（適用于新高考新教材）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(shí)規(guī)范練9二次函數(shù)與冪函數(shù)

課時(shí)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)=(3m22m)x3m的圖象不經(jīng)過坐標(biāo)原點(diǎn),則實(shí)數(shù)m的值為()A.13 B.1C.1 D.12.二次函數(shù)f(x)的圖象經(jīng)過(0,3),(2,3)兩點(diǎn),且f(x)的最大值是5,則該函數(shù)的解析式是()A.f(x)=2x28x+11 B.f(x)=2x2+8x1C.f(x)=2x24x+3 D.f(x)=2x2+4x+33.(2022河北衡水中學(xué)模擬)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(2x),且f(x)在區(qū)間[0,2]上單調(diào)遞增.若f(a)≥f(0),則實(shí)數(shù)a的取值范圍是()A.[0,+∞) B.(∞,0]C.[0,4] D.(∞,0]∪[4,+∞)5.已知函數(shù)f(x)=(m2m5)xm2-6是冪函數(shù),對任意的x1,x2∈(0,+∞),且x1≠x2,滿足f(x1)-f(x2)x1-x2>0,若A.恒大于0 B.恒小于0C.等于0 D.無法判斷6.已知函數(shù)f(x)=x4x2,則下列結(jié)論錯(cuò)誤的是()A.f(x)的圖象關(guān)于y軸對稱B.方程f(x)=0的解的個(gè)數(shù)為2C.f(x)在區(qū)間(1,+∞)上單調(diào)遞增D.f(x)的最小值為17.(2022浙江衢州五校聯(lián)考)若函數(shù)f(x)=x22ax+1a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)=|x22ax+b|(x∈R),下列說法正確的是()A.若a2b≤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a2b2>0,則函數(shù)h(x)=f(x)2有2個(gè)零點(diǎn)9.已知函數(shù)f(x)=(x22x3)(x2+ax+b)是偶函數(shù),則f(x)的值域是.

綜合提升組10.已知冪函數(shù)f(x)=xα滿足2f(2)=f(16),若a=f(log42),b=f(ln2),c=f(5-12),則a,b,cA.a>c>b B.a>b>cC.b>a>c D.b>c>a11.(2022湖南長沙高三檢測)已知冪函數(shù)f(x)=(m1)2xm2-4m+2在區(qū)間(0,+∞)上單調(diào)遞增,函數(shù)g(x)=2xa,?x1∈[1,5],?x2∈[1,5],使得f(x1)≥g(xA.a≥1 B.a≥23C.a≥31 D.a≥712.若函數(shù)f(x)=mx2+(n1)x+2(m>0,n>0)的單調(diào)遞增區(qū)間為12,+∞,則1m+1n的最小值為13.已知函數(shù)f(x)=x2+ax+1(a>0).(1)若f(x)的值域?yàn)閇0,+∞),求關(guān)于x的方程f(x)=4的解;(2)當(dāng)a=2時(shí),函數(shù)g(x)=[f(x)]22mf(x)+m21在區(qū)間[2,1]上有三個(gè)零點(diǎn),求實(shí)數(shù)m的取值范圍.創(chuàng)新應(yīng)用組14.已知f(x)=x22x,對任意的x1,x2∈[0,3],方程|f(x)f(x1)|+|f(x)f(x2)|=m在區(qū)間[0,3]上有解,則實(shí)數(shù)m的取值范圍是()A.[0,3] B.[0,4]C.{3} D.{4}

課時(shí)規(guī)范練9二次函數(shù)與冪函數(shù)1.B解析:由題意得3m22m=1,解得m=1或m=13,①當(dāng)m=1時(shí),f(x)=x3,函數(shù)圖象經(jīng)過原點(diǎn),不符合題意;②當(dāng)m=13時(shí),f(x)=x1,函數(shù)圖象不經(jīng)過原點(diǎn),符合題意,故m=2.D解析:因?yàn)槎魏瘮?shù)f(x)的圖象經(jīng)過(0,3),(2,3)兩點(diǎn),所以圖象的對稱軸為直線x=1.又由函數(shù)的最大值是5,可設(shè)f(x)=a(x1)2+5(a≠0).于是3=a+5,解得a=2,故f(x)=2(x1)2+5=2x2+4x+3.故選D.3.A解析:因?yàn)閮绾瘮?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.A解析:因?yàn)楹瘮?shù)f(x)=(m2m5)xm2-6是冪函數(shù),所以m2m5=1,解得m=2或m=3.因?yàn)閷θ我獾膞1,x2∈(0,+∞),且x1≠x2,滿足f(x1)-f(x2)x1-x2>0,所以函數(shù)f(x)在(0,+∞)上單調(diào)遞增,所以m26>0,所以m=3(m=2舍去),所以f(x)=x3.對任意的a,b∈R,a+b>0,即a>b,所以f(a)>f6.B解析:因?yàn)閒(x)=x4x2的定義域?yàn)镽,f(x)=(x)4(x)2=x4x2=f(x),所以f(x)是偶函數(shù),其圖象關(guān)于y軸對稱,故A正確;令f(x)=0,即x2(x+1)(x1)=0,解得x=0,x=1或x=1,方程f(x)=0的解的個(gè)數(shù)為3,故B錯(cuò)誤;令t=x2,g(t)=t2t=t12214,當(dāng)x>1時(shí),函數(shù)t=x2,g(t)=t2t都單調(diào)遞增,故f(x)在(1,+∞)單調(diào)遞增,故C正確;由當(dāng)t=12時(shí),g(t)取得最小值14,故f(x)的最小值是17.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(shí),函數(shù)f(x)在區(qū)間[0,2]上單調(diào)遞增,則f(x)min=f(0)=1a,由1a=1,得a=2,∵a≤0,∴a=2不符合;②當(dāng)0<a<2時(shí),f(x)min=f(a)=a22a2+1a=a2a+1,由a2a+1=1,得a=2或a=1,又0<a<2,∴a=1;③當(dāng)a≥2時(shí),函數(shù)f(x)在區(qū)間[0,2]上單調(diào)遞減,∴f(x)min=f(2)=44a+1a=55a,由55a=1,得a=65,又a≥2,∴a=65綜上可得a=1.8.AB解析:對于選項(xiàng)A,若a2b≤0,則f(x)=|(xa)2+ba2|=(xa)2+ba2在區(qū)間[a,+∞)上單調(diào)遞增,故A正確;對于選項(xiàng)B,當(dāng)a=0時(shí),f(x)=|x2+b|顯然是偶函數(shù),故B正確;對于選項(xiàng)C,取a=0,b=2,函數(shù)f(x)=|x22ax+b|可化為f(x)=|x22|,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(cuò)誤;對于選項(xiàng)D,如圖,a2b2>0,即a2b>2,則h(x)=|(xa)2+ba2|2有4個(gè)零點(diǎn),故D錯(cuò)誤.9.[16,+∞)解析:根據(jù)題意,函數(shù)f(x)=(x22x3)·(x2+ax+b)=x4+(a2)x3+(b2a3)x2(2b+3a)x3b.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)=x410x2+10.C解析:由2f(2)=f(16)可得2·2α=24α,即1+α=4α,得α=13,所以f(x)=x13,則函數(shù)f(x)在R上單調(diào)遞增.而log42=log22log24=12,ln2=lo又因?yàn)?5<12,所以5-故a,b,c的大小關(guān)系是b>a>c,故選C.11.A解析:由已知得(m1)2=1,解得m=0或m=2,當(dāng)m=0時(shí),f(x)=x2;當(dāng)m=2時(shí),f(x)=x2.因?yàn)閮绾瘮?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(shí),函數(shù)f(x)的值域?yàn)閇1,25].因?yàn)楹瘮?shù)g(x)=2xa在R上為增函數(shù),所以當(dāng)x∈[1,5]時(shí),函數(shù)g(x)的值域?yàn)閇2a,25a].因?yàn)?x1∈[1,5],?x2∈[1,5],使得f(x1)≥g(x2)成立,所以1≥2a,解得a≥1.12.4解析:函數(shù)f(x)圖象的對稱軸為直線x=n-12m=12,則m+n=1.已知m>0,n>0,所以1m+1n=1m+1n·(m+n)=2+13.解(1)因?yàn)閒(x)的值域?yàn)閇0,+∞),所以f(x)min=fa2=14a212a2+1=0.因?yàn)閍>0,所以a=2,所以f(x)=x2+2x+1.由f(x)=4,即x2+2x+1=4,即x2+2x3=0,解得x=3或x=1(2)g(x)=[f(x)]22mf(x)+m21在區(qū)間[2,1]上有三個(gè)零點(diǎn)等價(jià)于方程[f(x)]22mf(x)+m21=0在區(qū)間[2,1]上有三個(gè)不同的根.因?yàn)閇f(x)]22mf(x)+m21=0,所以f(x)=m+1或f(x)=m1.因?yàn)閍=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)]22mf(x)+m21=0在區(qū)間[2,1]上有三個(gè)不同的根,則f(x)=m+1在區(qū)間[2,1]上有一個(gè)實(shí)數(shù)根,f(x)=m1在區(qū)間[2,1]上有兩個(gè)不等實(shí)數(shù)根,即1<m+1≤4,0<14.D解