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1、,第二篇重點(diǎn)專題分層練,中高檔題得高分,第27練壓軸小題專練(1),明晰考情 高考題中填空題的最后2或3個(gè)小題,往往出現(xiàn)邏輯思維深刻,難度高檔的題目.,,欄目索引,核心考點(diǎn)突破練,,,高考押題沖刺練,考點(diǎn)一與函數(shù)有關(guān)的壓軸小題,,核心考點(diǎn)突破練,方法技巧本類壓軸題常以超越方程、分段函數(shù)、抽象函數(shù)等為載體,考查函數(shù)性質(zhì)、函數(shù)零點(diǎn)、參數(shù)的范圍和通過函數(shù)性質(zhì)求解不等式.解決該類問題的途徑往往是構(gòu)造函數(shù),進(jìn)而研究函數(shù)的性質(zhì),利用函數(shù)性質(zhì)去求解問題是常用方法,其間要注意導(dǎo)數(shù)的應(yīng)用.,1.偶函數(shù)f(x)滿足f(x1)f(x1),且當(dāng)x1,0時(shí),f(x)x2,若函數(shù)g(x)f(x)|lg x|,則g(x)在
2、(0,10)上的零點(diǎn)個(gè)數(shù)為________.,答案,解析,10,f(x1)f(x1),f(x)f(x2),故f(x)是周期函數(shù),且T2, 又函數(shù)f(x)是R上的偶函數(shù), f(1x)f(1x),f(x)的圖象關(guān)于x1對(duì)稱, 當(dāng)x0時(shí),在同一坐標(biāo)系中作出yf(x)和y|lg x|的圖象,如圖所示.,由圖象知函數(shù)g(x)的零點(diǎn)個(gè)數(shù)為10.,答案,解析,答案,解析,則實(shí)數(shù)t的取值范圍為________.,解析無論m1還是0
3、x23)ex,設(shè)關(guān)于x的方程f 2(x)af(x)0(aR)有4個(gè)不同的實(shí)數(shù)解,則a的取值范圍是 ______________.,解析由題意知,f(x)2xex(x23)ex ex(x22x3), 令f(x)0,解得x1或x3, 所以當(dāng)x1時(shí),f(x)0,當(dāng)3
4、法技巧數(shù)列與函數(shù)的交匯、數(shù)列與不等式的交匯問題是高考的熱點(diǎn).解決這類問題的關(guān)鍵在于利用數(shù)列與函數(shù)的對(duì)應(yīng)關(guān)系,將條件進(jìn)行準(zhǔn)確的轉(zhuǎn)化,確定數(shù)列的通項(xiàng)或前n項(xiàng)和,利用函數(shù)的性質(zhì)、圖象求解最值問題,不等關(guān)系或恒成立問題.,答案,解析,5.在公比為q的正項(xiàng)等比數(shù)列an中,a44,則當(dāng)2a2a6取得最小值時(shí), log2q_____.,答案,解析,因?yàn)閿?shù)列bn是單調(diào)遞增數(shù)列, 所以當(dāng)n2時(shí),由bn1bn, 得(n2)2n(n12)2n1,解得n21,,7.已知Sn和Tn分別為數(shù)列an與數(shù)列bn的前n項(xiàng)和,且a1e4,SneSn1e5,an ,則當(dāng)Tn取得最大值時(shí)n的值為________.,答案,解析,4或
5、5,解析由SneSn1e5,得Sn1eSne5(n2),,因?yàn)閍n ,所以bn5n.,所以當(dāng)n4或n5時(shí),Tn取得最大值.,,答案,解析,解得a1或a4.,當(dāng)a1時(shí),f(x)x29x10,數(shù)列an不是等差數(shù)列; 當(dāng)a4時(shí),f(x)x24x,Snf(n)n24n, a15,a27,an5(75)(n1)2n3,,1.(2018全國改編)已知f(x)是定義域?yàn)?,)的奇函數(shù),滿足f(1x)f(1x).若f(1)2,則f(1)f(2)f(3)f(50)________.,,高考押題沖刺練,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,2,1,2,3,4,5,6,7,8,9,10
6、,11,12,解析f(x)是奇函數(shù),f(x)f(x), f(1x)f(x1).f(1x)f(1x), f(x1)f(x1),f(x2)f(x), f(x4)f(x2)f(x)f(x), 函數(shù)f(x)是周期為4的周期函數(shù). 由f(x)為奇函數(shù)及其定義域?yàn)镽得f(0)0. 又f(1x)f(1x), f(x)的圖象關(guān)于直線x1對(duì)稱, f(2)f(0)0,f(2)0.,1,2,3,4,5,6,7,8,9,10,11,12,又f(1)2,f(1)2, f(1)f(2)f(3)f(4)f(1)f(2)f(1)f(0)20200, f(1)f(2)f(3)f(4)f(49)f(50)012f(49)f(50
7、)f(1)f(2)202.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,(,2,1,2,3,4,5,6,7,8,9,10,11,12,解析設(shè)mf(x),作出函數(shù)f(x)的圖象,如圖所示, 則當(dāng)m1時(shí),mf(x)有兩個(gè)根, 當(dāng)m<1時(shí),mf(x)有一個(gè)根. 若關(guān)于x的方程f 2(x)f(x)t0有三個(gè)不同的實(shí)根, 則等價(jià)為m2mt0有兩個(gè)不同的實(shí)數(shù)根m1,m2, 且m11,m21. 當(dāng)m1時(shí),t2, 此時(shí)由m2m20,解得m1或m2, f(x)1有兩個(gè)根,f(x)2有一個(gè)根,滿足條件;,1,2,3,4,5,6,7,8,9,10,11,12,則需h(1)<0即可,即11t<0
8、,解得t<2. 綜上,實(shí)數(shù)t的取值范圍為t2.,3.若存在兩個(gè)正實(shí)數(shù)x,y使等式2xm(y2ex)(ln yln x)0成立(其中e 2.718 28),則實(shí)數(shù)m的取值范圍是____________________.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,解析當(dāng)m0時(shí),不滿足題意,,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,設(shè)h(t)g(t),,則g(t)在(0,)上單調(diào)遞減, 當(dāng)te時(shí),g(t)0, 則當(dāng)t(0,e)時(shí),g(t)0,函數(shù)g(t)單調(diào)遞增, 當(dāng)t(e,)時(shí),g(t)<0,函數(shù)g(t)單
9、調(diào)遞減,,且當(dāng)t0時(shí),g(t),,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,因?yàn)榇嬖趚1,x20,1,使得f(x1)g(x2)成立,,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,5.設(shè)數(shù)列an的前n項(xiàng)和為Sn,若 為常數(shù),則稱數(shù)列an為“精致數(shù)列”.已知等差數(shù)列bn的首項(xiàng)為1,公差不為0,若數(shù)列bn為“精致數(shù)列”,則數(shù)列bn的通項(xiàng)公式為__________________.,bn2n1(nN*),解析設(shè)等差數(shù)列bn的公差為d,,即2(n1)d
10、4k2k(2n1)d,整理得(4k1)dn(2k1)(2d)0, 因?yàn)閷?duì)任意正整數(shù)n上式恒成立,,所以數(shù)列bn的通項(xiàng)公式為bn2n1(nN*).,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,4,解析依題意可得bn1pbn,則數(shù)列bn為等比數(shù)列.,當(dāng)且僅當(dāng)b8b922,即該數(shù)列為常數(shù)數(shù)列時(shí)取等號(hào).,答案,解析,7.當(dāng)n為正整數(shù)時(shí),定義函數(shù)N(n)表示n的最大奇因數(shù).如N(3)3,N(10)5,,S(n)N(1)N(2)N(3)N(2n),則S(5)________.,1,2,3,4,5,6,7,8,9,10,11,12,
11、342,1,2,3,4,5,6,7,8,9,10,11,12,解析N(2n)N(n),N(2n1)2n1, 而S(n)N(1)N(2)N(3)N(2n), S(n)N(1)N(3)N(5)N(2n1)N(2)N(4)N(2n), S(n)1352n1N(1)N(2)N(3)N(2n1),,即S(n)S(n1)4n1,又S(1)N(1)N(2)112, S(5)S(1)S(5)S(4)S(4)S(3)S(2)S(1) 4443424, S(5)24424344342.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,42,所以a2a4a6328242.,答案,解析,1,2,3,
12、4,5,6,7,8,9,10,11,12,解析設(shè)等比數(shù)列an的公比為q,依題意得2(a32)a2a4, 又S4a128,a2a3a428,得a38,,又a2a1,a12,q2,an2n,Sn2n12.,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,即Tn2n2Tn12n1, 故數(shù)列Tn2n2單調(diào)遞減,,又Tn2n2M恒成立,,10.已知數(shù)列an的奇數(shù)項(xiàng)是首項(xiàng)為1的等差數(shù)列,偶數(shù)項(xiàng)是首項(xiàng)為2的等比數(shù)列,數(shù)列an的前n項(xiàng)和為Sn,且滿足a4S3,a9a3a4,則使得 恰好為數(shù)列a
13、n的奇數(shù)項(xiàng)的正整數(shù)k的值為________.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,1,1,2,3,4,5,6,7,8,9,10,11,12,解析設(shè)等差數(shù)列的公差為d,等比數(shù)列的公比為q, 則a11,a22,a31d,a42q,a914d. 因?yàn)閍4S3,a9a3a4, 所以121d2q,14d1d2q, 解得d2,q3, 則對(duì)于nN*,有a2n12n1,a2n23n1, 所以S2n13(2n1)2(13323n1)3nn21, S2n1S2na2n3n1n21.,即(3m)3k1(m1)(k21). 當(dāng)k1時(shí),m3,滿足條件;,解得1
14、此時(shí)滿足條件的正整數(shù)k不存在. 綜上,k1.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,解析f(x)x2(ln 3x)22a(x3ln 3x)10a2(xa)2(ln 3x3a)2表示點(diǎn)M(x,ln 3x)與點(diǎn)N(a,3a)距離的平方,,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,12.(2018江蘇省海安高級(jí)中學(xué)月考)已知公比不為1的等比數(shù)列an中,a11,a2a,且an1k(anan2)對(duì)任意正整數(shù)n都成立,且對(duì)任意相鄰三項(xiàng)am,am1,am2按某順序排列后成等差數(shù)列,則滿足題意的 k的值為________.,1,2,3,4,5,6,7,8,9,10,11,12,所以amam1,am1am,am2am1. 若am1為等差中項(xiàng),則2am1amam2, 即2amam1am1,解得a1,不合題意. 若am為等差中項(xiàng),則2amam1am2, 即2am1amam1,化簡(jiǎn)得a2a20, 解得a2或a1(舍去).,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,若am2為等差中項(xiàng),則2am2am1am, 即2am1amam1,化簡(jiǎn)得2a2a10,,本課結(jié)束,