新教材2024-2025學(xué)年高中數(shù)學(xué)第一章數(shù)列1數(shù)列的概念及其函數(shù)特性1.2數(shù)列的函數(shù)特性分層作業(yè)北師大版選擇性必修第二冊(cè)

第一章1.2數(shù)列的函數(shù)特性A級(jí)必備學(xué)問(wèn)基礎(chǔ)練1.下列說(shuō)法不正確的是()A.數(shù)列可以用圖象來(lái)表示B.數(shù)列的通項(xiàng)公式不唯一C.數(shù)列中的項(xiàng)可以相等D.數(shù)列可分為遞增數(shù)列、遞減數(shù)列、常數(shù)列三類2.[2024河南焦作高二統(tǒng)考期中]已知數(shù)列{an}的通項(xiàng)公式為an=3n+kn-2,則“k>-1”是“{an}為遞增數(shù)列”的()A.充分不必要條件B.必要不充分條件C.既不充分也不必要的條件D.充要條件3.已知數(shù)列an=(n+1)-1011nA.{an}有最大項(xiàng),但沒(méi)有最小項(xiàng)B.{an}沒(méi)有最大項(xiàng),但有最小項(xiàng)C.{an}既有最大項(xiàng),又有最小項(xiàng)D.{an}既沒(méi)有最大項(xiàng),也沒(méi)有最小項(xiàng)4.在數(shù)列{an}中,a1=2,an=an+1-3,則14是{an}的第項(xiàng).

5.已知數(shù)列2a-1,a-3,3a-5為遞減數(shù)列,則a的取值范圍為.

6.依據(jù)數(shù)列的通項(xiàng)公式,寫(xiě)出數(shù)列的前5項(xiàng),并用圖象表示出來(lái).(1)an=(-1)n+2(n∈N+);(2)an=n+1n(n∈N+7.在數(shù)列{an}中,an=n(n-8)-20,n∈N+,請(qǐng)回答下列問(wèn)題:(1)這個(gè)數(shù)列共有幾項(xiàng)為負(fù)?(2)這個(gè)數(shù)列從第幾項(xiàng)起先遞增?(3)這個(gè)數(shù)列中有無(wú)最小值?若有,求出最小值;若無(wú),請(qǐng)說(shuō)明理由.B級(jí)關(guān)鍵實(shí)力提升練8.已知an=n2-21n2,則數(shù)列{A.第9項(xiàng),第10項(xiàng) B.第10項(xiàng),第11項(xiàng)C.第11項(xiàng),第12項(xiàng) D.第12項(xiàng),第13項(xiàng)9.已知數(shù)列{an}滿意a1=a,an+1=an2-2an+1(n∈N+).若數(shù)列{A.-2 B.-1 C.0 D.(-1)n10.下列說(shuō)法正確的是()A.數(shù)列1,0,-1,-2與數(shù)列-2,-1,0,1是相同的數(shù)列B.數(shù)列0,2,4,6,8,…可記為{2n},n∈N+C.數(shù)列n+1n的第k項(xiàng)為1D.數(shù)列2,6,11.[2024江西南昌第十九中學(xué)階段練習(xí)]已知數(shù)列{an}滿意an=1n+1·20232022n,則當(dāng)aA.2024 B.2023或2022C.2022 D.2022或202112.(多選題)[2024湖南長(zhǎng)沙雅禮中學(xué)統(tǒng)考期末]下列數(shù)列{an}中是遞增數(shù)列的是()A.an=(n-3)2 B.an=-1C.an=tann D.an=lnn13.(多選題)已知函數(shù)f(x)=-x2+2x+1,設(shè)數(shù)列{an}的通項(xiàng)公式為an=f(n)(n∈N+),則此數(shù)列()A.圖象是二次函數(shù)y=-x2+2x+1的圖象B.是遞減數(shù)列C.從第3項(xiàng)往后各項(xiàng)均為負(fù)數(shù)D.有兩項(xiàng)為114.已知數(shù)列{an}的通項(xiàng)公式為an=9n(n+1)115.已知在數(shù)列{an}中,an+1=2anan+2對(duì)隨意正整數(shù)n都成立,且a7=1216.數(shù)列4n-73n17.設(shè)an=-n2+10n+11,則數(shù)列{an}中第項(xiàng)的值最大.

18.已知數(shù)列{an}的通項(xiàng)公式是an=9n2-9n+29(1)推斷98101是不是數(shù)列{an}中的項(xiàng)(2)試推斷數(shù)列{an}中的項(xiàng)是否都在區(qū)間(0,1)內(nèi).(3)在區(qū)間13,23內(nèi)有沒(méi)有數(shù)列{C級(jí)學(xué)科素養(yǎng)創(chuàng)新練19.已知數(shù)列{an}的通項(xiàng)公式為an=n2-11n+an,a5是數(shù)列{an}的最小項(xiàng),則實(shí)數(shù)a的取值范圍是(A.[-40,-25] B.[-40,0]C.[-25,25] D.[-25,0]20.已知數(shù)列{an}的通項(xiàng)公式為an=n-254n-255(n(1)探討數(shù)列{an}的增減性;(2)求數(shù)列{an}的最大項(xiàng)和最小項(xiàng).

參考答案1.2數(shù)列的函數(shù)特性1.D2.A由題意得數(shù)列{an}為遞增數(shù)列等價(jià)于“對(duì)隨意n∈N+,an+1-an>0恒成立”,得3n+1+k(n+1)-2-(3n+kn-2)>0,即k>-2·3n對(duì)隨意n∈N+恒成立,故k>(-2·3n)max=-6,所以“k>-1”是“{an}為遞增數(shù)列”的充分不必要條件.故選A.3.C當(dāng)n=2k,k∈N+時(shí),a2k=(2k+1)10112k,a2(k+1)=(2k+a2(k+1)-a2k=-42當(dāng)k≤4時(shí),a2(k+1)-a2k>0;當(dāng)k≥5時(shí),a2(k+1)-a2k<0,故a10最大.當(dāng)n=2k-1,k∈N+時(shí),a2k-1=-2k·10112k-1,a2(k+1)-1=-a2k+1-a2k-1=42k當(dāng)k≤4時(shí),a2k+1-a2k-1<0;當(dāng)k≥5時(shí),a2k+1-a2k-1>0,故a9最小.綜上,{an}既有最大項(xiàng),又有最小項(xiàng).故選C.4.5a1=2,a2=a1+3=5,a3=a2+3=8,a4=a3+3=11,a5=a4+3=14.5.(-2,1)∵數(shù)列2a-1,a-3,3a-5為遞減數(shù)列,∴2a-1>a-6.解(1)a1=1,a2=3,a3=1,a4=3,a5=1.圖象如圖1.(2)a1=2,a2=32,a3=43,a4=54,a5=67.解(1)因?yàn)閍n=n(n-8)-20=(n+2)(n-10),所以當(dāng)0<n<10,n∈N+時(shí),an<0,所以數(shù)列{an}共有9項(xiàng)為負(fù).(2)因?yàn)閍n+1-an=2n-7,所以當(dāng)an+1-an>0時(shí),n>72,n∈N+故數(shù)列{an}從第4項(xiàng)起先遞增.(3)an=n(n-8)-20=(n-4)2-36,依據(jù)二次函數(shù)的性質(zhì)知,當(dāng)n=4時(shí),an取得最小值-36,即這個(gè)數(shù)列有最小值,最小值為-36.8.B假設(shè)an=an+1,則有n2-21n9.A∵數(shù)列{an}滿意a1=a,an+1=an2-2an∴a2=a2-2a+1.∴a=a2-2a+1,解得10.C對(duì)于A,數(shù)列是有依次的一列數(shù),數(shù)列1,0,-1,-2與數(shù)列-2,-1,0,1依次不同,故A錯(cuò)誤;對(duì)于B,當(dāng)n=1時(shí),a1=2,不符合a1=0,故B錯(cuò)誤;對(duì)于C,數(shù)列n+1n的第k項(xiàng)為k+1k=對(duì)于D,數(shù)列2,6,12,…,110故選C.11.D令bn=1an,則bn+1b∴當(dāng)n>2024時(shí),bn+1bn<1,{bn當(dāng)n<2024時(shí),bn+1bn>1,{bn當(dāng)n=2024時(shí),bn+1bn=1,即b2024=b2024,a故當(dāng)n=2024或n=2024時(shí),{an}取得最小值,最小值為a2024=a2024=20232021故選D.12.BD對(duì)于A,結(jié)合對(duì)應(yīng)函數(shù)y=(x-3)2在(-∞,3)內(nèi)單調(diào)遞減,在(3,+∞)內(nèi)單調(diào)遞增,可知數(shù)列{an}不為遞增數(shù)列;對(duì)于B,結(jié)合對(duì)應(yīng)函數(shù)y=-13x在R上單調(diào)遞增,可知數(shù)列{a對(duì)于C,結(jié)合對(duì)應(yīng)函數(shù)y=tanx的單調(diào)遞增區(qū)間為-π2+kπ,π2對(duì)于D,由于an=lnnn+1=ln1-1n+1,結(jié)合對(duì)應(yīng)函數(shù)y=ln1-故選BD.13.BC∵函數(shù)f(x)=-x2+2x+1,數(shù)列{an}的通項(xiàng)公式為an=f(n)(n∈N+),∴an=-n2+2n+1,對(duì)于選項(xiàng)A,數(shù)列{an}的圖象是當(dāng)n取正整數(shù)時(shí)f(n)=-n2+2n+1的圖象上的對(duì)應(yīng)點(diǎn)的坐標(biāo),∴此數(shù)列圖象不是二次函數(shù)y=-x2+2x+1的圖象,故A錯(cuò)誤;對(duì)于選項(xiàng)B,an=-n2+2n+1=-(n-1)2+2,∴此數(shù)列是遞減數(shù)列,故B正確;對(duì)于選項(xiàng)C,an=-n2+2n+1=-(n-1)2+2,a1=2,a2=1,a3=-2,此數(shù)列是遞減數(shù)列,∴從第3項(xiàng)往后各項(xiàng)均為負(fù)數(shù),故C正確;對(duì)于選項(xiàng)D,an=-n2+2n+1=-(n-1)2+2,a1=2,a2=1,a3=-2,且此數(shù)列是遞減數(shù)列,此數(shù)列有一項(xiàng)為1,故D錯(cuò)誤.故選BC.14.a8和a9∵數(shù)列{an}的通項(xiàng)公式為an=9n∴an+1-an=9n+1(n+2)10n+1-9∴當(dāng)n<8時(shí),an+1>an;當(dāng)n>8時(shí),an+1<an;當(dāng)n=8時(shí),an+1=an,故數(shù)列{an}中的最大項(xiàng)為a8和a9.15.1由已知a7=2a6a6+2=又因?yàn)閍6=2a5a5+2=16.527設(shè)an=4則an+1-an=4(∴當(dāng)n≥3時(shí),an+1<an,當(dāng)n<3時(shí),an+1>an,∴數(shù)列4n-73n中的最大項(xiàng)為a317.5依據(jù)題意,an=-n2+10n+11=-(n-5)2+36,當(dāng)n=5時(shí),an取得最大值.18.解(1)∵an=9n∴由an=3n-23∵1003不是正整數(shù),∴98101不是數(shù)列{an(2)∵an=3n-23n+1=3n+1-33n+1=1-33∴數(shù)列{an}中的項(xiàng)都在區(qū)間(0,1)內(nèi).(3)令13<an<23,即則3n+1<9n-6又n∈N+,∴n=2.故在區(qū)間13,23內(nèi)有數(shù)列{an}中的項(xiàng),且只有一項(xiàng),是第2項(xiàng),a19.D由條件可知,對(duì)隨意的n∈N+,都有an≥a5恒成立,即n2-11n+an≥a5-30,整理得(n-5)(當(dāng)n≤4時(shí),不等式化簡(jiǎn)為a≥5n(n-6)恒成立,當(dāng)n=1時(shí),5n(n-6)取得最大值-25,所以a≥-25,當(dāng)n≥6時(shí),不等式化簡(jiǎn)