列{an}滿足an≤m(n∈N*),求實(shí)數(shù)m的取值范圍.19.(本小題滿分12分)已知函數(shù)f(x)=ex2x.(1)求曲線y=f(x)在點(diǎn)(0,1)處的切線方程;(2)求函數(shù)f(x)在[0,2]上的最大值與最小值.20.(本小題滿分12分)(2021陜西西安檢測(cè))已知函數(shù)f(x)=2lnx+ax(1)若函數(shù)f(x)有極值,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a=1時(shí),若函數(shù)f(x)在x=x1,x=x2(x1≠x2)處導(dǎo)數(shù)相等,證明:f(x1)+f(x2)>2.21.(本小題滿分12分)在數(shù)列{an}中,前n項(xiàng)和為Sn,且Sn=n(n+1)2.記Tn為等比數(shù)列{bn}的前n項(xiàng)和,且b2+b4=20,(1)求數(shù)列{an}和{bn}的通項(xiàng)公式.(2)記a1b1+a2b2+…+anbn=Hn,是否存在m,n∈N*,使得Hn=am?22.(本小題滿分12分)已知函數(shù)f(x)=lnx2ax,a∈R.(1)討論函數(shù)f(x)的單調(diào)性;(2)若關(guān)于x的不等式f(x)+ex≥e2a在[1,+∞)上恒成立,求實(shí)數(shù)a的取值范圍.參考答案綜合測(cè)評(píng)1.D由通項(xiàng)公式,得a9=2×9+1=19,故選D.2.Dan=3+(n1)d,令2019=3+(n1)d,則n=2016d+1,∵n∈N*,d∈N*,∴d是2016的約數(shù),故d不可能是5,3.D∵S2=S7,∴S7S2=a3+a4+a5+a6+a7=5a5=0,∴a5=0,∴S4=S5,S9=9(a1+a9)又d<0,∴當(dāng)n=4或n=5時(shí),Sn最大,即S2+S7<S4+S5.故選D.4.Ay'=lnx+1,則曲線在點(diǎn)(e,e)處的切線斜率為lne+1=2,所以切線方程為ye=2(xe),即y=2xe,故選A.5.C設(shè)等比數(shù)列{an}的公比為q.由題意知q≠1,則a1(所以an=14×2n1=2n3,所以bn=n所以數(shù)列{bn}的前10項(xiàng)和T10=10(b1+b10)2=5×(26.B∵f(x)=2xf'(1)+lnx,∴f'(x)=2f'(1)+1x,令x=1,得f'(1)=2f'(1)+1,∴f'(1)=17.A由y=x3x,得y'=3x21.令f(x)=y'=3x21,則f'(x)=6x,由f'(x)=6x>0,得x>0.所以函數(shù)f(x)=y'=3x21的單調(diào)遞增區(qū)間為(0,+∞).8.A構(gòu)造函數(shù)g(x)=f(x)-1ex,則g'(x)=f'(x)-f(x)由于函數(shù)y=f(x)2021為R上的奇函數(shù),則f(0)2021=0,則f(0)=2021,所以g(0)=f(0)-1e由f(x)2020ex<1,得f(x)1<2020ex,即f(x)-1ex<2020,所以g由于函數(shù)y=g(x)在R上單調(diào)遞減,因此x>0,故選A.9.ABD∵a3>0,a7>0,∴6=2a3+3a7≥22a3·3a又a5>0,∴上式可化為a5≥2,當(dāng)且僅當(dāng)3a3=2a7時(shí),等號(hào)成立.故選ABD.10.BD對(duì)于A,求導(dǎo)得y'=1+1x2>0,函數(shù)在(∞,0)和(0,+∞)上單調(diào)遞增,所以函數(shù)無極值點(diǎn);對(duì)于B,x=0是函數(shù)的極小值點(diǎn);對(duì)于C,求導(dǎo)得y'=6x21<0恒成立,函數(shù)在R上單調(diào)遞減,所以函數(shù)無極值點(diǎn);對(duì)于D,求導(dǎo)得y'=1+lnx,當(dāng)x∈0,1e時(shí),y'<0,當(dāng)x∈1e,+∞時(shí),y'>0,當(dāng)x=1e時(shí),y'=0,所以x=1e11.CD對(duì)于A,∵an+1an=3(n+1)3n=3,∴數(shù)列{an}不為“差遞減數(shù)列”;對(duì)于B,∵an+1an=(n+1)2+1n21=2n+1,∴{an}不為“差遞減數(shù)列”;對(duì)于C,∵an+1an=n+1-n=1n+1+n,∴數(shù)列{an}為“差遞減數(shù)列”;對(duì)于D,an+1an=lnn+1n∴數(shù)列{an}為“差遞減數(shù)列”.故選CD.12.CD當(dāng)0<x<1時(shí),lnx<0,則lnx+1lnx<0,故A錯(cuò)誤;要使f(x)=lg(ax+1)有意義,當(dāng)a=0時(shí),f(x)=0的定義域?yàn)镽,當(dāng)a≠0時(shí),則ax+1>0,分a>0和a<0兩種情況,求得x>1a或x<1a,故B錯(cuò)誤;f(x)=exx2,則f'(x)=xex(2x),令f'(x)=0,解得x=0或x=2,所以當(dāng)x<0時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)0<x<2時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x>2時(shí),f'(x)<0,f(x)單調(diào)遞減,故函數(shù)f(x)=exx2在x=2處取得極大值,故C正確;要證明圓上任一點(diǎn)M關(guān)于直線對(duì)稱點(diǎn)M'也在圓上,即要證直線過圓心,x2+y210x+4y5=0可化為(x5)2+(y+2)2=34,圓心為(5,2),代入直線方程得5a+25a2=0,圓心在直線上,故D正確13.18f'(x)=3x2+2ax+b,由題意得f(1)=10,f'(1)=0,即a2+a+b+1=10,2a+b+3=0,解得a=414.π+2f(x)=x2cosx,則f'(x)=1+2sinx.∵x∈[0,π],∴1+2sinx>0,∴函數(shù)f(x)在[0,π]上單調(diào)遞增,∴f(x)的最大值為f(π)=π+2.15.1+52設(shè){an}的公比為q,則q>由a3,12a5,a4成等差數(shù)列可得a5=a3+a4,即a1q4=a1q2+a1q3,即q2q1=0,解得q=1+52(負(fù)數(shù)舍去則a4+a16.40易知a1=1,a2=2,a3=3,a4=4,∴a1+a2+a3+a4=0.又sinnπ2+cosn∴a4n+1+a4n+2+a4n+3+a4n+4=0,∴S100=0.17.解若選①a5=6,a1+S3=50,設(shè)等差數(shù)列{an}的公差為d,則a5=a1+4d=6所以前n項(xiàng)和為Sn=14nn(n1)=n2+15n,所以當(dāng)n=7或8時(shí),Sn取得最大值.若選②S12>S9,a2+a21<0,由S12S9=a10+a11+a12=3a11>0,解得a11>0;由a2+a21=a11+a12<0,所以a12<0,所以等差數(shù)列{an}的公差d=a12a11<0,所以n≤11時(shí),an>0,n≥12時(shí),an<0,所以n=11時(shí),Sn取得最大值.若選③S9>0,S10<0,由S9=9(a1+a9)2=9a由S10=10(a1+a10)2=得a5+a6<0,所以a6<0.所以等差數(shù)列{an}的公差d=a6a5<0,所以當(dāng)n≤5時(shí),an>0,當(dāng)n≥6時(shí),an<0,所以n=5時(shí),Sn取得最大值.18.(1)證明由題知,2an+22an+1=an+1an,即bn+1=12bn,且b1=a2a1=53=則數(shù)列{bn}是以2為首項(xiàng),12為公比的等比數(shù)列(2)解由(1)知bn=an+1an=12則當(dāng)n≥2時(shí),數(shù)列{bn}的前n1項(xiàng)和Sn1=a2a1+a3a2+…+anan1=ana1=2-12n則an=712n-3,n≥2,當(dāng)n=1時(shí),a1則由指數(shù)函數(shù)單調(diào)性知,an=712n若滿足an≤m(n∈N*),則m≥7,即實(shí)數(shù)m的取值范圍是[7,+∞).19.解(1)函數(shù)f(x)=ex2x的導(dǎo)數(shù)為f'(x)=ex2,可得y=f(x)在點(diǎn)(0,1)處的切線的斜率為12=1,則曲線y=f(x)在點(diǎn)(0,1)處的切線方程為y+x1=0.(2)令f'(x)=ex2=0,得x=ln2,則當(dāng)0<x<ln2時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)ln2<x<2時(shí),f'(x)>0,f(x)單調(diào)遞增.因此x=ln2為f(x)的極小值點(diǎn),也是最小值點(diǎn),又f(0)=1,f(2)=e24,f(ln2)=22ln2,所以f(x)在[0,2]上的最小值為22ln2,最大值為e24.20.(1)解f(x)=2lnx+ax,x∈(0,+∞),f'(x)=2x-ax2=2x-ax2,當(dāng)a≤0時(shí),f'(x)>0恒成立,即f(x)在(0,+∞)上單調(diào)遞增,當(dāng)a>0時(shí),當(dāng)x∈a2,+∞時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x∈0,a2時(shí),f'(x)<0,f(x)單調(diào)遞減,∴當(dāng)x=a2時(shí),函數(shù)f(x)取得極小值,∴a>0(2)證明當(dāng)a=1時(shí),f(x)=2lnx+1x,f'(x)=2x-1x2,f'(x1∴2x1-1x1∵x1≠x2,化簡(jiǎn)可得2x1x2(x1+x2)=0,f(x1)+f(x2)=2lnx1+1x1+2lnx2+1x2=2ln(x2x2)+x1+x2x∵x1+x2=2x1x2,由x1·x2>0且x1≠x2可得x1+x2>2x1x2,∴2x1x2>2x1x2,即∴f(x1)+f(x2)>2.21.解(1)當(dāng)n=1時(shí),a1=S1=1;當(dāng)n≥2時(shí),an=SnSn1=n(n+1)2-n(n-1)2=n,對(duì)n=設(shè)等比數(shù)列{bn}的公比為q,由b2+b4=20,T4=30,可得q≠1,則b1+b3=10,q=b2+又b1q+b1q3=20,解得b1=2,所以bn=2n.(2)存在.由(1)得,anbn=n·12則Hn=1×12+2×14+3×18+…+n1212Hn=1×14+2×18+3×116+…+n12兩式相減可得12Hn=12+14+18+…+12nn12n+1=12(1可得Hn=2(n+2)·12n<2.假設(shè)存在m,n∈N*,使得Hn=am,可得2(n+2)·12n=m,則m=1,解得n=2.故當(dāng)m=1,n=2時(shí),Hn=am.22.解(1)依題意,f(x)定義域?yàn)?0,+∞),f'(x)=1x2a=1-2axx,當(dāng)a≤0時(shí),12ax>0,f'函數(shù)f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),函數(shù)f(x)在0,12a上單調(diào)遞增,在12a,+∞上單調(diào)遞減.(2)由題意得,當(dāng)x≥1時(shí),lnx+ex2ax+2ae≥0恒成立.令h(x)=lnx+ex2ax+2ae,則h'(x)=1x+ex2a令φ(x)=1x+ex2a,則φ'(x)=ex1因?yàn)閤≥1,所以ex≥e,1x2≤1,所以φ'(x)所以φ(x)在[1,+∞)上單