適用于新高考新教材廣西專版2025屆高考數(shù)學(xué)一輪總復(fù)習(xí)第三章函數(shù)與基本初等函數(shù)課時規(guī)范練9二次函數(shù)與冪函數(shù)

課時規(guī)范練9二次函數(shù)與冪函數(shù)基礎(chǔ)鞏固組1.若冪函數(shù)f(x)=(3m2-2m)x3m的圖象不經(jīng)過坐標(biāo)原點,則實數(shù)m的值為()A.13 B.-1C.-1 D.12.(2023湖南聯(lián)考模擬)已知f(x)=(x-2)(x+a)是偶函數(shù),則a=()A.-1 B.1 C.-2 D.23.已知冪函數(shù)f(x)=x2+m是定義在區(qū)間[-1,m]上的奇函數(shù),則f(m+1)=()A.8 B.4 C.2 D.14.已知二次函數(shù)f(x)滿足f(2+x)=f(2-x),且f(x)在區(qū)間[0,2]上單調(diào)遞增.若f(a)≥f(0),則實數(shù)a的取值范圍是()A.[0,+∞) B.(-∞,0]C.[0,4] D.(-∞,0]∪[4,+∞)5.(2023海南統(tǒng)考模擬)已知f(x)=(m2+m-5)xm為冪函數(shù),則()A.f(x)在(-∞,0)上單調(diào)遞增B.f(x)在(-∞,0)上單調(diào)遞減C.f(x)在(0,+∞)上單調(diào)遞增D.f(x)在(0,+∞)上單調(diào)遞減6.(多選)(2023全國模擬)已知二次函數(shù)f(x)=mx2-4mx+12m-3(m<0),若對任意x1≠x2,則()A.當(dāng)x1+x2=4時,f(x1)=f(x2)恒成立B.當(dāng)x1+x2>4時,f(x1)<f(x2)恒成立C.?x0使得f(x0)≥0成立D.對任意x1,x2,均有f(xi)≤8m-3(i=1,2)恒成立7.若函數(shù)f(x)=x2-2ax+1-a在區(qū)間[0,2]上的最小值為-1,則a=()A.2或65 B.1或C.2 D.18.(多選)已知函數(shù)f(x)=|x2-2ax+b|(x∈R),下列說法正確的是()A.若a2-b≤0,則f(x)在區(qū)間[a,+∞)上單調(diào)遞增B.存在a∈R,使得f(x)為偶函數(shù)C.若f(0)=f(2),則f(x)的圖象關(guān)于x=1對稱D.若a2-b-2>0,則函數(shù)h(x)=f(x)-2有2個零點9.已知函數(shù)f(x)=(x2-2x-3)(x2+ax+b)是偶函數(shù),則f(x)的值域是.

綜合提升組10.(2023河北滄州三模)已知f(x)為奇函數(shù),當(dāng)0≤x≤2時,f(x)=2x-x2,當(dāng)x>2時,f(x)=|x-3|-1,則()A.-f(-26)>f(20.3)>f(30.3)B.f(20.3)>f(30.3)>-f(-26)C.-f(-26)>f(30.3)>f(20.3)D.f(30.3)>f(20.3)>-f(-26)11.已知冪函數(shù)f(x)=(m-1)2xm2-4m+2在區(qū)間(0,+∞)上單調(diào)遞增,函數(shù)g(x)=2x-a,?x1∈[1,5],?x2∈[1,5],使得f(x1)≥g(x2)成立,A.a≥1 B.a≥-23C.a≥31 D.a≥712.(2023北京海淀一模)已知二次函數(shù)f(x),對任意的x∈R,有f(2x)<2f(x),則f(x)的圖象可能是()13.已知函數(shù)f(x)=x2+ax+1(a>0).(1)若f(x)的值域為[0,+∞),求關(guān)于x的方程f(x)=4的解;(2)當(dāng)a=2時,函數(shù)g(x)=[f(x)]2-2mf(x)+m2-1在區(qū)間[-2,1]上有三個零點,求實數(shù)m的取值范圍.創(chuàng)新應(yīng)用組14.(2023江西名校聯(lián)考)已知函數(shù)f(x)=x2-2x,g(x)=ax+2,若對任意的x1∈[-1,2],存在x2∈[-1,2],使f(x2)=g(x1),則a的取值范圍是()A.0,12 B.-1,12C.[-1,0] D.(0,3]

課時規(guī)范練9二次函數(shù)與冪函數(shù)1.B解析由題意得3m2-2m=1,解得m=1或m=-13,①當(dāng)m=1時,f(x)=x3,函數(shù)圖象經(jīng)過原點,不符合題意;②當(dāng)m=-13時,f(x)=x-1,函數(shù)圖象不經(jīng)過原點,符合題意,故m=-2.D解析(方法1)∵f(x)=x2+(a-2)x-2a,∴f(-x)=x2-(a-2)x-2a,由f(-x)=f(x),得x2-(a-2)x-2a=x2+(a-2)x-2a,解得a=2.(方法2)f(x)=x2+(a-2)x-2a,∵f(x)是偶函數(shù),∴f(x)的圖象關(guān)于直線x=0對稱,即-a-22=0,解得a=2.3.A解析因為冪函數(shù)在區(qū)間[-1,m]上是奇函數(shù),所以m=1,所以f(x)=x2+m=x3,所以f(m+1)=f(1+1)=f(2)=23=8.4.C解析由題意可知二次函數(shù)f(x)的圖象開口向下,對稱軸為直線x=2(如圖).若f(a)≥f(0),從圖象觀察可知0≤a≤4.5.B解析由題意m2+m-5=1,解得m=2或m=-3,∴f(x)=x2或f(x)=x-3,對于f(x)=x2,f(x)在(0,+∞)上單調(diào)遞增,在(-∞,0)上單調(diào)遞減;對于f(x)=x-3,f(x)在(0,+∞)上單調(diào)遞減,在(-∞,0)上單調(diào)遞減.故只有B選項符合.故選B.6.AD解析f(x)=mx2-4mx+12m-3(m<0)的圖象是開口向下的拋物線,其對稱軸為直線x=2,對于A,由x1+x2=4,得x1,x2關(guān)于直線x=2對稱,則f(x1)=f(x2)恒成立,∴A正確.對于B,當(dāng)x1+x2>4,若x1>x2,則有x1-2>2-x2,∴f(x1)<f(x2);若x1<x2,則不等式可化為x2-2>2-x1,∴f(x2)<f(x1),∴B錯誤.對于C,∵m<0,∴Δ=(-4m)2-4m(12m-3)=-32m2+12m<0,∴f(x)的圖象在x軸下方,不存在x0使得f(x0)≥0成立,∴C錯誤.對于D,f(x)max=f(2)=4m-8m+12m-3=8m-3,∴?x1,x2,均有f(xi)≤8m-3(i=1,2)恒成立,∴D正確.故選AD.7.D解析由題意知,二次函數(shù)f(x)的圖象的對稱軸為直線x=a,且圖象開口向上.①當(dāng)a≤0時,函數(shù)f(x)在區(qū)間[0,2]上單調(diào)遞增,則f(x)min=f(0)=1-a,由1-a=-1,得a=2,∵a≤0,∴a=2不符合;②當(dāng)0<a<2時,f(x)min=f(a)=a2-2a2+1-a=-a2-a+1,由-a2-a+1=-1,得a=-2或a=1,又0<a<2,∴a=1;③當(dāng)a≥2時,函數(shù)f(x)在區(qū)間[0,2]上單調(diào)遞減,∴f(x)min=f(2)=4-4a+1-a=5-5a,由5-5a=-1,得a=65,又a≥2,∴a=65不符合.綜上可得a=8.AB解析對于選項A,若a2-b≤0,則f(x)=|(x-a)2+b-a2|=(x-a)2+b-a2在區(qū)間[a,+∞)上單調(diào)遞增,故A正確;對于選項B,當(dāng)a=0時,f(x)=|x2+b|顯然是偶函數(shù),故B正確;對于選項C,取a=0,b=-2,函數(shù)f(x)=|x2-2ax+b|可化為f(x)=|x2-2|,滿足f(0)=f(2),但f(x)的圖象不關(guān)于直線x=1對稱,故C錯誤;對于選項D,如圖,a2-b-2>0,即a2-b>2,則h(x)=|(x-a)2+b-a2|-2有4個零點,故D錯誤.9.[-16,+∞)解析根據(jù)題意,函數(shù)f(x)=(x2-2x-3)·(x2+ax+b)=x4+(a-2)x3+(b-2a-3)x2-(2b+3a)x-3b.由f(x)是偶函數(shù),知f(-x)=f(x),則必有a-2=0,2b+3a=0,解得a=2,b=-3,所以f(x)=x4-10x2+9=(x2-5)2-1610.A解析當(dāng)0≤x≤2時,f(x)=2x-x2,則f(x)在(0,1)內(nèi)單調(diào)遞增,在[1,2]上單調(diào)遞減,當(dāng)x>2時,f(x)=|x-3|-1,則f(x)在(2,3)內(nèi)單調(diào)遞減,在[3,+∞)上單調(diào)遞增,且f(2)=0=|2-3|-1,∴f(x)在(0,1)內(nèi)單調(diào)遞增,在[1,3)上單調(diào)遞減,在[3,+∞)上單調(diào)遞增.∵函數(shù)f(x)為奇函數(shù),∴-f(-26)=f(26)>f(5)=1=f(1),又1<20.3<30.3<3,則f(1)>f(20.3)>f(30.3),∴-f(-26)>f(20.3)>f(30.3).故選A.11.A解析由已知得(m-1)2=1,解得m=0或m=2,當(dāng)m=0時,f(x)=x2;當(dāng)m=2時,f(x)=x-2.因為冪函數(shù)f(x)在區(qū)間(0,+∞)上單調(diào)遞增,所以f(x)=x2.所以當(dāng)x∈[1,5]時,函數(shù)f(x)的值域為[1,25].因為函數(shù)g(x)=2x-a在R上為增函數(shù),所以當(dāng)x∈[1,5]時,函數(shù)g(x)的值域為[2-a,25-a].因為?x1∈[1,5],?x2∈[1,5],使得f(x1)≥g(x2)成立,所以1≥2-a,解得a≥1.12.A解析∵對任意的x∈R,有f(2x)<2f(x),令x=0,則f(0)<2f(0),∴f(0)>0,即f(0)=c>0,排除C,D.設(shè)二次函數(shù)f(x)=ax2+bx+c(a≠0),∴f(2x)=4ax2+2bx+c,2f(x)=2ax2+2bx+2c,由f(2x)<2f(x),得4ax2+2bx+c<2ax2+2bx+2c,則2ax2-c<0,∴c>2ax2對任意的x∈R恒成立,∵c>0,∴2a<0,故排除B.故選A.13.解(1)因為f(x)的值域為[0,+∞),所以f(x)min=f-a2=14a2-12a2+1=0.因為a>0,所以a=2,所以f(x)=x2+2x+1.由f(x)=4,即x2+2x+1=4,即x2+2x-3=0,解得x=-3或x=1(2)g(x)=[f(x)]2-2mf(x)+m2-1在區(qū)間[-2,1]上有三個零點等價于方程[f(x)]2-2mf(x)+m2-1=0在區(qū)間[-2,1]上有三個不同的根.因為[f(x)]2-2mf(x)+m2-1=0,所以f(x)=m+1或f(x)=m-1.因為a=2,所以f(x)=x2+2x+1.結(jié)合f(x)在[-2,1]上的圖象(圖略)可知,要使方程[f(x)]2-2mf(x)+m2-1=0在區(qū)間[-2,1]上有三個不同的根,則f(x)=m+1在區(qū)間[-2,1]上有一個實數(shù)根,f(x)=m-1在區(qū)間[-2,1]上有兩個不等實數(shù)根,即1<m+1≤4,0<m-114.B解析函數(shù)f(x)=x2-2x=(x-1)2-1,當(dāng)x∈[-1,2]時,-2≤x-1≤1,則0≤(x-1)2≤4,則f(x)=(x-1)2-1∈[-1,3].函數(shù)g(x)=ax+2在[-1,2]的值域記為