2025屆高三一輪復(fù)習(xí)數(shù)學(xué)試題(人教版新高考新教材)考點(diǎn)規(guī)范練8　冪函數(shù)

考點(diǎn)規(guī)范練8冪函數(shù)一、基礎(chǔ)鞏固1.已知冪函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)(4,2),則f(x)的單調(diào)遞增區(qū)間為()A.(-∞,+∞) B.(-∞,0]C.[0,+∞) D.(1,+∞)答案:C解析:設(shè)f(x)=xα,由圖象經(jīng)過(guò)點(diǎn)(4,2),得4α=2,即22α=2,得α=12,所以f(x)=x12,單調(diào)遞增區(qū)間為[0,+2.下面四個(gè)冪函數(shù)的圖象中,是函數(shù)y=x-23答案:B解析:由冪函數(shù)的性質(zhì)可知,函數(shù)y=x-23的圖象在區(qū)間(0,+∞)內(nèi)單調(diào)遞減,則AC錯(cuò)誤;令f(x)=x-23=1x213,x∈(-∞,0)∪(0,+∞),因?yàn)閒(-x)=1(-3.已知冪函數(shù)f(x)=x3m-7(m∈N)的圖象關(guān)于y軸對(duì)稱,且與x軸、y軸均無(wú)交點(diǎn),則m的值為()A.-1 B.0 C.1 D.2答案:C解析:由題意可得,3m-7<0,解得m<73,且3m-7為偶數(shù),m∈N,故m=14.若a<0,則0.5a,5a,5-a的大小關(guān)系是()A.5-a<5a<0.5a B.5a<0.5a<5-aC.0.5a<5-a<5a D.5a<5-a<0.5a答案:B解析:由于5-a=15a.因?yàn)閍<0,所以函數(shù)y=xa在區(qū)間(0,+∞)內(nèi)單調(diào)遞減.又15<0.5<5,所以5a<0.5a<5.如圖,函數(shù)y=1x,y=x的圖象和直線y=1,x=1將平面直角坐標(biāo)系的第一象限分成八個(gè)部分:①②③④⑤⑥⑦⑧.若冪函數(shù)f(x)的圖象經(jīng)過(guò)的部分是④⑧,則f(x)可能是(A.f(x)=x2 B.f(x)=1C.f(x)=x12 D.f(x)=x答案:B解析:由圖象知,冪函數(shù)f(x)的定義域?yàn)?0,+∞).當(dāng)0<x<1時(shí),f(x)>1,且f(x)<1x當(dāng)x>1時(shí),0<f(x)<1,且f(x)>1x所以f(x)可能是f(x)=1x6.已知函數(shù)f(x)=x2+(2a-1)x-3,當(dāng)a=2,x∈[-2,3]時(shí),函數(shù)f(x)的值域?yàn)?若函數(shù)f(x)在區(qū)間[-1,3]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是.

答案:-214,解析:(1)當(dāng)a=2時(shí),f(x)=x2+3x-3,其圖象的對(duì)稱軸為直線x=-32∈[-故f(x)max=f(3)=15,f(x)min=f-32=-214,故函數(shù)f(x)(2)因?yàn)楹瘮?shù)f(x)在區(qū)間[-1,3]上單調(diào)遞增,所以-2a-12≤-1,故7.已知函數(shù)f(x)同時(shí)滿足:①f(0)=0;②在區(qū)間[1,3]上單調(diào)遞減;③f(1+x)=f(1-x).該函數(shù)的表達(dá)式可以是f(x)=.

答案:2x-x2(答案:不唯一)解析:由f(1+x)=f(1-x)可知,y=f(x)的圖象關(guān)于直線x=1對(duì)稱;可設(shè)f(x)為二次函數(shù),又f(0)=0且f(x)在區(qū)間[1,3]上單調(diào)遞減,所以可設(shè)f(x)=2x-x2,符合題意.二、綜合應(yīng)用8.已知f(x)=x3,若當(dāng)x∈[1,2]時(shí),f(x2-ax)+f(1-x)≤0,則實(shí)數(shù)a的取值范圍是()A.(-∞,1] B.[1,+∞)C.[32,+∞] D.(-∞,3答案:C解析:∵f(-x)=-f(x),f'(x)=3x2≥0,∴f(x)在R上為奇函數(shù)且單調(diào)遞增.由f(x2-ax)+f(1-x)≤0,得f(x2-ax)≤f(x-1),∴x2-ax≤x-1,即x2-(a+1)x+1≤0.設(shè)g(x)=x2-(a+1)x+1,則有g(shù)(1)=1-a≤0,g(2)=3-2a≤0,9.若x2>x13成立,則x的取值范圍是答案:(-∞,0)∪(1,+∞)解析:如圖所示,分別作出函數(shù)y=x2與y=x13的圖象,由于兩函數(shù)的圖象都過(guò)點(diǎn)(1,1),由圖象可知不等式x2>x13的解集為(-∞,0)∪(1,10.已知冪函數(shù)f(x)=x-12,若f(a+1)<f(10-2a),則實(shí)數(shù)a的取值范圍是答案:(3,5)解析:∵f(x)=x-12∴f(x)是定義在區(qū)間(0,+∞)內(nèi)的減函數(shù).又f(a+1)<f(10-2a),∴a+1>0,10-2a>011.設(shè)二次函數(shù)f(x)=ax2+2ax+1在區(qū)間[-3,2]上有最大值4,則實(shí)數(shù)a的值為.

答案:38或-解析:由題意可知f(x)的圖象的對(duì)稱軸為直線x=-1.當(dāng)a>0時(shí),f(2)=4a+4a+1=8a+1,f(-3)=3a+1.可知f(2)>f(-3),即f(x)max=f(2)=8a+1=4.故a=38.當(dāng)a<0時(shí),f(x)max=f(-1)=a-2a+1=-a+1=4,即a=-3.綜上所述,a=38或a=-三、探究創(chuàng)新12.已知函數(shù)f(x)=(m2-m-1)xm2+m-1是冪函數(shù),且在區(qū)間(0,+∞)內(nèi)單調(diào)遞增,若a,b∈R,且a+b>0,ab<0,則f(a)A.恒等于0 B.恒小于0C.恒大于0 D.無(wú)法判斷答案:C解析:由于函數(shù)f(x)=(m2-m-1)xm2+m-1是冪函數(shù),則m2-m-1=1,解得當(dāng)m=-1時(shí),f(x)=x-1,在區(qū)間(0,+∞)內(nèi)單調(diào)