n與an-1(n≥2(的關(guān)系；11若bn=求數(shù)列{bn{的前n項(xiàng)和Tn.4.若存在非零常數(shù)t，使得數(shù)列{an{滿足an+1-a1a2a3?an=t(n≥1,n∈N(，則稱數(shù)列{an{為“H(22*.-2k；337.已知冪函數(shù)f(x(=(m2+3m+3(x3m-1為偶函數(shù).(1)求f(x(的解析式；(2)若f(a-1(≥f(1+2a(，求實(shí)數(shù)a的取值范圍.8.已知定義在R上的函數(shù)f(x(=滿足5f(2(=9f(1(，且f(0(=0.(2)證明：f(x(在R上為增函數(shù).9.已知定義在R上的函數(shù)f(x(滿足f(x(-2f(-x(=3x2-5x+2.(1)求f(x(的解析式；若g(x(=-f(x(-2x2+在區(qū)間[a,a+1[內(nèi)有最小值2，求實(shí)數(shù)a的值.4410.已知函數(shù)f(x(=a-11.函數(shù)f(x(=lnx-(2)若函數(shù)f(x(有兩個(gè)極值點(diǎn)x1、x2，曲線y=f(x(上兩點(diǎn)(x1,f(x1((、(x2,f(x2((連線斜率記為k，求5512.若將對(duì)于任意x、y∈R總有f(x+y(+f(x-y(=2n=2f(n+1(-f(n((n∈N*(，求log2+log2+?+log2的值；66+C(=(b-c)?(sinB+sinC),b=、7716.如圖所示，已知函數(shù)f(x(=Asin(ωx+φ((ω>0,|φ|<的圖象(1)求函數(shù)f(x(的解析式；(2)求函數(shù)g(x(=f(x(+2cos(ωx+φ(在區(qū)間內(nèi)的值域.8817.已知函數(shù)f(x(=2sin2+sinπx-1，將函數(shù)f(x(的所有正的零點(diǎn)從小到大排列組成數(shù)列{an{.記數(shù)列{cn{滿足cn=且數(shù)列{cn{的前n項(xiàng)和為Sn，求證：S2n<ln2.99(2)求二面角A-PC-B的大小.20.如圖所示，在三棱柱ABC-A1B1C1中，H是正方形AA1B1B的中心，AA1AA1B1BH=.(1)求異面直線AC與A1B1夾角的余弦值；(2)求平面AA1C1與平面A1C1B1夾角的正弦值.,N是AB的中點(diǎn).斯圓．如圖，在棱長(zhǎng)為3的正方體ABCD-AlBlClDl中，點(diǎn)M是BC的中點(diǎn)，點(diǎn)P是正方體表面DCClDl上一動(dòng)點(diǎn)(包括邊界)，且兩直線APAMN截正方體ABCD-AlBlClDl的截面周長(zhǎng)，若無(wú)，說(shuō)明理由.24.在平面直角坐標(biāo)系xOy中有兩個(gè)定點(diǎn)A(-3,0(，B(3,0(，已知?jiǎng)狱c(diǎn)M在平面xOy中且M到A，B兩i和國(guó)家體育總局體育科學(xué)研究所牽頭組織編制的《中國(guó)人群身體活動(dòng)指南(2021)》(以下簡(jiǎn)稱《指Dn-1(n≥2(之間的遞推關(guān)系，并證明：數(shù)列{Dn-nDn-1{(n≥2(是等比數(shù)列；x=1+x++??)會(huì)知識(shí)宣講團(tuán).①求應(yīng)從[80,90(和[90,100[學(xué)生中分別抽取的學(xué)生人數(shù)；Pn(1≤n≤60(，試證明數(shù)列{Pn-Pn-1{是等比數(shù)列(2≤n≤59(，求出數(shù)列{Pn{(1≤n≤60(的通項(xiàng)公參考數(shù)據(jù)：若隨機(jī)變量ξ服從正態(tài)分布N(μ,σ2(，則P(μ-σ<ξ<μ+σ(=0.6827，P(μ-2σ<ξ<μ+2σ(=0.9545，P(μ-3σ<ξ<μ+3σ(=0.99731.31.已知?jiǎng)狱c(diǎn)P與兩定點(diǎn)A(2,0)，B(-2,0)連線的斜率之積為(2)求△ABD面積的最大值.同的兩點(diǎn)M、N.若|MN|=求直線MN的方程.(3)點(diǎn)M是線段AB的中點(diǎn)，過(guò)點(diǎn)F且與直線l垂直的直線m交直線OM于點(diǎn)P，求三角形PAB面積35.已知點(diǎn)F1(-2,0),F2(2,0)分別為橢圓=1的左、右焦點(diǎn)，經(jīng)過(guò)點(diǎn)F1且傾斜角為量. n與an-1(n≥2(的關(guān)系；an-1+(n≥2(n=an-1+(n≥2(.n=an-1+(n≥2(，設(shè)an+x=(an-1+x(，可得an-1-所以可得，n-(an-1-((n≥2(，且a1-11n-(n-1，即an=-(n-1+是數(shù)列{an{的第10項(xiàng).所以是數(shù)列{an{的第10項(xiàng).3.已知等差數(shù)列{an{滿足a3>a1,a1+a3=10,a1,a2-1,a3成等比數(shù)列.若bn=求數(shù)列{bn{的前n項(xiàng)和Tn.n=3n-122n=利用裂項(xiàng)相消法運(yùn)算求解即可.2-1所以數(shù)列{an{的通項(xiàng)公式an=2+3(n-1(=3n-1.n=4.若存在非零常數(shù)t，使得數(shù)列{an{滿足an+1-a1a2a3?an=t(n≥1,n∈N(，則稱數(shù)列{an{為“H(t(數(shù)(3)要證t>Sn+1-Sn-eS-n等價(jià)于a1a2?an<eS-n，即lna1+lna2+?+lnan<a1+a2+?+an-n，構(gòu)造函數(shù)f(x(=lnx-x+1，利用其單調(diào)性可證明結(jié)論.+1-a1a2a3?an=2，兩式作差可得a+1=a1a2a3?an(an+1-1(+log2bn+1-log2bn，即a+1=a1a2a3?an(an+1-1(+log2q，所以a+1=(an+1-t((an+1-1(+log2q，33即(t+1(an+1=t+log2bn+1-log2bn對(duì)于n≥1,n∈N恒成立，1=1n=2n-1，故所求的t=-1，數(shù)列{bn{的通項(xiàng)公式bn=2n-1.(3)設(shè)函數(shù)f(x(=lnx-x+1，則fI(x(=當(dāng)x>1時(shí)，fI(x(<0，則f(x(=lnx-x+1在區(qū)間(1,+∞(單調(diào)遞減，且f(1(=ln1-1+1=0，又由{an{是“H(t(數(shù)列”，n+1-a1a2a3?an=t，對(duì)于n≥1,n∈N恒成立，2=a1+t>1，再結(jié)合a1>1,t>0,a2>1，反復(fù)利用an+1=a1a2a3?an+t，可得對(duì)于任意的n≥1,n∈N,an>1，則f(an(<f(1(=0，即lnan-an+1<0，則lnan<an-1，即lna1<a1-1,lna2<a2-1,?,lnan<an-1，相加可得lna1+lna2+?+lnan<a1+a2+?+an-n，則ln(a1a2?an(<Sn-n，又因?yàn)閥=lnx在x∈(0,+∞(上單調(diào)遞增，所以a1a2?an<eS-n，又an+1-a1a2a3?an=t，所以an+1-t<eS-n，n+1-Sn-t<eS-n，故t>Sn+1-Sn-eS-n.用單調(diào)性來(lái)證明lnan-an+1<0，進(jìn)而得到ln(a1a2?an(<Sn-n，得證.2=43=9n=n2n+1=an+2n+1，∴an+1-an=2n+1當(dāng)n≥2時(shí)，an=an-an-1+an-1-an-2+?+a2-a1+a1，n=2(n-1(+1+2(n-2(+1+?+2×1+1+1=n2，當(dāng)n=1時(shí)，a1=12=1也滿足an=n2，n=n2，2=43=9.n=令f(n(=44f∵f(4(=70=f(5(-2k；對(duì)任意的n∈N*，-n≠n，都有f(-n(=|-n|=|n|=f(n(，故數(shù)列{n{，{-n{為“f(x(下的一對(duì)孿生數(shù)列”.nn=k.n+bn<-2.由h(x(=，得h(x(的定義域?yàn)閧x|x≠0{，(x(<0，所以函數(shù)h(x(在(-∞,0(，(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，55由于h(cn(=h(dn(，不妨令h(x1(=h(x2(，x1<x2，則0<x1<即證x1+x2>2，即證x2>2-x1.又h(x(在(1,+∞(上單調(diào)遞增，所以只需證h(x2(>h(2-x1(，(提示：因?yàn)?<x1<1，所以2-x1>1)又h(x1(=h(x2(，所以只需證h(x1(>h(2-x1(.令F(x(=h(x(-h(2-x((0<x<1(，則F(x(=(x-1(2[(2-x(ex-xe2-x[x(2-x(只需證F(x(>0.(0<x<1(，令m(x(=(2-x(ex-xe2-x(0<x<1(，因?yàn)楫?dāng)0<x<1時(shí)，x(2-x(>0，(x-1(2>0，所以只需證m(x(>0.易知m(x(=(1-x((ex-e2-x(，因?yàn)?<x<1，所以m(x(<0，故F(x(>0，即h(x1(>h(2-x1(，故x1+x2>2，故cn+dn>2，則(c1+d1(+(c2+d2(+?+(c2025+d2025(>2×2025=4050所以數(shù)列{cn+dn{的前2025項(xiàng)和大于4050.7.已知冪函數(shù)f(x(=(m2+3m+3(x3m-1為偶函數(shù).(2)若f(a-1(≥f(1+2a(，求實(shí)數(shù)a的取值范圍.-1或m=-2.當(dāng)m=-1時(shí)，f(x(=x-4，滿足f(-x(=f(x(，此時(shí)f(x(為偶函數(shù)，符合題意；當(dāng)m=-2時(shí)，f(x(=x-7，不滿足f(-x(=f(x(，此時(shí)f(x(不是偶函數(shù)，不符合題意.綜上可得，f(x(=x-4.66由得f(x(=x-4=，所以f(x(在(0,+∞(上單調(diào)遞減，在(-∞,0(上單調(diào)遞增且f(x(為偶函數(shù)，因?yàn)閒(a-1(≥f(1+2a(，解得a≤-2或0≤a<1或a>1.故實(shí)數(shù)a的取值范圍為(-∞,-2[∪[0,1(∪(1,+∞(.8.已知定義在R上的函數(shù)f(x(=滿足5f(2(=9f(1(，且f(0(=0.(2)證明：f(x(在R上為增函數(shù).f(x(=奇函數(shù).f(x(=又函數(shù)f(x(的定義域?yàn)镽，f(-x(==-f(x(，故函數(shù)f(x(為奇函數(shù).,x21<x2，xx>0,2x>0,2x-2x>0,f(x2(-f(x1(>0，即f(x2(>f(x1(，故f(x(在R上為增函數(shù).9.已知定義在R上的函數(shù)f(x(滿足f(x(-2f(-x(=3x2-5x+2.(1)求f(x(的解析式；若g(x(=-f(x(-2x2+在區(qū)間[a,a+1[內(nèi)有最小值2，求實(shí)數(shù)a的值.f(x(=-3x2-(2)a=-3或a=0【分析】(1)將f(x(-2f(-x(=3x2-5x+2式中x換成-x，聯(lián)立兩式即可求解；將f(x(代入g(x(=-f(x(-2x2+x，可知g(x(的圖象的對(duì)稱軸，分類討論區(qū)間[a,a+1[和對(duì)稱軸之【詳解】(1)因?yàn)閒(x(-2f(-x(=3x2-5x+2①,用-x替換x得，f(-x(-2f(x(=3x2+5x+2②,77聯(lián)立①②,解得f(x(=-3x2-x-2；(2)由題可得，g(x(=-f(x(-2x2+x=x2+2x+2，對(duì)稱軸為x=-1，則g(x(在區(qū)間(-∞,-1(上單調(diào)遞減，在區(qū)間(-1,+∞(上單調(diào)遞增.解得a=-3或a=-1(舍)；10.已知函數(shù)f(x(=a-f(x(=或f(x(=-2--則則f(x(=-=0，則a=-2，則f(x(=-2-，綜上所述，f(x(=或f(x(=-2-=(-x-a(2+|11.函數(shù)f(x(=lnx-(2)若函數(shù)f(x(有兩個(gè)極值點(diǎn)x1、x2，曲線y=f(x(上兩點(diǎn)(x1,f(x1((、(x2,f(x2((連線斜率記為k，求88當(dāng)a=3時(shí)，f(x(=lnx-該由fI(x(<0可得2-3<x<2+3；由fI(x(>0可得0<x<2-3或x>2+3，fI(x(=對(duì)于方程x2+(2-2a(x+1=0，Δ=(2-2a(2-4=4(a2-2a(，因?yàn)楹瘮?shù)f(x(有兩個(gè)極值點(diǎn)x1、x2，則方程x2+(2-2a(x+1=0在(0,+∞(上有兩個(gè)不等的實(shí)根，所以4a2-8a>0，解得a>2,ln-ln-==x1-x2x1-x2t(=lnt-，則hI(t(=2-12<902，99t(=lnt-19>219>e2，19<12.若將對(duì)于任意x、y∈R總有f(x+y(+f(x-y(=2n=2f(n+1(-f(n((n∈N*(，求log2+log2+?+log2的值；f(1(=令x=y=1，則f(2(+f(0(=2f(1(f(1(，(2)令x=n，y=1，n∈N?,則f(n+1(+f(n-1(=2f(n(f(1(=f(n(，2f(n+1(-f(n(=2[2f(n(-f(n-1([，即an=2an-1，則log2+log2+?+log2=0+1+2+?+99=×100=4950.即g(y(=g(-y(，故g(x(是偶函數(shù).因?yàn)間(x+y(+g(x-y(=2g(x(g(y(，所以g(x+y(+g(x-y(>2g(y(，即Cn+1+Cn-1>2Cn，Cn+1>2Cn-Cn-1=Cn+(Cn-Cn-1(>Cn，又g(x(是偶函數(shù)，所以有g(shù)(x2(>g(x1(. ==(3-22(cos2+(3+22(sin2(3-22(+(3+22=、2|AC|化簡(jiǎn)得x2+y2+6x+1=0，即A的軌跡方程為(x-3(2+y2=8(y≠0(，由cosA+cosC=及A+B+C=π,得=-cos(B+C(-cos(B+A(=sinBsinC-cosBcosC+sinAsinB-cosAcosB=sinB(sinA+sinC(-cosB(cosA+cosC(=、2sin2B-cosBcosB=或cosB=-1所以ac=a(c2-a2+2(+c(a2-c2+2(=(c2-a2((a-c(+2(a+c(=-(a+c((a-c(2+2(a+c(=-2(a-c(2+4=-2(a+c(2+8ac+4=8ac-4，解得所以cosB=設(shè)B(x0,y0(，其中-2<x0<2，=4-從而sinB=+C(=(b-c)?(sinB+sinC),b=、所以a2-ac=b2-c2，即a2+c2-b2=ac.由余弦定理cosB=所以.2+c2+ac=9.S=2sin2θ令f(x(=sinx-sinxcosx，則fI(x(=cosx-(cos2x-sin2x(=-2cos2x+cosx+1=(2cosx+1((-cosx+1(.所以當(dāng)-<cosx<1時(shí)，fI(x(>0，f(x(單調(diào)遞增.則0<f(x(<所以0<S<.即sin2θ-cos2θ=sinθ-cosθ:sin(2θ-(=sin(θ-將點(diǎn)P(x,y(繞坐標(biāo)原點(diǎn)O逆時(shí)針旋轉(zhuǎn)角θ至點(diǎn)PI(xI,yI(.將①代入方程，得(xIcosθ+yIsinθ(2+、3(xIcosθ+yIsinθ((-xIsinθ+yIcosθ(=6，故曲線CI是以F1/(-4,0(和F2/(4,0(為焦點(diǎn)的雙曲線.(1)求函數(shù)f(x(的解析式；(2)求函數(shù)g(x(=f(x(+2cos(ωx+φ(在區(qū)間內(nèi)的值域.f(x(=2sin(f(x(=2sin(17.已知函數(shù)f(x(=2sin2+sinπx-1，將函數(shù)f(x(的所有正的零點(diǎn)從小到大排列組成數(shù)列{an{.記數(shù)列{cn{滿足cn=且數(shù)列{cn{的前n項(xiàng)和為Sn，求證：S2n<ln2.n=n<求解n的范圍可得；由cn=代入所證不等式，先通過(guò)化簡(jiǎn)變形將所證不等式轉(zhuǎn)化為++?++則an=n-，n∈N?,所以bn=[an+1[=故數(shù)列{bn{的通項(xiàng)公式為bn=n.綜上所述，存在正整數(shù)N≥7，當(dāng)n≥N時(shí)，恒有Pn<.故N的最小值為7.由cn=構(gòu)造函數(shù)-lnx,x>1，f<0，則f在(1,+∞(上單調(diào)遞減，由x>1，則f(x)<f(1)=0，所以任意x>1，1--lnx<0，即1-<lnx恒成立.所以有<ln(n+1(-lnn，<ln(n+2(-ln(n+1(，?,<ln(2n(-ln(2n-1(.以上各式相加得，++?++<ln(n+1)-lnn+ln(n+2)-ln(n+1)+?+ln(2n)-ln(2n-1)=ln(2n)-lnn=ln2-lnx,x>1，利用導(dǎo)數(shù)的單調(diào)性證明1-<lnx,x>1，賦值可得=ln(n+1(-lnn，222>(2)求二面角A-PC-B的大小.又AB=BC=1，所以AC2=AB2+又PA∩AB=A，PA,ABC平面PAB，所以BC丄平面PAB.則A(0,1,0(,B(0,0,0(,C(1,0,0(,P(0,1,1(，所以AP=(0,0,1(,AC=(1,-1,0(設(shè)平面APC的一個(gè)法向量為=(x,y,z(，設(shè)平面BPC的一個(gè)法向量為=(a,b,c(，-1(，由圖可知二面角A-PC-B為銳角，設(shè)二面角A-PC-B的大小為θ,即二面角A-PC-B的大小為.20.如圖所示，在三棱柱ABC-A1B1C1中，H是正方形AA1B1B的中心，AA1=22，C1H⊥平面AA1B1B(1)求異面直線AC與A1B1夾角的余弦值；(2)求平面AA1C1與平面A1C1B1夾角的正弦值.所以A=(-2,-2,5)，A=(-22,0,0)，所以異面直線AC與A1B1夾角的余弦值為.設(shè)平面AA1C1的法向量為1,y1,z1)，設(shè)平面A1B1C1的法向量為2,y2,z2)，所以平面AA1C1與平面A1C1B1夾角的正弦值為.弦值.“二面角D-AB-E為直二面角，:BC丄平面AEB，設(shè)E(x0,0,0)(x0>0)，“F為CE上的點(diǎn)，E=(-x0,1,2)，:設(shè)E=λE=(-λx0,λ,2λ)，:F((1-λ)x0,λ,2λ)，0=所以=0，:AE丄BE，“BC丄平面AEB，AEC平面AEB，:BC丄AE，又BC∩BE=B，BC、BEC平面BCE，:AE丄平面BCE；設(shè)面BDF的法向量為=(0,-2,2(，:0,1,1)，:.O=0，:平面BDF丄平面ABCD.,N是AB的中點(diǎn).2=所以當(dāng)C1M=C1B1時(shí)，AM⊥A1N.斯圓．如圖，在棱長(zhǎng)為3的正方體ABCD-AIBICIDI中AMN截正方體ABCD-AIBICIDI的截面周長(zhǎng)，若無(wú)，說(shuō)明理由.【詳解】(1)由于ABCD-AIBICIDI是正方體，由于∠ADP=∠MCP=90°,tan∠APD=tan∠MPC，建立平面直角坐標(biāo)系D-xy，設(shè)點(diǎn)P(x,y(.“DP=2CP，D(0,0(、C(3,0(，所以EF這段弧的長(zhǎng)度為=P.P=(-x,-y(.(3-x,-y(=x2-3x+y2，:PD.PC∈[-2,3[，故PA.PC的取值范圍[-2,3[.以下證明此時(shí)D'P即D/OO?平面AMN，SR?平面AMN，所以D/O所以所求的截面五邊形AMQTN的周長(zhǎng)24.在平面直角坐標(biāo)系xOy中有兩個(gè)定點(diǎn)A(-3,0(，B(3,0(，已知?jiǎng)狱c(diǎn)M在平面xOy中且M到A，B兩點(diǎn)的斜率乘積為為定點(diǎn)(-1,0(解得點(diǎn)M的軌跡方程為=4，所以(-3+1(2+02+(t-0(2=4，解得t=23.設(shè)向量C與向量夾角則cosθ=2=6-與點(diǎn)M的軌跡方程聯(lián)立，得(3k2+2(x2+12kx+54=0，有x1+x2=，x1x2=直線方程為(x-x1(+y1，令x=0，得y=所以直線PH過(guò)定點(diǎn)K(0,，所以PH⊥平面CAK，又因?yàn)锳K?平面CAK，所以PH⊥AK，此時(shí)kAK=，kPH=又直線HP與平面ABC夾角為銳角，所以直線HP與平面ABC夾角的正切值為6iP-ln(Pi+1(+k≥0成立，求整數(shù)k的最小值.所以乙贏的概率為P=1-i-1i-1+P-ln(Pi+1(+k≥0即k≥ln(Pi+1(-eP，所以k≥ln(1+=ln(1+>-2(∵e<2(，所以-2<ln(1+<-1，所以滿足k≥ln(1+的整數(shù)k的最小值為-1.訪的18~64歲的市民數(shù)為隨機(jī)變量X(X≥2(，且該市隨機(jī)抽取的18~64歲的市民是達(dá)標(biāo)成年人的概2+?-12第n個(gè)人為達(dá)標(biāo)成年人，n-1，依題意，可得隨機(jī)變量X(X≥2(的分布列如下表所示：X234?nP2?=1+2×+3×2+?+(n-1(×n-2，由①-②,可得S=1+(2+?+((n-2-(n-1(((n-1， Dn-1(n≥2(之間的遞推關(guān)系，并證明：數(shù)列{Dn-nDn-1{(n≥2(是等比數(shù)列；x=1+x++??)n+1=n(Dn+Dn-1(，n≥2，證明見(jiàn)解析法數(shù)得Dn+1=n(Dn+Dn-1(，將其化為=-1，可證得等比數(shù)列{Dn-nDn-1{；即得由參考公式取x=-1即可證明.2=1. 所以Dn+1=n(Dn+Dn-1(，n≥2，又D2-2D1=1，所以數(shù)列{Dn-nDn-1{，n≥2是首項(xiàng)為1，公比為-1的等比數(shù)列當(dāng)n無(wú)窮大時(shí)=1-1++?=e-1=，得證.比數(shù)列{Dn-nDn-1{，為第三題證明結(jié)論做好知識(shí)鋪墊.會(huì)知識(shí)宣講團(tuán).①求應(yīng)從[80,90(和[90,100[學(xué)生中分別抽取的學(xué)生人數(shù)；所以{Pn-Pn-1{是以首項(xiàng)為公比為-的等比數(shù)列，n-2(n≥2(成立，則有-3+?+(0+(-(+(-(2+?+(-(n-2n-1Pn(1≤n≤60(，試證明數(shù)列{Pn-Pn-1{是等比數(shù)列(2≤n≤59(，求出數(shù)列{Pn{(1≤n≤60(的通項(xiàng)公參考數(shù)據(jù)：若隨機(jī)變量ξ服從正態(tài)分布N(μ,σ2(，則P(μ-σ<ξ<μ+σ(=0.6827，P(μ-2σ<ξ<μ+2σ(=0.9545，P(μ-3σ<ξ<μ+3σ(=0.99731.當(dāng)3≤n≤59時(shí)，-2，則-1=--2(，又所以n-2n-2n-1+PA=，kPB=PB=-得整理得(2)求△ABD面積的最大值.-t2-t3(0<t<1)，根據(jù)導(dǎo)數(shù)判斷函數(shù)f(t(的單調(diào)性并求出最大值即可求出△ABD面積的最大值.，B(x2,y2(，-2k,mk=4-2k2，由Δ=(2km-8)2-4k2m2=32(2-km(=64(k2-1(>0，得k<-1或k>1，所以AB的垂直平分線方程為：y-2k-m=-(x-2(， 令=t(0<t<1(，設(shè)f(t(=1+t-t2-