高中數(shù)學(xué)學(xué)考復(fù)習(xí)優(yōu)化練習(xí)5冪函數(shù)含答案

優(yōu)化集訓(xùn)5冪函數(shù)基礎(chǔ)鞏固1.已知y=x2,y=12x,y=4x2,y=x5+1,y=(x-1)2,y=x,y=ax(a>1),上述函數(shù)是冪函數(shù)的個(gè)數(shù)是(A.0 B.1 C.2 D.32.(2023浙江諸暨期末)已知冪函數(shù)f(x)=xα的圖象過(guò)點(diǎn)(2,4),若f(m)=4,則實(shí)數(shù)mA.2 B.±2 C.4 D.±43.(2023浙江溫州期末)已知冪函數(shù)f(x)=xα,則“α>0”是“此冪函數(shù)圖象過(guò)點(diǎn)(1,1)”的()A.充分不必要條件B.必要不充分條件C.充分必要條件D.既不充分也不必要條件4.函數(shù)y=ax2+a與y=ax(a≠0)在同一坐標(biāo)系中的圖象可能是(5.(2021浙江杭州期末測(cè)試)已知函數(shù)f(x)=x-2,若f(2a2-5a+4)<f(a2+a+4),則實(shí)數(shù)a的取值范圍是(A.(-∞,12)∪(2,+∞) B.C.(0,12]∪[2,6) D.6.(多選)下列關(guān)于冪函數(shù)描述正確的有()A.冪函數(shù)的圖象必過(guò)定點(diǎn)(0,0)和(1,1)B.冪函數(shù)的圖象不可能過(guò)第四象限C.當(dāng)冪指數(shù)α=-1,12,3時(shí),冪函數(shù)y=xαD.當(dāng)冪指數(shù)α=12,3時(shí),冪函數(shù)y=xα7.(多選)已知冪函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)P(4,2),則()A.f(x)=(2)xB.f(x)的定義域?yàn)閇0,+∞)C.f(x)的值域?yàn)閇0,+∞)D.f(x)>x2的解集為(0,1)8.若函數(shù)y=xα的圖象經(jīng)過(guò)點(diǎn)(2,16)與(3,m),則m的值為.

9.已知函數(shù)f(x)=x2-ax+2,若函數(shù)在[1,+∞)上有兩個(gè)零點(diǎn),則a的取值范圍是.

10.若冪函數(shù)f(x)=(m2-3m-3)·xm在(0,+∞)上為增函數(shù),則實(shí)數(shù)m=.

11.若(a+1)-1<(3-2a)-1,則實(shí)數(shù)a的取值范圍是.

12.已知冪函數(shù)f(x)=x(m2+m)-1(m∈N*),經(jīng)過(guò)點(diǎn)(2,2),若f(2-a)>f13.給出封閉函數(shù)的定義:若對(duì)于定義域D內(nèi)的任意一個(gè)自變量x0,都有函數(shù)值f(x0)∈D,則稱(chēng)函數(shù)y=f(x)在D上封閉.若定義域D=(0,1),則函數(shù)①f1(x)=3x-1;②f2(x)=-12x2-12x+1;③f3(x)=1-x;④f4(x)=x12,其中在D上封閉的是14.已知冪函數(shù)f(x)=(m-1)2xm2-4m(1)求m的值;(2)當(dāng)x∈[1,2]時(shí),記f(x)的值域?yàn)榧螦,若集合B=[2-k,4-k],且A∩B=?,求實(shí)數(shù)k的取值范圍.15.已知函數(shù)f(x)=x+ax(a>0),具有如下性質(zhì):在(0,a]上單調(diào)遞減,在[a,+∞)上單調(diào)遞增(1)若函數(shù)y=x+2bx(x>0)的值域?yàn)閇6,+∞),求(2)已知函數(shù)f(x)=4x2-12x-32x16.已知函數(shù)f(x)=(a2-a-1)xa(a是常數(shù))為冪函數(shù),且f(x)在(0,+∞)上單調(diào)遞增.(1)求f(x)的表達(dá)式;(2)判斷函數(shù)g(x)=f(x)+4x17.已知冪函數(shù)f(x)=(k2-4k+5)x-m2+4m(m∈Z)的圖象關(guān)于y(1)求m和k的值;(2)求滿(mǎn)足不等式(2a-1)-3<(a+2)-能力提升18.已知a=243,b=425,c=A.b<a<c B.a<b<cC.b<c<a D.c<a<b19.若點(diǎn)(m,81)在冪函數(shù)f(x)=(m-2)xn的圖象上,則函數(shù)g(x)=n-x+x-A.[0,2] B.[1,2]C.[2,2] D.[2,3]20.(多選)已知a>0,函數(shù)f(x)=(2-aA.若f(x)有最小值,則a≥2B.存在正實(shí)數(shù)a,使得f(x)是R上的減函數(shù)C.存在實(shí)數(shù)a,使得f(x)的值域?yàn)镽D.若a>2,則存在x0∈(1,+∞),使得f(x0)=f(2-x0)21.已知函數(shù)f(x)=x2+(1-2a)x-2a.(1)討論關(guān)于x的不等式f(x)>0的解集;(2)若f(1)=6,求函數(shù)y=f(x)x-1在x22.集合A是由具備下列性質(zhì)的函數(shù)f(x)組成的:①函數(shù)f(x)的定義域是[0,+∞);②函數(shù)f(x)的值域是[-2,4);③函數(shù)f(x)在[0,+∞)上是增函數(shù).試分別探究下列兩小題:(1)判斷函數(shù)f1(x)=x-2(x≥0)及f2(x)=4-6·(12)x(x≥0)是否屬于集合A?并簡(jiǎn)要說(shuō)明理由(2)對(duì)于(1)中你認(rèn)為屬于集合A的函數(shù)f(x),不等式f(x)+f(x+2)<2f(x+1)是否對(duì)于任意的x≥0恒成立?若不成立,為什么?若成立,請(qǐng)說(shuō)明你的結(jié)論.23.已知冪函數(shù)f(x)=x-3n2+9(n∈N(1)求函數(shù)y=f(x)的解析式;(2)設(shè)函數(shù)g(x)=3f(x)+2tx+3,求函數(shù)y=g(x)在區(qū)間[2,6]上的最小值G

優(yōu)化集訓(xùn)5冪函數(shù)基礎(chǔ)鞏固1.C解析形如y=xα(α∈R)的函數(shù)是冪函數(shù),冪函數(shù)的系數(shù)為1,指數(shù)α是常數(shù).故選C.2.D解析由題意冪函數(shù)f(x)=xα的圖象過(guò)點(diǎn)(2,4),則2α=4,解得α=2,則f(x)=x2,由f(m)=4得m2=4,∴m2=16,解得m=±43.A解析由題知,冪函數(shù)f(x)=xα,根據(jù)冪函數(shù)圖象性質(zhì)特點(diǎn)知,冪函數(shù)圖象恒過(guò)點(diǎn)(1,1),所以當(dāng)α>0時(shí),冪函數(shù)圖象過(guò)點(diǎn)(1,1),冪函數(shù)圖象過(guò)點(diǎn)(1,1)時(shí),α>0,也可以α<0.4.D解析當(dāng)a>0時(shí),二次函數(shù)y=ax2+a的圖象開(kāi)口向上,且對(duì)稱(chēng)軸為直線(xiàn)x=0,頂點(diǎn)坐標(biāo)為(0,a),排除A,C;當(dāng)a<0時(shí),二次函數(shù)y=ax2+a的圖象開(kāi)口向下,且對(duì)稱(chēng)軸為直線(xiàn)x=0,頂點(diǎn)坐標(biāo)為(0,a),函數(shù)y=ax的圖象在第二、四象限.故選D5.C解析由題意,f(x)在[2,+∞)上單調(diào)遞增,∵f(2a2-5a+4)<f(a2+a+4),即2≤2a2-5a+4<a2+a+4,∴a2-6a<0且2a2-5a+2≥0,可得2≤a<6或0<a≤12.故選C6.BD解析選項(xiàng)A,冪函數(shù)的圖象必定過(guò)定點(diǎn)(1,1),不一定過(guò)(0,0),如y=x-1,故A錯(cuò)誤;選項(xiàng)B,冪函數(shù)的圖象不可能過(guò)第四象限,正確;選項(xiàng)C,當(dāng)冪指數(shù)α=12時(shí),冪函數(shù)y=xα不是奇函數(shù),故C錯(cuò)誤;選項(xiàng)D,當(dāng)冪指數(shù)α=12,3時(shí),冪函數(shù)y=xα是增函數(shù),7.BCD解析設(shè)f(x)=xα,因?yàn)閒(x)的圖象經(jīng)過(guò)點(diǎn)P(4,2),所以4α=2,解得α=12,所以f(x)=x12=x,顯然A不正確;因?yàn)橹挥蟹秦?fù)實(shí)數(shù)有算術(shù)平方根,所以f(x)的定義域?yàn)閇0,+∞),因此B符合題意;因?yàn)閤≥0,所以有f(x)≥0,因此C符合題意;由f(x)>x2?x>x2?x≥0,x8.81解析由題意函數(shù)y=xα的圖象經(jīng)過(guò)點(diǎn)(2,16)與(3,m),則16=2α,解得α=4,則y=x4,故m=34=81.9.(22,3]解析f(x)=x2-ax+2=0即a=x+2x,x∈[1,+∞),考慮函數(shù)y=a,y=x+2x,x∈[1,+∞)的圖象有兩個(gè)交點(diǎn),即a的取值范圍是(2210.4解析f(x)是冪函數(shù),所以m2-3m-3=1,m2-3m-4=0,解得m=4或m=-1.當(dāng)m=4時(shí),f(x)=x4,在(0,+∞)上單調(diào)遞增,符合題意.當(dāng)m=-1時(shí),f(x)=1x,在(0,+∞)上單調(diào)遞減,不符合題意.綜上所述,m的值為411.(-∞,-1)∪(23,32)解析由題意得a+1>0,3-即實(shí)數(shù)a的取值范圍為(-∞,-1)∪(23,12.[1,32)解析∵冪函數(shù)f(x)經(jīng)過(guò)點(diǎn)(2,2),∴2=2(m2+m)-1,即212=2(m2+m)-1.∴m2+m=2,解得m=1或m=-2.又m∈N*,∴m=1.∴f(x)=x12,則函數(shù)的定義域?yàn)閇0,+∞),并且在定義域上為增函數(shù)13.②③④解析函數(shù)f1(x)=3x-1在(0,1)內(nèi)單調(diào)遞增,則f1(x)∈(-1,2),則f1(x)不是封閉函數(shù);f2(x)=-12x2-12x+1在(0,1)內(nèi)單調(diào)遞減,則f2(x)∈(0,1),則f2(x)是封閉函數(shù);f3(x)=1-x在(0,1)內(nèi)單調(diào)遞減,則f3(x)∈(0,1),則f3(x)是封閉函數(shù);f4(x)=x12在(0,1)內(nèi)單調(diào)遞增,則f4(x)∈(0,1),則f4(14.解(1)由題意得(m-1)2=1,∴m=0或2.當(dāng)m=0時(shí),f(x)=x2在(0,+∞)上單調(diào)遞增,滿(mǎn)足題意.當(dāng)m=2時(shí),f(x)=x-2在(0,+∞)上單調(diào)遞減,不滿(mǎn)足題意,舍去.∴m=0.(2)由(1)知,f(x)=x2.∵f(x)在[1,2]上單調(diào)遞增,∴A=[1,4].易知B≠?,要滿(mǎn)足A∩B=?,只需4-k<1或2-k>4,解得k>3或k<-2,即k的取值范圍為(-∞,-2)∪(3,+∞).15.解(1)對(duì)于函數(shù)y=x+2bx(x>0),因?yàn)槠渲涤驗(yàn)閇6,+∞),即最小值為6=22b,解得b=log(2)令t=2x+1,因?yàn)閤∈[0,1],所以t∈[1,3].所以y=4x2-12x由題可知函數(shù)y=t+4t-8在[1,2]上單調(diào)遞減,在[2,3]上單調(diào)遞增,則函數(shù)f(x)=4x2-12x-32x+1在所以函數(shù)f(x)的值域?yàn)閇-4,-3].16.解(1)由題意可得a2-a-1=1,a>0,解得a=(2)∵g(x)=f(x)∴g(x)在(2,+∞)上單調(diào)遞增.證明如下:任取2<x1<x2,則g(x1)-g(x2)=(x1+4x1)-(x2+4x2)=(x1-x2)+(∵2<x1<x2,∴x1-x2<0,x1x2-4>0,x1x2>0,∴(x1-x2)(x1x2-4)x1即g(x1)<g(x2),∴g(x)在(2,+∞)上單調(diào)遞增.17.解∵冪函數(shù)f(x)=(k2-4k+5)x-∴k2-4k+5=1,解得k=2.又冪函數(shù)f(x)在(0,+∞)上單調(diào)遞增,∴-m2+4m>0,解得0<m<4.∵m∈Z,∴m=1或m=2或m=3.當(dāng)m=1或m=3時(shí),f(x)=x3,圖象關(guān)于原點(diǎn)對(duì)稱(chēng),不合題意;當(dāng)m=2時(shí),f(x)=x4,圖象關(guān)于y軸對(duì)稱(chēng),符合題意.綜上,m=2,k=2.(2)由(1)知m=2,∴原不等式即為(2a-1)-3<(a+2)-3.而函數(shù)y=x-3在(-∞,0)和(0,+∞)上均單調(diào)遞減,且當(dāng)x>0時(shí),y=x-3>0,當(dāng)x<0時(shí),y=x-3<0,∴滿(mǎn)足不等式的條件為0<a+2<2a-1,或a+2<2a-1<0,或2a-1<0<a+2,解得-2<a<12,或a>故滿(mǎn)足不等式(2a-1)-3<(a+2)-3m2的a的取值范圍為(-2,12)∪能力提升18.A解析因?yàn)閍=243=423>425=b19.B解析由已知可得m-2=1,f(m)=mn=81,解得m=3,n=4,故g(x)=4-x+x-3,要使函數(shù)有意義,有4-x≥0,x-3≥0,解得3≤x≤4,故函數(shù)g(x)的定義域?yàn)閇3,4],且g(x)=4-x+x-3>0.因?yàn)閇g(x)]2=(4-x+x-20.AC解析對(duì)于A,當(dāng)a>0時(shí),f(x)在(1,+∞)上單調(diào)遞增,若f(x)有最小值,則2-a≤0,2-a≤1,解得a≥2,A符合題意;對(duì)于B,當(dāng)x>1時(shí),f(x)=xa,由冪函數(shù)性質(zhì)知,當(dāng)a>0時(shí),f(x)單調(diào)遞增,B不符合題意;對(duì)于C,∵f(x)在(1,+∞)上單調(diào)遞增,∴當(dāng)x>1時(shí),f(x)>a,若f(x)的值域?yàn)镽,則2-a>0,2-a≥1,解得0<a≤1,C符合題意;對(duì)于D,當(dāng)x0∈(1,+∞)時(shí),2-x0∈(-∞,1),由f(x0)=f(2-x0),得x0a=(2-a)(2-x0)=(a-2)x0+2(2-a);當(dāng)a>2,x0>1時(shí),x0a-(a-2)x0-2(2-a)>x∴x0(x0a-1-a+2)>0,∴x0a-(a-2)x0-2(2-a)>0恒成立,∴x0a=(a-2)x0+2(2-a)在(1,+∞)內(nèi)無(wú)解,即不存在x0∈(1,+∞),使得f(x0)=f(221.解(1)由f(x)>0,得x2+(1-2a)x-2a>0,即(x-2a)(x+1)>0,當(dāng)a=-12時(shí),不等式(x+1)2>0,解得x≠-1,不等式的解集為{x|x≠-1};當(dāng)-1<2a,即a>-12時(shí),不等式的解集為{x|x<-1或x>2a};當(dāng)-1>2a,即a<-12時(shí),不等式的解集為{x|x>-1或x<2綜上所述,當(dāng)a=-12時(shí),不等式的解集為{x|x≠-當(dāng)a>-12時(shí),不等式的解集為{x|x<-1或x>2a當(dāng)a<-12時(shí),不等式的解集為{x|x>-1或x<2a}(2)由f(1)=6,得f(1)=12+(1-2a)×1-2a=6,解得a=-1,所以f(x)=x2+3x+2,因?yàn)閤>1,所以x-1>0,y=f(x)x-1=x2+3x+2x-1=x-當(dāng)且僅當(dāng)x-1=6x-1,即x=6+1時(shí)所以當(dāng)x=6+1時(shí),函數(shù)y=f(x)x-1在(1,+∞)22.解(1)函數(shù)f1(x)=x-2不屬于集合A.因?yàn)閒1(x)的值域是[-2,+∞),所以函數(shù)f1(x)=x-2不屬于集合A.函數(shù)f2(x)=4-6·(12)x(x≥0)屬于集合A因?yàn)棰俸瘮?shù)f2(x)的定義域是[0,+∞);②f2(x)的值域是[-2,4);③函數(shù)f2(x)在[0,+∞)上是增函數(shù),所以f2(x)=4-6·12x(x≥0)屬于集合(2)由f(x)+f(x+2)-2f(x+1)=6·12x·(-14)<0,所以不等式f(x)+f(x+2)<2f(x+1)對(duì)任意的x≥23.解(1)因?yàn)閮绾瘮?shù)f(x)=x-3n2+9(n∈N)在區(qū)間(0,+∞)上單調(diào)遞增,所以-3n2+9>0(n∈N)