新教材高中數(shù)學(xué)第4章冪函數(shù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)1.3冪函數(shù)提升訓(xùn)練含解析湘教版必修第一冊(cè)

PAGEPAGE4冪函數(shù)基礎(chǔ)過關(guān)練題組一冪函數(shù)的概念1.下列函數(shù)是冪函數(shù)的是 ()A.y=2x2 B.y=x3+x C.y=3x D.y=x2.已知冪函數(shù)y=f(x)的圖象經(jīng)過點(diǎn)4,14,則f(2)=A.12 B.2 C.22 3.函數(shù)f(x)=(1-x)-12+(2x-1)0的定義域是A.(-∞,1] B.-∞,12C.(-∞,-1) D.14.(2021安徽合肥八中高一上期中)已知點(diǎn)(m,8)在冪函數(shù)f(x)=(m-1)xn的圖象上,則n-m= ()A.19 B.18 C.85.已知函數(shù)f(x)=(m2+2m)·xm2+m-1,m為何值時(shí),函數(shù)f(x)是(1)題組二冪函數(shù)的圖象及其應(yīng)用6.函數(shù)y=x43的圖象是 (7.如圖所示,曲線C1和C2分別是函數(shù)y=xm和y=xn在第一象限內(nèi)的圖象,則下列結(jié)論正確的是 ()A.n<m<0 B.m<n<0C.n>m>0 D.m>n>08.(2020湖南衡陽一中高一上期中)函數(shù)y=x12-1的圖象關(guān)于x軸對(duì)稱的圖象大致是 (題組三冪函數(shù)的性質(zhì)及其應(yīng)用9.下列函數(shù)中,是偶函數(shù)且在(0,+∞)上為減函數(shù)的是 ()A.y=x2 B.y=x3C.y=x-2 D.y=-x310.(2020山西長治二中高一上期末)已知α∈{-1,1,2,3},則使函數(shù)y=xα的值域?yàn)镽,且為奇函數(shù)的所有α的值為 ()A.1,3 B.-1,1 C.-1,3 D.-1,1,311.如果冪函數(shù)f(x)=xα的圖象過點(diǎn)(-2,4),那么f(x)的單調(diào)遞增區(qū)間是 ()A.(-∞,+∞) B.[0,+∞)C.(-∞,0] D.(-∞,0)∪(0,+∞)12.(2020安徽安慶高一上期末)已知冪函數(shù)f(x)=(a2-2a-2)·xa在區(qū)間(0,+∞)上是增函數(shù),則a的值為 ()A.3 B.-1 C.-3 D.113.若冪函數(shù)f(x)的圖象過點(diǎn)(3,27),則滿足不等式f(a-3)>f(1-a)的實(shí)數(shù)a的取值范圍是.

14.(2021河北衡水武邑中學(xué)高一上期中)已知冪函數(shù)f(x)=(m2-5m+7)x-m-1(m∈R)為偶函數(shù).(1)求f12的值(2)若f(2a+1)=f(a),求實(shí)數(shù)a的值.答案全解全析基礎(chǔ)過關(guān)練1.Dy=2x2,y=x3+x,y=3x均不是冪函數(shù),y=x122.A設(shè)冪函數(shù)為f(x)=xα,∵冪函數(shù)的圖象經(jīng)過點(diǎn)4,14,∴1∴α=-1,∴f(x)=x-1,∴f(2)=2-1=123.B依題意得1-x>0,2x-1≠0,解得x<1,且x4.A由冪函數(shù)的定義可知,m-1=1,∴m=2,∴點(diǎn)(2,8)在冪函數(shù)f(x)=xn的圖象上,∴2n=8,∴n=3,∴n-m=3-2=19,故選A5.解析(1)若函數(shù)f(x)為正比例函數(shù),則m∴m=1.(2)若函數(shù)f(x)為反比例函數(shù),則m2+m-(3)若函數(shù)f(x)為冪函數(shù),則m2+2m=1,∴m=-1±2.6.A∵y=x43=又∵43>1,∴在第一象限內(nèi),y=x43的圖象與y=x7.A由題中圖象可知,兩函數(shù)在第一象限內(nèi)單調(diào)遞減,故m<0,n<0.由冪函數(shù)圖象的特點(diǎn)知n<m,故n<m<0.8.By=x12-1的定義域?yàn)閇0,+∞),且為增函數(shù),所以函數(shù)圖象從左到右是上升的,所以y=x19.Cy=x2是偶函數(shù),在(0,+∞)上為增函數(shù),故A不正確;y=x3是奇函數(shù),故B不正確;y=x-2是偶函數(shù),且在(0,+∞)上為減函數(shù),故C正確;y=-x3是奇函數(shù),故D不正確.10.A當(dāng)α=-1時(shí),y=xα的值域不是R,當(dāng)α=2時(shí),y=xα是偶函數(shù),當(dāng)α=1,3時(shí),y=xα的值域?yàn)镽,且為奇函數(shù),因此選項(xiàng)A正確,故選A.11.B依題意得(-2)α=4=(-2)2,即α=2,∴f(x)=x2,∴f(x)的單調(diào)遞增區(qū)間是[0,+∞),故選B.12.A由題意知a2-2a-2=1,解得a=3或a=-1,又f(x)在區(qū)間(0,+∞)上是增函數(shù),所以a=3,故選A.13.答案(2,+∞)解析設(shè)冪函數(shù)為f(x)=xα,因?yàn)槠鋱D象過點(diǎn)(3,27),所以27=3α,解得α=3,所以f(x)=x3.因?yàn)閒(x)=x3在R上為增函數(shù),所以由f(a-3)>f(1-a),得a-3>1-a,解得a>2.所以滿足不等式f(a-3)>f(1-a)的實(shí)數(shù)a的取值范圍是(2,+∞).14.解析(1)由題意知m2-5m+7=1,解得m=2或m=3,當(dāng)m=2時(shí),f(x)=x-3,為奇函數(shù),不滿足題意;當(dāng)m=3