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第三部分 考前臨門一腳(一)巧用性質(zhì)妙解函數(shù) 速解技法學(xué)一招函數(shù)性質(zhì)主要指函數(shù)的單調(diào)性、奇偶性、周期性、對(duì)稱性,要深刻理解并加以巧妙地運(yùn)用以對(duì)稱性為例,若函數(shù)f (x)滿足f (ax)f (bx),則函數(shù)圖象關(guān)于直線x對(duì)稱;若函數(shù)f (x)滿足f (ax)f (bx)c,則函數(shù)圖象關(guān)于點(diǎn)對(duì)稱定義在R上的奇函數(shù)f (x)滿足f (x2)f (x),且在0,1上是增函數(shù),則有()Af f f Bf f f Cf f f Df f f 解析由題設(shè)知f (x)f (x2)f (2x),所以函數(shù)f (x)的圖象關(guān)于直線x1對(duì)稱由于奇函數(shù)f (x)在0,1上是增函數(shù),故f (x)在1,0上也是增函數(shù),綜上,函數(shù)f (x)在1,1上是增函數(shù),在1,3上是減函數(shù)又f f f ,所以f f f f .答案B已知函數(shù)f (x)x3sin x的定義域?yàn)?,1,若f (log2m)f (log4(m2)成立,則實(shí)數(shù)m的取值范圍為_解析由f (x)x3sin x的定義域?yàn)?,1,易知f (x)在1,1上單調(diào)遞增,由f (log2m)f (log4(m2),可得解得故m2.綜上可知,實(shí)數(shù)m的取值范圍為.答案經(jīng)典好題練一手1已知定義在R上的函數(shù)f (x)滿足f (2x)f (2x),當(dāng)x2時(shí),f (x)單調(diào)遞增,如果x1x24,且(x12)(x22)0,則f (x1)f (x2)的值為()A可正可負(fù)B可能為0C恒大于0 D恒小于0解析:選D由f (2x)f (2x)可知,函數(shù)f (x)的圖象關(guān)于點(diǎn)(2,0)中心對(duì)稱因?yàn)閤2時(shí),f (x)單調(diào)遞增因?yàn)閤1x24且(x12)(x22)0,設(shè)x12x2,則x24x1,所以f (x2)f (4x1)又因?yàn)閒 (4x1)f (x1),所以f (x2)f (x1),即f (x1)f (x2)0且a1,函數(shù)f (x)4loga,其中x,則函數(shù)f (x)的最大值與最小值之和為_解析:依題意知,f (x)44loga,令g(x)4loga,其定義域?yàn)?,可知g(x)4logag(x),函數(shù)g(x)的圖象關(guān)于原點(diǎn)對(duì)稱,從而可知函數(shù)f (x)的圖象關(guān)于點(diǎn)(0,4)對(duì)稱,故函數(shù)f (x)的最大值與最小值之和為8.答案:8常用結(jié)論記一番1函數(shù)的單調(diào)性在公共定義域內(nèi):(1)若函數(shù)f (x)是增函數(shù),函數(shù)g(x)是增函數(shù),則f (x)g(x)是增函數(shù);(2)若函數(shù)f (x)是減函數(shù),函數(shù)g(x)是減函數(shù),則f (x)g(x)是減函數(shù);(3)若函數(shù)f (x)是增函數(shù),函數(shù)g(x)是減函數(shù),則f (x)g(x)是增函數(shù);(4)若函數(shù)f (x)是減函數(shù),函數(shù)g(x)是增函數(shù),則f (x)g(x)是減函數(shù)提示在利用函數(shù)單調(diào)性解不等式時(shí),易忽略函數(shù)定義域這一限制條件2函數(shù)的奇偶性(1)判斷函數(shù)的奇偶性有時(shí)可以用定義的等價(jià)形式:f (x)f (x)0,1;(2)設(shè)f (x),g(x)的定義域分別是D1,D2,那么在它們的公共定義域上:奇奇奇,奇奇偶,偶偶偶,偶偶偶,奇偶奇3有關(guān)函數(shù)f (x)周期性的常用結(jié)論(1)若f (xa)f (xa),則函數(shù)f (x)的周期為2|a|;(2)若f (xa)f (x),則函數(shù)f (x)的周期為2|a|;(3)若f (xa),則函數(shù)f (x)的周期為2|a|;(4)若f (xa),則函數(shù)f (x)的周期為2|a|.(二)最值函數(shù)大顯身手 速解技法學(xué)一招對(duì)于任意xR,函數(shù)f (x)表示yx3,yx,yx24x3中的最大者,則f (x)的最小值是()A2 B3C8 D1解析如圖,分別畫出函數(shù)yx3,yx,yx24x3的圖象,得到三個(gè)交點(diǎn)A(0,3),B(1,2),C(5,8)由圖象可得函數(shù)f (x)的表達(dá)式為f (x)所以f (x)的圖象是圖中的實(shí)線部分,圖象的最低點(diǎn)是B(1,2),所以函數(shù)f (x)的最小值是2.答案A已知函數(shù)f (x)x2xm,g(x)log2x,minm,n表示m,n中的最小值,設(shè)函數(shù)h(x)minf (x),g(x)(x0),則當(dāng)函數(shù)h(x)有三個(gè)零點(diǎn)時(shí),實(shí)數(shù)m的取值范圍為()A. B.C. D.解析在同一直角坐標(biāo)系中,作出函數(shù)yf (x)和yg(x)的圖象如圖所示當(dāng)兩函數(shù)圖象交于點(diǎn)A(1,0)時(shí),即有11m0,解得m,所以當(dāng)函數(shù)h(x)有三個(gè)零點(diǎn)時(shí),即為點(diǎn)A和yf (x)與x軸的兩個(gè)交點(diǎn),若滿足條件,則需解得m0f (x)為增函數(shù);f (x)0f (x)為減函數(shù);f (x)0f (x)為常數(shù)函數(shù)2求函數(shù)f (x)極值的方法求函數(shù)的極值應(yīng)先確定函數(shù)的定義域,解方程f (x)0,再判斷f (x)0的根是否是極值點(diǎn),可通過(guò)列表的形式進(jìn)行分析,若遇極值點(diǎn)含參數(shù)不能比較大小時(shí),則需分類討論若函數(shù)f (x)2sin x(x0,)的圖象在切點(diǎn)P處的切線平行于函數(shù)g(x)2的圖象在切點(diǎn)Q處的切線,則直線PQ的斜率為()A. B2C. D.解析由題意得f (x)2cos x,g(x)xx.設(shè)P(x1,f (x1),Q(x2,g(x2),又f (x1)g(x2),即2cos x1x2x2,故4cos2x1x2x2,所以44cos2x1x2x2,即4sin2x1(x2x2)2,所以sin x10,x10,x2x2,x21,故P(0,0),Q,故kPQ.答案A求曲線的切線方程時(shí),要注意是在點(diǎn)P處的切線還是過(guò)點(diǎn)P的切線,前者點(diǎn)P為切點(diǎn),后者點(diǎn)P不一定為切點(diǎn)技法領(lǐng)悟 已知函數(shù)f (x)(xR)滿足f (1)1,且f (x)的導(dǎo)數(shù)f (x),則不等式f (x2)的解集為_解析設(shè)F(x)f (x)x,F(xiàn)(x)f (x),f (x),F(xiàn)(x)f (x)0,即函數(shù)F(x)在R上單調(diào)遞減f (x2),f (x2)f (1),F(xiàn)(x2)1,即x(,1)(1,)答案(,1)(1,)已知函數(shù)f (x)(axb)ln xbx3在(1,f (1)處的切線方程為y2.(1)求a,b的值;(2)求函數(shù)f (x)的極值;(3)若g(x)f (x)kx在(1,3)上是單調(diào)函數(shù),求k的取值范圍解(1)因?yàn)閒 (1)b32,所以b1.又f (x)aln xabaln xa1,而函數(shù)f (x)在(1,f (1)處的切線方程為y2,所以f (1)1a10,所以a0.(2)由(1)得f (x)ln xx3,f (x)1(x0)令f (x)0,得x1.當(dāng)0x0;當(dāng)x1時(shí),f (x)0),g(x)k1,又g(x)在x(1,3)上是單調(diào)函數(shù),若g(x)為增函數(shù),有g(shù)(x)0,即g(x)k10,即k1在x(1,3)上恒成立又1,所以k.若g(x)為減函數(shù),有g(shù)(x)0,即g(x)k10,即k1在x(1,3)上恒成立,又1,所以k0.綜上,k的取值范圍為(,0.技法領(lǐng)悟 破解此類問(wèn)題需注意兩點(diǎn):(1)求函數(shù)的單調(diào)區(qū)間時(shí)應(yīng)優(yōu)先考慮函數(shù)的定義域;(2)求得函數(shù)在多個(gè)區(qū)間單調(diào)性相同時(shí),區(qū)間之間用“,”分割,或用“和”相連,不能用“”相連經(jīng)典好題練一手1已知直線2xy10與曲線yaexx相切(其中e為自然對(duì)數(shù)的底數(shù)),則實(shí)數(shù)a的值是()A.B1C2De解析:選B由題意知yaex12,則a0,xln a,代入曲線方程得y1ln a,所以切線方程為y(1ln a)2(xln a),即y2xln a12x1a1.2若函數(shù)f (x)axx2ln x存在極值,且這些極值的和不小于4ln 2,則a的取值范圍為()A2,) B2,)C2,) D4,)解析:選Cf (x)a2x(x0),因?yàn)閒 (x)存在極值,所以f (x)0在(0,)上有根,即2x2ax10在(0,)上有根,所以a280,顯然當(dāng)0時(shí),f (x)無(wú)極值,不合題意,所以a280,即a2或a0,則f (x1),f (x2)為f (x)的極值,所以f (x1)f (x2)(ax1xln x1)(ax2xln x2)a(x1x2)(xx)(ln x1ln x2)ln 24ln 2,所以a2.綜上,a的取值范圍為2,)3是圓周率,e是自然對(duì)數(shù)的底數(shù),在3e,e3,e,3,3,e六個(gè)數(shù)中,最小的數(shù)與最大的數(shù)分別是()A3e,3 B3e,eCe3,3 De,3解析:選Ae3,eln 3eln ,ln eln 3,即ln 3eln e,ln eln 3.又函數(shù)yln x,yex,yx在定義域上單調(diào)遞增,故3ee3,e3e0,即0xe時(shí),函數(shù)f (x)單調(diào)遞增;當(dāng)f (x)e時(shí),函數(shù)f (x)單調(diào)遞減故函數(shù)f (x)的單調(diào)遞增區(qū)間為(0,e),單調(diào)遞減區(qū)間為(e,)由e3及函數(shù)f (x)的單調(diào)性,得f ()f (3)f (e),即.由,得ln 33,在3e,e3,e,3,3,e六個(gè)數(shù)中的最大的數(shù)是3,同理得最小的數(shù)為3e.4已知函數(shù)f (x)1ln xa2x2ax(aR)(1)討論函數(shù)f (x)的單調(diào)性;(2)若a0且x(0,1),求證:x21.解:(1)函數(shù)f (x)的定義域?yàn)?0,),f (x)2a2xa.若a0,則f (x)0,則當(dāng)x時(shí),f (x)0,當(dāng)0x時(shí),f (x)時(shí),f (x)0,故f (x)在上單調(diào)遞減,在上單調(diào)遞增若a0,則當(dāng)x時(shí),f (x)0,當(dāng)0x時(shí),f (x)時(shí),f (x)0.故f (x)在上單調(diào)遞減,在上單調(diào)遞增(2)證明:若a0且x(0,1),則f (x)1ln x,x(0,1)欲證x21,只需證x21,即證x(1ln x)0,故函數(shù)g(x)在(0,1)上單調(diào)遞增,所以g(x)g(1)1.設(shè)函數(shù)h(x)(1xx3)ex,則h(x)(2x3x2x3)ex.設(shè)函數(shù)p(x)2x3x2x3,則p(x)16x3x2.當(dāng)x(0,1)時(shí),p(0)p(1)8p(0)2,當(dāng)x(x0,1)時(shí),p(x0)p(1)0,當(dāng)x(x1,1)時(shí),h(x)h(0)1,所以x(1ln x)(1xx3)ex,x(0,1),即x20,0)的圖象相鄰兩條對(duì)稱軸的距離為,且f (0)1.(1)求函數(shù)f (x)的解析式;(2)設(shè),f ,f ,求tan(22)的值解:(1)函數(shù)f (x)Acos(A0,0)的圖象相鄰兩條對(duì)稱軸的距離為,2,又f (0)1,A1,A2,f (x)2cos.(2),f 2cos2cos(2)2cos 2,cos 2,sin 2,則tan 2.,f 2cos2cos 2,cos 2,sin 2,則tan 2.tan(22).常用結(jié)論記一番三角公式中常用的變形(1)對(duì)于含有sin cos ,sin cos 的問(wèn)題,利用(sin cos )212sin cos ,建立sin cos 與sin cos 的關(guān)系(2)對(duì)于含有sin ,cos 的齊次式,利用tan 轉(zhuǎn)化為含tan 的式子(3)對(duì)于形如cos2sin 與cos2sin cos 的變形,前者用平方關(guān)系sin2cos21化為二次型函數(shù),而后者用降冪公式化為一個(gè)角的三角函數(shù)(4)含tan tan 與tan tan 時(shí)考慮tan().(五)正弦余弦相得益彰 速解技法學(xué)一招邊角互化的技巧:若要把“邊”化為“角”, 常利用“a2Rsin A,b2Rsin B,c2Rsin C”,,若要把“角”化為“邊”,常利用 (R為ABC外接圓的半徑)等.在ABC中,角A,B,C所對(duì)的邊分別是a,b,c,且.(1)證明:sin Asin Bsin C;(2)若b2c2a2bc,求tan B.解(1)證明:根據(jù)正弦定理,可設(shè)k(k0)則aksin A,bksin B,cksin C.代入中,有,變形可得sin Asin Bsin Acos Bcos Asin Bsin(AB)在ABC中,由ABC,有sin(AB)sin(C)sin C,所以sin Asin Bsin C.(2)由已知,b2c2a2bc,根據(jù)余弦定理,有cos A.所以sin A.由(1),sin Asin Bsin Acos Bcos Asin B,所以sin Bcos Bsin B,故tan B4.如圖,在ABC中,B,AB8,點(diǎn)D在邊BC上,且CD2,cosADC.(1)求sinBAD;(2)求BD,AC的長(zhǎng)解(1)在ADC中,cosADC,sinADC ,則sinBADsin(ADCB)sinADCcosBcosADCsinB.(2)在ABD中,sinADBsin(ADC)sinADC.由正弦定理得BD3,在ABC中,BCBDDC5,由余弦定理得AC2AB2BC22ABBCcosB825228549,即AC7. 經(jīng)典好題練一手1已知ABC的三個(gè)內(nèi)角A,B,C的對(duì)邊分別為a,b,c,若,則該三角形的形狀是()A直角三角形B等腰三角形C等邊三角形 D鈍角三角形解析:選A因?yàn)?,由正弦定理得,所以sin 2Asin 2B.由,可知ab,所以AB.又A,B(0,),所以2A2B,即AB,所以C,于是ABC是直角三角形2在ABC中,內(nèi)角A,B,C的對(duì)邊分別為a,b,c,若acos Cccos A2bsin A,則A的值為()A. B.C. D.或解析:選D由acos Cccos A2bsin A,結(jié)合正弦定理可得sin Acos Csin Ccos A2sin Bsin A,即sin(AC)2sin Bsin A,故sin B2sin Bsin A又sin B0,可得sin A,故A或.3非直角ABC中,內(nèi)角A,B,C的對(duì)邊分別是a,b,c,已知c1,C.若sin Csin(AB)3sin 2B,則ABC的面積為()A. B.C.或 D.解析:選D因?yàn)閟in Csin(AB)sin(AB)sin(AB)2sin Acos B6sin Bcos B,因?yàn)锳BC非直角三角形,所以cos B0,所以sin A3sin B,即a3b.又c1,C,由余弦定理得a2b2ab1,結(jié)合a3b,可得b2,所以Sabsin Cb2sin.故選D.4在ABC中,角A,B,C的對(duì)邊分別為a,b,c,面積為S,已知2acos22ccos2b.(1)求證:2(ac)3b;(2)若cos B,S,求b.解:(1)證明:由已知得,a(1cos C)c(1cos A)b.整理得acacos Cccos Ab.在ABC中,由余弦定理,得acos Cccos Aacb.acb,即2(ac)3b.(2)cos B,sin B.Sacsin Bac,ac8.又b2a2c22accos B(ac)22ac(1cos B),2(ac)3b,b216,解得b216,b4.常用結(jié)論記一番1解三角形中常用結(jié)論:(1)三角形中正弦、余弦、正切滿足的關(guān)系式有:2R,c2a2b22abcos C,tan Atan Btan Ctan Atan Btan C,abABsin Asin Bcos Ac2(c為最大邊);鈍角三角形a2b2,CB,AC;任意角的正弦值都大于其他角的余弦值(4)在ABC中,A,B,C成等差數(shù)列B60;在ABC中,A,B,C成等差數(shù)列,且a,b,c成等比數(shù)列三角形為等邊三角形2設(shè)ABC的內(nèi)角A,B,C的對(duì)邊分別為a,b,c,其面積為S.(1)Sahabhbchc(ha,hb,hc分別表示a,b,c邊上的高)(2)Sabsin Cbcsin Acasin B.(3)Sr(abc)(r為三角形ABC內(nèi)切圓的半徑)(六)向量小題三招搞定速解技法學(xué)一招如圖, 在直角梯形ABCD中, , 2,且rs,則2r3s()A1B2C3 D4解析法一:根據(jù)圖形,由題意可得 ( )() .因?yàn)閞s,所以r,s,則2r3s123.法二:因?yàn)?,所以2(),整理得(),則r,s,2r3s3.法三:如圖,延長(zhǎng)AD,BC交于點(diǎn)P,則由得DCAB,且AB4DC,又2,所以E為PB的中點(diǎn),且.于是,().則r,s,2r3s3.法四:如圖,建立平面直角坐標(biāo)系xAy,依題意可設(shè)點(diǎn)B(4m,0),D(3m,3h),E(4m,2h),其中m0,h0.由rs,得(4m,2h)r(4m,0)s(3m,3h),所以解得所以2r3s123.答案C技法領(lǐng)悟解決平面向量問(wèn)題的常用方法(1)求解有關(guān)平面向量的問(wèn)題時(shí),若能靈活利用平面向量加、減法運(yùn)算及其幾何意義進(jìn)行分析,則有利于問(wèn)題的順利獲解這種解題思路,我們不妨稱之為按“圖”處理(2)建系法:處理有關(guān)平面圖形的向量問(wèn)題時(shí),若能靈活建立平面直角坐標(biāo)系,則可借助向量的坐標(biāo)運(yùn)算巧解題,這也體現(xiàn)了向量的代數(shù)化手段的重要性(3)基底法:求解有關(guān)平面向量的問(wèn)題時(shí),若能靈活地選取基底,則有利于問(wèn)題的快速獲解理論依據(jù):適當(dāng)選取一組基底e1,e2,利用平面向量基本定理及相關(guān)向量知識(shí),可將原問(wèn)題轉(zhuǎn)化為關(guān)于e1,e2的代數(shù)運(yùn)算問(wèn)題經(jīng)典好題練一手1已知0,|1,|2, 0,則|的最大值為()A.B2C. D2解析:選C由0可知,.故以B為坐標(biāo)原點(diǎn),分別以BA,BC所在的直線為x軸,y軸建立如圖所示的平面直角坐標(biāo)系,則由題意,可得B(0,0),A(1,0),C(0,2)設(shè)D(x,y),則(x1,y),(x,2y)由0,可得(x1)(x)y(2y)0,整理得2(y1)2.所以點(diǎn)D在以E為圓心,半徑r的圓上因?yàn)閨表示B,D兩點(diǎn)間的距離,而|,所以|的最大值為| |r.2已知向量a,b滿足a(a2b)0,|a|b|1,且|ca2b|1,則|c|的最大值為()A2 B4C.1 D.1解析:選D設(shè)a,a2b,c,且設(shè)點(diǎn)A在x軸上,則點(diǎn)B在y軸上,由|ca2b|1,可知|c(a2b)|1,所以點(diǎn)C在以B為圓心,1為半徑的圓上,如圖所示法一:因?yàn)閍(a2b)0,所以2ab|a|2.又|a|b|1,所以|a2b|,所以|c|max|1|a2b|11.法二:連接AB,因?yàn)閍2b,所以2b.因?yàn)閨a|b|1,所以|2,|1,所以|,所以|c|max|11.3在RtABC中,CA4,CB3,M,N是斜邊AB上的兩個(gè)動(dòng)點(diǎn),且MN2,則的取值范圍為()A. B4,6C. D.解析:選C設(shè)MN的中點(diǎn)為E,則有2,所以()2()2|2| |2|21.易知|的最小值等于點(diǎn)C到斜邊AB的距離,即,所以的最小值為21.當(dāng)點(diǎn)M(或點(diǎn)N)與點(diǎn)A重合時(shí),|最大,此時(shí)|21242214,所以的最大值為1.綜上,的取值范圍是.4.如圖,在梯形ABCD中,ABCD,CD2,BAD,若2,則_.解析:法一:因?yàn)?,所以,所以.因?yàn)锳BCD,CD2,BAD,所以2|cos,化簡(jiǎn)得|2.故()|2(2)222cos12.法二:如圖,建立平面直角坐標(biāo)系xAy.依題意,可設(shè)點(diǎn)D(m,m),C(m2,m),B(n,0),其中m0,n0,則由2,得(n,0)(m2,m)2(n,0)(m,m),所以n(m2)2nm,化簡(jiǎn)得m2.故(m,m)(m2,m)2m22m12.答案:12常用結(jié)論記一番1在四邊形ABCD中:(1),則四邊形ABCD為平行四邊形;(2)且()()0,則四邊形ABCD為菱形;(3)且|,則四邊形ABCD為矩形;(4)若 (0,1),則四邊形ABCD為梯形2設(shè)O為ABC所在平面上一點(diǎn),內(nèi)角A,B,C所對(duì)的邊長(zhǎng)分別為a,b,c,則(1)O為ABC的外心222.(2)O為ABC的重心0.(3)O為ABC的垂心.(4)O為ABC的內(nèi)心abc0.(5)O為ABC的A的旁心abc.(七)玩轉(zhuǎn)通項(xiàng)搞定數(shù)列速解技法學(xué)一招幾種常見的數(shù)列類型及通項(xiàng)的求法遞推公式解法an1anf (n)轉(zhuǎn)化為an1anf (n),利用累加法(逐差相加法)求解an1f (n)an轉(zhuǎn)化為f (n),利用累乘法(逐商相乘法)求解an1panq轉(zhuǎn)化為特殊數(shù)列ank的形式求解an1panf (n)利用待定系數(shù)法,構(gòu)造數(shù)列bn,消去f (n)帶來(lái)的差異已知數(shù)列an滿足a1,an1an,求an(nN*)解由條件知,分別令n1,2,3,(n1),代入上式得(n1)個(gè)等式累乘,即.又a1,an.技法領(lǐng)悟 累加、累乘法起源于等差、等比數(shù)列通項(xiàng)公式的求解使用過(guò)程中要注意賦值后得到(n1)個(gè)式子,若把其相加或相乘,等式的左邊得到的結(jié)果是ana1或,添加首項(xiàng)后,等式的左邊累加或累乘的結(jié)果才為an.已知數(shù)列an的首項(xiàng)a11,an1(nN*),求數(shù)列的前10項(xiàng)和解因?yàn)閍n1,所以2,即2,所以是首項(xiàng)為1,公差為2的等差數(shù)列,所以2n1,所以an,而,所以.經(jīng)典好題練一手1在數(shù)列an中,a12,an1anlg,則an()A2lg nB2(n1)lg nC2nlg n D1nlg n解析:選A由an1anlgan1anlg,那么ana1(a2a1)(anan1)2lg 2lg lg lg 2lg22lg n.2已知數(shù)列an滿足a11,anan11(n2,nN*),則數(shù)列an的通項(xiàng)公式an_.解析:由anan11(n2,nN*),得an2(an12),而a12121,數(shù)列an2是首項(xiàng)為1,公比為的等比數(shù)列an2n1,an2n1.答案:2n13設(shè)數(shù)列an是首項(xiàng)為1的正項(xiàng)數(shù)列,且aanannan10(n2,nN*),則數(shù)列an的通項(xiàng)公式an_.解析:由題設(shè)得(anan1)(anan1n)0,由an0,an10知anan10,于是anan1n,所以ana1(a2a1)(a3a2)(anan1)123n.答案:4在數(shù)列an中,已知a11,an12an43n1(nN*),求通項(xiàng)公式an.解:原遞推式可化為an13n2(an3n1),即an133n12an23n1,比較系數(shù)得4,即an143n2(an43n1),則數(shù)列an43n1是首項(xiàng)為a143115,公比為2的等比數(shù)列,故an43n152n1,即an43n152n1.常用結(jié)論記一番等差(比)數(shù)列的重要結(jié)論(1)數(shù)列an是等差數(shù)列數(shù)列c是等比數(shù)列;數(shù)列an是等比數(shù)列,則數(shù)列l(wèi)oga|an|是等差數(shù)列(2)an,bn是等差數(shù)列,Sn,Tn分別為它們的前n項(xiàng)和,若bm0,則.(3)首項(xiàng)為正(或?yàn)樨?fù))遞減(或遞增)的等差數(shù)列前n項(xiàng)和最大(或最小)問(wèn)題轉(zhuǎn)化為解不等式,也可化為二次型函數(shù)SnAn2Bn來(lái)分析,注意nN*.(4)等差(比)數(shù)列中,Sm,S2mSm,S3mS2m,(各項(xiàng)均不為0)仍是等差(比)數(shù)列(八)掌握規(guī)律巧妙求和速解技法學(xué)一招求數(shù)列的前n項(xiàng)和的主要方法(1)公式法:對(duì)于等差數(shù)列或等比數(shù)列可用公式法(2)裂項(xiàng)相消法:將數(shù)列的每一項(xiàng)分解為兩項(xiàng)的差,在求和時(shí)中間的一些項(xiàng)可以相互抵消,從而累加相消(3)錯(cuò)位相減法:若an為等差數(shù)列,bn為等比數(shù)列,則對(duì)于數(shù)列anbn的前n項(xiàng)和可用錯(cuò)位相減法(4)倒序相加法:如果一個(gè)數(shù)列an中與首、末兩端等“距離”的兩項(xiàng)的和等于同一個(gè)常數(shù),那么求這個(gè)數(shù)列前n項(xiàng)和即可用倒序相加法(5)分組求和法:將原數(shù)列分解成可用公式法求和的若干個(gè)數(shù)列已知各項(xiàng)均為正數(shù)的等差數(shù)列an滿足:a42a2,且a1,4,a4成等比數(shù)列,設(shè)an的前n項(xiàng)和為Sn.(1)求數(shù)列an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和為Tn.解(1)根據(jù)題意,設(shè)等差數(shù)列an的公差為d,a42a2,且a1,4,a4成等比數(shù)列,a10,解得a12,d2,數(shù)列an的通項(xiàng)公式為an2n.(2)由(1)知a1d2,則Sn2n2n2n,設(shè)bn,則bn.Tn,Tn,兩式相減得,Tn,Tn223.技法領(lǐng)悟 利用錯(cuò)位相減法求和的3個(gè)注意點(diǎn)(1)判斷模型,即判斷數(shù)列an,bn中一個(gè)為等差數(shù)列,一個(gè)為等比數(shù)列;(2)錯(cuò)開位置,一般先乘以公比,再把前n項(xiàng)和退后一個(gè)位置來(lái)書寫,這樣避免兩式相減時(shí)看錯(cuò)列;(3)相減,相減時(shí)定要注意式中最后一項(xiàng)的符號(hào),考生常在此處出錯(cuò),一定要細(xì)心已知數(shù)列an滿足a1,an1aan,bn(nN*),Snb1b2bn,Pnb1b2bn,求2PnSn的值解因?yàn)閍1,an1aan,nN*,所以an1an0,an1an(an1),所以bn.Pnb1b2bn,Snb1b2bn2,故2PnSn2.技法領(lǐng)悟 利用裂項(xiàng)相消法求和的2個(gè)注意點(diǎn)(1)抵消后并不一定只剩下第一項(xiàng)和最后一項(xiàng),也有可能前面剩兩項(xiàng),后面也剩兩項(xiàng);(2)將通項(xiàng)裂項(xiàng)后,有時(shí)需要調(diào)整前面的系數(shù),使裂開的兩項(xiàng)之差和系數(shù)之積與原通項(xiàng)相等如:若an是等差數(shù)列,則,. 經(jīng)典好題練一手1設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,Sm113,Sm0,Sm115,其中mN*且m2.則數(shù)列的前n項(xiàng)和的最大值為()A.B.C. D.解析:選D因?yàn)镾m113,Sm0,Sm115,所以amSmSm101313,am1Sm1Sm15015,因?yàn)閿?shù)列an為等差數(shù)列,所以公差dam1am15(13)2,所以解得a113.所以an132(n1)152n,當(dāng)an0時(shí),n7.5,當(dāng)an10時(shí),n6.5,又nN*,所以數(shù)列的前6項(xiàng)為正數(shù),又因?yàn)椋詳?shù)列的前n項(xiàng)和的最大值為1.2數(shù)列1,3,5,7,的前n項(xiàng)和Sn_.解析:利用分組求和法,可得Sn(1352n1)n21.答案:n213已知數(shù)列an的前n項(xiàng)和為Sn,Sn2an1(nN*),數(shù)列bn是等差數(shù)列,且b1a1,b4a3.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)若cn,求數(shù)列cn的前n項(xiàng)和Tn.解:(1)Sn2an1,Sn12an11, 兩式相減,得Sn1Sn2an12an,an12an.又當(dāng)n1時(shí),S1a12a11,a11.數(shù)列an是以1為首項(xiàng),2為公比的等比數(shù)列,an2n1,b1a11,b4a34.數(shù)列bn為等差數(shù)列,bnn.(2)an2n1,bnn,cn2n12n1,Tn4.4在公差不為0的等差數(shù)列an中,a1,a4,a8成等比數(shù)列(1)若數(shù)列an的前10項(xiàng)和為45,求數(shù)列an的通項(xiàng)公式;(2)若bn,且數(shù)列bn的前n項(xiàng)和為Tn,若Tn,求數(shù)列an的公差解:設(shè)數(shù)列an的公差為d(d0),由a1,a4,a8成等比數(shù)列,可得aa1a8,即(a13d)2a1(a17d),得a19d.(1)由數(shù)列an的前10項(xiàng)和為45,得10a145d45,即90d45d45,所以d,a13.故數(shù)列an的通項(xiàng)公式為an3(n1).(2)因?yàn)閎n,所以數(shù)列bn的前n項(xiàng)和Tn,即Tn,因此1,解得d1或1.故數(shù)列an的公差為1或1. 常用結(jié)論記一番常用裂項(xiàng)公式(1);(2);(3)an(anan1)(an1an2)(a2a1)a1a1;(4)n(n1)n(n1)(n2)(n1)n(n1);(5);(6)1.(九)求得通項(xiàng)何愁放縮 速解技法學(xué)一招已知數(shù)列an滿足a18,(n1)an1(n3)an8n8(nN*)(1)求an;(2)求證:.解(1)由已知(n1)an1(n3)an8n8,兩邊同除以(n1)(n2)(n3),得,即8.利用累加法,可得8,化簡(jiǎn)求得an14(n1)(n2),所以an4n(n1)(2)證明:法一:,通過(guò)計(jì)算,當(dāng)n4時(shí),.法二:.當(dāng)n3時(shí),.設(shè)數(shù)列an的前n項(xiàng)和為Sn,滿足2Snan12n11(nN*),且a1,a25,a3成等差數(shù)列(1)求數(shù)列an的通項(xiàng)公式;(2)求證:對(duì)一切正整數(shù)n,有.解(1)由2Snan12n11,得2Sn1an22n21,兩式相減得an23an12n1,2S1a23a22a13,a33a246a113,由a1,a25,a3成等差數(shù)列a1a32(a25)a11.an13an2nan12n13(an2n),數(shù)列an2n為首項(xiàng)是a123,公比是3的等比數(shù)列則an2n3n,an3n2n.(2)證明:法一:當(dāng)n1時(shí),123n22nan2n.11.由上式得:對(duì)一切正整數(shù)n,有.法二:an3n2n(32)(3n13n223n3222n1)3n1,1. 經(jīng)典好題練一手已知數(shù)列an滿足a12且an1(nN*)(1)求證:數(shù)列為等比數(shù)列,并求數(shù)列an的通項(xiàng)公式;(2)求證:n2(nN*)證明:(1)由題意得,即1,故21,又1,所以數(shù)列是以為首項(xiàng),為公比的等比數(shù)列1n1n,ann.(2)由(1)知1111,nnn2n1n2.常用結(jié)論記一番常見幾種放縮形式(1);(2);(3)2;(4);(5);(6)(n1);(7)(n2)(十)繞過(guò)通項(xiàng)也可放縮 速解技法學(xué)一招有一類數(shù)列不等式問(wèn)題,數(shù)列的通項(xiàng)雖然很難求得,,但可借助遞推關(guān)系變形后達(dá)到放縮的目的.常用的放縮變形參考后面的“常用結(jié)論記一番”.已知數(shù)列an的首項(xiàng)為a11,且an1(nN*)(1)求a2,a3的值,并證明:a2n1a2n12;(2)令bn|a2n12|,Snb1b2bn.求證:Sn0,所以0,即an12與an2異號(hào),故an22與an2同號(hào),于是a2n12與a2n12同號(hào)又a1210,所以a2n12.另一方面,a2n1a2n1a2n1a2n1a2n1.由a2n10,即a2n1a2n1.綜上所述,a2n1a2n12.(2)證明:a2n12.由bn|a2n12|,知.又1a2n1a2n12,所以n1.故Snb1b2bn12n1,綜上所述,Sn0,nN*.當(dāng)n1時(shí),12(1),命題成立當(dāng)n2時(shí),由an1ann1得anan1n,所以an(an1an1)1,an1an1.從而有(ak1ak1)an1an2222(1)常用結(jié)論記一番經(jīng)常用到的幾種放縮(1);(2)2.(十一)線性規(guī)劃布線行針 速解技法學(xué)一招解不含實(shí)際背景的線性規(guī)劃問(wèn)題的一般步驟(1)畫出可行域;(2)根據(jù)線性目標(biāo)函數(shù)的幾何意義確定其取得最優(yōu)解的點(diǎn);- 1.請(qǐng)仔細(xì)閱讀文檔,確保文檔完整性,對(duì)于不預(yù)覽、不比對(duì)內(nèi)容而直接下載帶來(lái)的問(wèn)題本站不予受理。
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