PAGEPAGE11綜合拔高練五年高考練考點(diǎn)1函數(shù)的概念與表示1.(2019江蘇,4,5分,)函數(shù)y=7+6x-x2.(2016浙江,12,6分,)設(shè)函數(shù)f(x)=x3+3x2+1.已知a≠0,且f(x)-f(a)=(x-b)(x-a)2,x∈R,則實(shí)數(shù)a=,b=.

考點(diǎn)2分段函數(shù)的應(yīng)用3.(2019課標(biāo)全國Ⅱ,12,5分,)設(shè)函數(shù)f(x)的定義域?yàn)镽,滿足f(x+1)=2f(x),且當(dāng)x∈(0,1]時(shí),f(x)=x(x-1).若對(duì)任意x∈(-∞,m],都有f(x)≥-89,則m的取值范圍是 ()A.-∞,94 B.-∞,73C.4.(2018天津,14,5分,)已知a∈R,函數(shù)f(x)=x2+2x+a-2,x≤0,-x2+2x-2a,考點(diǎn)3函數(shù)基本性質(zhì)的綜合運(yùn)用5.(2020天津,3,5分,)函數(shù)y=4xx2+1的圖象大致為6.(2018課標(biāo)全國Ⅱ,11,5分,)已知f(x)是定義域?yàn)?-∞,+∞)的奇函數(shù),滿足f(1-x)=f(1+x).若f(1)=2,則f(1)+f(2)+f(3)+…+f(50)= ()A.-50 B.0C.2 D.507.(2020全國新高考Ⅰ,8,5分,)若定義在R的奇函數(shù)f(x)在(-∞,0)單調(diào)遞減,且f(2)=0,則滿足xf(x-1)≥0的x的取值范圍是()A.[-1,1]∪[3,+∞) B.[-3,-1]∪[0,1]C.[-1,0]∪[1,+∞) D.[-1,0]∪[1,3]8.(2019浙江,16,4分,)已知a∈R,函數(shù)f(x)=ax3-x.若存在t∈R,使得|f(t+2)-f(t)|≤23,則實(shí)數(shù)a的最大值是.

考點(diǎn)4冪函數(shù)及其應(yīng)用9.(2020全國Ⅱ文,10,5分,)設(shè)函數(shù)f(x)=x3-1x3,則f(x)(A.是奇函數(shù),且在(0,+∞)單調(diào)遞增 B.是奇函數(shù),且在(0,+∞)單調(diào)遞減C.是偶函數(shù),且在(0,+∞)單調(diào)遞增 D.是偶函數(shù),且在(0,+∞)單調(diào)遞減10.(2017山東,10,5分,)已知當(dāng)x∈[0,1]時(shí),函數(shù)y=(mx-1)2的圖象與y=x+m的圖象有且只有一個(gè)交點(diǎn),則正實(shí)數(shù)m的取值范圍是 ()A.(0,1]∪[23,+∞) B.(0,1]∪[3,+∞)C.(0,2]∪[23,+∞) D.(0,2]∪[3,+∞)11.(2020江蘇,7,5分,)已知y=f(x)是奇函數(shù),當(dāng)x≥0時(shí),f(x)=x23,則f(-8)的值是三年模擬練應(yīng)用實(shí)踐1.(2021北京房山高一上期中,)已知函數(shù)f(x)=x,x≥a,x2,0<x<a,若對(duì)任意的x1,x2∈(0,+∞),且x1A.(0,+∞) B.(0,1] C.(1,+∞) D.[1,+∞)2.(2020黑龍江大慶實(shí)驗(yàn)中學(xué)高一上月考,)設(shè)f(x)=x,0<x<1,2(x-1),x≥1,若fA.8 B.6 C.4 D.23.(2020山東德州高一上期中,)已知函數(shù)f(x)是定義在R上的單調(diào)函數(shù),A(0,1),B(2,-1)是其圖象上的兩點(diǎn),則不等式|f(x-1)|>1的解集為 ()A.(-1,1) B.(-∞,-1)∪(1,+∞)C.(1,3) D.(-∞,1)∪(3,+∞)4.(多選)(2020山東菏澤高一上期末聯(lián)考,)下列關(guān)于函數(shù)f(x)=x2-x4|xA.f(x)的定義域?yàn)閇-1,0)∪(0,1]B.f(x)的值域?yàn)?-1,1)C.f(x)在定義域上是增函數(shù)D.f(x)的圖象關(guān)于原點(diǎn)對(duì)稱5.(多選)(2021山東省實(shí)驗(yàn)中學(xué)高一上期中,)對(duì)于定義在R上的函數(shù)f(x),下列說法正確的是 ()A.若f(x)是奇函數(shù),則f(x-1)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱B.若對(duì)x∈R,有f(x+1)=f(x-1),則f(x)的圖象關(guān)于直線x=1對(duì)稱C.若函數(shù)f(x+1)的圖象關(guān)于直線x=-1對(duì)稱,則f(x)為偶函數(shù)D.若f(1+x)+f(1-x)=2,則f(x)的圖象關(guān)于點(diǎn)(1,1)對(duì)稱6.(多選)(2020山東淄博高一上期中,)我們把定義域?yàn)閇0,+∞)且同時(shí)滿足以下兩個(gè)條件的函數(shù)f(x)稱為“Ω函數(shù)”:(1)對(duì)任意的x∈[0,+∞),總有f(x)≥0;(2)若x≥0,y≥0,則有f(x+y)≥f(x)+f(y)成立.下列判斷正確的是 ()A.若f(x)為“Ω函數(shù)”,則f(0)=0B.若f(x)為“Ω函數(shù)”,則f(x)在[0,+∞)上為增函數(shù)C.函數(shù)g(x)=0,x∈Q,D.函數(shù)g(x)=x2+x在[0,+∞)上是“Ω函數(shù)”7.(2021天津第二南開學(xué)校高一上期中,)已知f(x)=12x+1,x≤0,-(x-1)28.(2020天津六校高一上期中聯(lián)考,)已知函數(shù)f(x)=x2-4x+10(x∈[m,n])的值域?yàn)閇3m,3n],則2m+n=.

9.(2021北京人大附中高一上期中,)設(shè)函數(shù)f(x)=x,(1)若存在x∈R,使得f(1+x)=f(1-x)成立,則實(shí)數(shù)a的取值范圍是;

(2)若函數(shù)f(x)為R上的單調(diào)函數(shù),則實(shí)數(shù)a的取值范圍是.

10.(2020湖南衡陽一中高一上期中,)已知函數(shù)f(x)對(duì)任意的實(shí)數(shù)a,b都有f(a+b)=f(a)+f(b),且當(dāng)x>0時(shí),有f(x)>0.(1)求證:f(x)是R上的增函數(shù);(2)求證:f(x)是R上的奇函數(shù);(3)若f(1)=1,解不等式f(x2)-f(x+2)>4.遷移應(yīng)用11.(2020山東煙臺(tái)高一上期中,)經(jīng)過函數(shù)性質(zhì)的學(xué)習(xí),我們知道“函數(shù)y=f(x)的圖象關(guān)于y軸成軸對(duì)稱圖形”的充要條件是“y=f(x)為偶函數(shù)”.(1)若f(x)為偶函數(shù),且當(dāng)x≤0時(shí),f(x)=2x-1,求f(x)的解析式,并求不等式f(x)>f(2x-1)的解集;(2)某數(shù)學(xué)學(xué)習(xí)小組針對(duì)上述結(jié)論進(jìn)行探究,得到一個(gè)真命題:“函數(shù)y=f(x)的圖象關(guān)于直線x=a成軸對(duì)稱圖形”的充要條件是“y=f(x+a)為偶函數(shù)”.若函數(shù)g(x)的圖象關(guān)于直線x=1對(duì)稱,且當(dāng)x≥1時(shí),g(x)=x2-1x①求g(x)的解析式;②求不等式g(x)>g(3x-1)的解集.

答案全解全析五年高考練1.答案[-1,7]解析由題意可得7+6x-x2≥0,即x2-6x-7≤0,解得-1≤x≤7,故該函數(shù)的定義域是[-1,7].2.答案-2;1解析f(x)-f(a)=x3-a3+3(x2-a2)=(x-a)[x2+ax+a2+3(x+a)]=(x-a)[x2+(a+3)·x+a2+3a]=(x-a)(x-a)(x-b),則x2+(a+3)x+a2+3a=x2-(a+b)x+ab,即a+3=-(3.B由題可知,當(dāng)x∈(0,1]時(shí),f(x)=x(x-1)=x2-x,則當(dāng)x=12時(shí),f(x)min=-14,且當(dāng)x=13時(shí),f(x)=-29.當(dāng)x∈(1,2]時(shí),x-1∈(0,1],則f(x)=2f(x-1).當(dāng)x∈(-1,0]時(shí),x+1∈(0,1],則f(x)=12∴若x∈(1,2],則當(dāng)x=32時(shí),f(x)min=-12,且x=43時(shí),f(x同理,若x∈(2,3],則當(dāng)x=52時(shí),f(x)min=-1,且x=73時(shí),f(x)=-∴函數(shù)f(x)的大致圖象如圖所示.∵f(x)≥-89對(duì)任意x∈(-∞,m]恒成立,∴當(dāng)x∈(-∞,m]時(shí),f(x)min≥-89,由圖可知m≤73.4.答案1解析當(dāng)x>0時(shí),f(x)=-x2+2x-2a,此時(shí)只需-x2+2x-2a≤x恒成立,即2a≥-x2+x恒成立,因?yàn)閤>0時(shí),y=-x2+x的最大值為14所以a≥18當(dāng)-3≤x≤0時(shí),f(x)=x2+2x+a-2,此時(shí)只需x2+2x+a-2≤-x恒成立,即a≤-x2-3x+2恒成立,因?yàn)?3≤x≤0時(shí),y=-x2-3x+2的最小值為2,所以a≤2.故a的取值范圍為185.A設(shè)y=f(x)=4xx2+1,易知f(x)的定義域?yàn)镽,f(-x)=-4xx2+1=-f(x),∴函數(shù)f(x)=4xx2+1是奇函數(shù),∴y=f(x)的圖象關(guān)于原點(diǎn)對(duì)稱6.C因?yàn)閒(x)是定義在(-∞,+∞)上的奇函數(shù),所以f(-x)=-f(x)①,且f(0)=0.又因?yàn)閒(1-x)=f(1+x),所以f(-x)=f(2+x)②.由①②可得f(x+2)=-f(x),則有f(x+4)=f(x).由f(1)=2,得f(-1)=-2,于是有f(2)=f(0)=0,f(3)=f(-1)=-2,f(4)=f(0)=0,f(5)=f(1)=2,f(6)=f(2)=0,……,所以f(1)+f(2)+f(3)+…+f(50)=12×[f(1)+f(2)+f(3)+f(4)]+f(49)+f(50)=12×0+f(1)+f(2)=2+0=2.7.D∵f(x)是定義在R上的奇函數(shù),∴f(x-1)的圖象關(guān)于點(diǎn)(1,0)中心對(duì)稱,又∵f(x)在(-∞,0)上單調(diào)遞減,∴f(x-1)在(-∞,1)上單調(diào)遞減,在(1,+∞)上也單調(diào)遞減,且過(-1,0)和(3,0),f(x-1)的大致圖象如圖:若xf(x-1)≥0,則x≥0,解得1≤x≤3或-1≤x≤0.綜上,滿足xf(x-1)≥0的x的取值范圍是[-1,0]∪[1,3].故選D.8.答案4解析|f(t+2)-f(t)|=|a(t+2)3-(t+2)-(at3-t)|=|a(6t2+12t+8)-2|.令m=6t2+12t+8=6(t+1)2+2,則m∈[2,+∞),設(shè)g(m)=f(t+2)-f(t)=am-2,則|g(m)|≤23有解當(dāng)a=0時(shí),g(m)=-2,不符合題意;當(dāng)a>0時(shí),g(m)∈[2a-2,+∞),∵|g(m)|≤23有解,∴2a-2≤23,得0<a≤當(dāng)a<0時(shí),g(m)∈(-∞,2a-2],∵|g(m)|≤23有解,∴2a-2≥-23,得a≥23,與a綜上可知,0<a≤43,即a的最大值為49.A由函數(shù)y=x3和y=-1x3都是奇函數(shù),知函數(shù)f(x)=x3-1x3是奇函數(shù).由函數(shù)y=x3和y=-1x3都在區(qū)間(0,+∞)上單調(diào)遞增,知函數(shù)f(x)=x3-1x3在區(qū)間(0,+∞)上單調(diào)遞增,故函數(shù)f(x)=x3-1x10.B①當(dāng)0<m≤1時(shí),在同一平面直角坐標(biāo)系中作出函數(shù)y=(mx-1)2與y=x+m的圖象,如圖.易知此時(shí)兩函數(shù)圖象在x∈[0,1]上有且只有一個(gè)交點(diǎn);②當(dāng)m>1時(shí),在同一平面直角坐標(biāo)系中作出函數(shù)y=(mx-1)2與y=x+m的圖象,如圖.要滿足題意,則(m-1)2≥1+m,解得m≥3或m≤0(舍去),∴m≥3.綜上,正實(shí)數(shù)m的取值范圍為(0,1]∪[3,+∞).方法總結(jié)已知函數(shù)有零點(diǎn)(方程有根或圖象有交點(diǎn))求參數(shù)的值或取值范圍常用的方法:①直接法:直接根據(jù)題設(shè)條件構(gòu)建關(guān)于參數(shù)的方程或不等式,再通過解方程或不等式確定參數(shù)的值或取值范圍.②分離參數(shù)法:先將參數(shù)分離,轉(zhuǎn)化成求函數(shù)最值問題加以解決.③數(shù)形結(jié)合法:在同一平面直角坐標(biāo)系中畫出函數(shù)的圖象,然后數(shù)形結(jié)合求解.11.答案-4解析由函數(shù)f(x)是奇函數(shù)得f(-8)=-f(8)=-823三年模擬練1.B根據(jù)題意,對(duì)任意的x1,x2∈(0,+∞),且x1<x2都有f(x2)-f(x1)x又函數(shù)f(x)=x,x≥a,x2,0<x<a,所以a易錯(cuò)警示解決分段函數(shù)的單調(diào)性,一方面要考慮每一段函數(shù)的單調(diào)性,另一方面要考慮函數(shù)在分界點(diǎn)處函數(shù)值之間的大小關(guān)系,解題時(shí)要防止遺漏某一方面導(dǎo)致解題錯(cuò)誤.2.C由題意知,當(dāng)a∈(0,1)時(shí),若f(a)=f(a+1),則a=2a,所以a=14,則f1a-當(dāng)a∈[1,+∞)時(shí),若f(a)=f(a+1),則2(a-1)=2a,顯然無解.綜上可得f1a-1=4,3.D由題意可知f(0)=1,f(2)=-1,又知f(x)是定義在R上的單調(diào)函數(shù),所以f(x)在R上單調(diào)遞減.由|f(x-1)|>1得f(x-1)>1或f(x-1)<-1,即f(x-1)>f(0)或f(x-1)<f(2),所以x-1<0或x-1>2,解得x<1或x>3,故選D.4.ABD由x2-x4≥0,|x-1|-1≠0,得-1≤x≤1且x≠0,此時(shí)f(x)=x2-x4-(x-1)-1=x2-x4-x=|x|1-x2-x,因此A正確;當(dāng)0<x≤1時(shí),f(x)=-1-x2∈(-1,0],當(dāng)-1≤x<0時(shí),f(x)=1-x2∈[0,1),故f(x)的值域?yàn)?-1,1),因此5.ACD對(duì)于A,將f(x)的圖象向右平移1個(gè)單位長度得到函數(shù)f(x-1)的圖象,若f(x)為奇函數(shù),則其圖象關(guān)于點(diǎn)(0,0)對(duì)稱,則函數(shù)f(x-1)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱,A正確;對(duì)于B,若對(duì)x∈R,有f(x+1)=f(x-1),即f(x-2)=f(x),函數(shù)f(x)的圖象不一定關(guān)于直線x=1對(duì)稱,B錯(cuò)誤;對(duì)于C,將f(x+1)的圖象向右平移1個(gè)單位長度得到函數(shù)f(x)的圖象,若函數(shù)f(x+1)的圖象關(guān)于直線x=-1對(duì)稱,則f(x)的圖象關(guān)于直線x=0對(duì)稱,即f(x)為偶函數(shù),C正確;對(duì)于D,若f(1+x)+f(1-x)=2,即f(1+x)-1=-[f(1-x)-1],則f(x)的圖象關(guān)于點(diǎn)(1,1)對(duì)稱,D正確.故選ACD.6.AD對(duì)于選項(xiàng)A,由條件(1)知,f(x)≥0,則f(0)≥0,由條件(2)知,f(0+0)≥f(0)+f(0),即f(0)≤0,所以f(0)=0,A正確;對(duì)于選項(xiàng)B,當(dāng)f(x)=0(x∈[0,+∞))時(shí),符合條件(1),(2),f(x)是“Ω函數(shù)”,但f(x)在[0,+∞)上不是增函數(shù),B錯(cuò)誤;對(duì)于選項(xiàng)C,取x=2-2,y=2+2,則g(2-2)=1,g(2+2)=1,g[(2-2)+(2+2)]=g(4)=0,不滿足g(x+y)≥g(x)+g(y),所以g(x)不是“Ω函數(shù)”,C錯(cuò)誤;對(duì)于選項(xiàng)D,g(x)=x2+x在[0,+∞)上單調(diào)遞增,所以g(x)≥g(0)=0,滿足條件(1),g(x+y)-g(x)-g(y)=[(x+y)2+(x+y)]-(x2+x)-(y2+y)=2xy,當(dāng)x≥0,y≥0時(shí),2xy≥0,此時(shí)g(x+y)≥g(x)+g(y),滿足條件(2),D正確.故選AD.7.答案[-4,2]解析∵f(x)≥-1,∴x≤0,∴-4≤x≤0或0<x≤2,即-4≤x≤2.∴使f(x)≥-1成立的x的取值范圍是[-4,2].解題模板解與分段函數(shù)有關(guān)的不等式,應(yīng)在不同的區(qū)間上分類求解,再求它們的并集.8.答案9解析∵f(x)=x2-4x+10=(x-2)2+6≥6,∴3m≥6,∴m≥2,又函數(shù)f(x)圖象的對(duì)稱軸為直線x=2,∴函數(shù)f(x)在[m,n]上單調(diào)遞增.∴f(m)=3m,f(n)=3n,即m2-4m+10=3m,n2-4n+10=3n,解得m=2或m=5,n=2或n=5,又m<n,∴m=2,n=5,∴2m+n=4+5=9,故答案為9.9.答案(1)a>1(2)a≤0或a=1解析(1)若f(1+x)=f(1-x),則函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱.在同一直角坐標(biāo)系中畫出函數(shù)y=x和y=-x2+2x的圖象,如圖所示.若存在x∈R,使得f(1+x)=f(1-x),則a>1.(2)結(jié)合圖象知,若f(x)為R上的單調(diào)函數(shù),則a≤0或a=1.10.解析(1)證明:任取x1,x2∈R,且x1<x2,則f(x2)-f(x1)=f(x2-x1+x1)-f(x1),∵對(duì)任意的實(shí)數(shù)a,b都有f(a+b)=f(a)+f(b),∴f(x2-x1+x1)=f(x2-x1)+f(x1),∴f(x2)-f(x1)=f(x2-x1).∵當(dāng)x>0時(shí),f(x)>0,且x2-x1>0,∴f(x2-x1)>0,∴f(x2)>f(x1),即f(x)在R上為增函數(shù).(2)證明:∵對(duì)任意的實(shí)數(shù)a,b都有f(a+b)=f(a)+f(b),∴令a=b=0,則f(0)=f(0)+f(0)=2f(0),∴f(0)=0,令a=x,b=-x,則f(x-x)=f(x)+f(-x)=f(0)=0,∴f(-x)=-f(x),即函數(shù)