資源描述:
《高考數(shù)學(xué)新題型.docx》由會(huì)員上傳分享,免費(fèi)在線閱讀,更多相關(guān)內(nèi)容在教育資源-天天文庫(kù)。
1、高考數(shù)學(xué)新題型選編(共70個(gè)題)1、(Ⅰ)已知函數(shù):f(x)2n1(xna)(xa)n,(x[0,),nN)求函數(shù)f(x)的最小值;(Ⅱ)證明:anbn(ab)n(a0,b0,nN);22均為正數(shù),則有a1na2na3nakn(Ⅲ)定理:若a,a2,aLakL(a1a2a3Lak)n成立13kk(其中k2,kN,k為常數(shù)).請(qǐng)你構(gòu)造一個(gè)函數(shù)g(x),證明:當(dāng)a1,a2,a3,L,ak,ak1均為正數(shù)時(shí),a1na2na3nLakn1(a1a2a3Lak1)n.k1k1解:(Ⅰ)令f'(x)2n1nxn1n(ax)n10得(2x)n1(ax)n
2、12xaxxa?2分當(dāng)0xa時(shí),2xxaf'(x)0故f(x)在[0,a]上遞減.當(dāng)xa,f'(x)0故f(x)在(a,)上遞增.所以,當(dāng)xa時(shí),f(x)的最小值為f(a)0.?.4分(Ⅱ)由b0,有f(b)f(a)0即f(b)2n1(anbn)(ab)n0故anbn(ab)n(a0,b0,nN).???????????????5分22a1na2na3nakn(Ⅲ)證明:要證:1L1(a1a2a31Lak1)nkk只要證:(k1)n1(a1na2na3nLakn1)(a1a2a3Lak1)n設(shè)g(x)(k1)n1(a1na2na3nLxn)
3、(a1a2a3Lx)n???????7分則g'(x)(k1)n1nxn1n(a1a2Lakx)n1令g'(x)0得xa1a2Lak????????????????????.8分k當(dāng)0xa1a2Lak時(shí),g'(x)n[(kxx]n1n(a1a2Lakx)n1kx)n1x)n1n(a1a2Lakn(a1a2Lak0故g(x)在[0,a1a2Lak]上遞減,類似地可證g(x)在(a1a2Lak,)遞增kk所以當(dāng)xa1a2Lak時(shí),g(x)的最小值為g(a1a2Lak)??????10分kk而g(a1a2Lak)(k1)n1[a1na2nLakn(
4、a1a2Lak)n](a1a2Laka1a2Lak)nkkk=(k1)n1[kn(a1na2nLakn)(a1a2Kak)n(k1)(a1a2Lak)n]kn(k1)n1[knnnLnk(a1a2Ln]=(k1)n1n1nnLn(a1a2n=n(a1a2ak)ak)n1[k(a1a2ak)Lak)]kk故g(a1a2Lak)由定理知:kn1(a1na2nLakn)(a1a2Lak)n00g(a1a2Lak)kQak1[0,)g(ak1)0k故(k1)n1(a1na2na3nLakn1)(a1a2a3Lak1)n即:anananLanaaaL
5、an.??????????..14分1231k1(1231k1)kk2、用類比推理的方法填表等差數(shù)列an中等比數(shù)列bn中a3=a2db3b2?qa3a4a2a5b3?b4b2?b5a1a2a3a4a55a3答案:b1?b2?b3?b4?b5b353、10.定義一種運(yùn)算“*”:對(duì)于自然數(shù)n滿足以下運(yùn)算性質(zhì):(i)1*1=1,(ii)(n+1)*1=n*1+1,則n*1等于A.nB.n+1C.n-1D.n2答案:D4、若f(n)為n21(nN*)的各位數(shù)字之和,如:1421197,19717,則f(14)17;記f1(n)f(n),f2(n)f
6、(f1(n)),,fk1(n)f(fk(n)),kN*,則f2008(8)____答案:55、下面的一組圖形為某一四棱錐S-ABCD的側(cè)面與底面。aaaaa2aaaaa(1)請(qǐng)畫出四棱錐S-ABCD的示意圖,是否存在一條側(cè)棱垂直于底面?如果存在,請(qǐng)給出證明;如果不存在,請(qǐng)說明理由;(2)若SA面ABCD,E為AB中點(diǎn),求二面角E-SC-D的大??;(3)求點(diǎn)D到面SEC的距離。(1)存在一條側(cè)棱垂直于底面(如圖)??????3分證明:SAAB,SAAD,且AB、AD是面ABCD內(nèi)的交線SA底面ABCD????????(2)分別取SC、SD的中
7、點(diǎn)G、F,連GE、GF、FA,S則GF//EA,GF=EA,AF//EG而由SA面ABCD得SACD,2aa5分又ADCD,CD面SAD,CDAF又SA=AD,F是中點(diǎn),AFSDAF面SCD,EG面SCD,面SEC面SCD所以二面角E-SC-D的大小為90????10分(3)作DHSC于H,面SEC面SCD,DH面SEC,DH之長(zhǎng)即為點(diǎn)D到面SEC的距離,12分FGADEH在RtSCD中,DHSDDC2aa6aSC3a3BC答:點(diǎn)D到面SEC的距離為6a?????????14分36、一個(gè)計(jì)算裝置有一個(gè)入口A和一輸出運(yùn)算結(jié)果的出口B,將自
8、然數(shù)列n(n1)中的各數(shù)依次輸入Aan,結(jié)果表明:①?gòu)腁口輸入n1;②當(dāng)n2時(shí),口,從B口得到輸出的數(shù)列1時(shí),從B口得a13從A口輸入n,從B口得到的結(jié)果an是將前一結(jié)果an1先