2024秋新教材高中數(shù)學(xué)第四章指數(shù)函數(shù)與對(duì)數(shù)函數(shù)4.2.2指數(shù)函數(shù)的圖象和性質(zhì)分層演練含解析新人教A版必修第一冊(cè)

4.2.2指數(shù)函數(shù)的圖象和性質(zhì)分層演練綜合提升A級(jí)基礎(chǔ)鞏固1.若12x+1>1,則x的取值范圍是 ()A.(-1,1) B.(-1,+∞)C.(0,1)∪(1,+∞) D.(-∞,-1)答案:D2.若1>n>m>0,則指數(shù)函數(shù)①y=mx,②y=nx的圖象為 ()ABCD答案:C3.函數(shù)f(x)=2x在區(qū)間[-1,3]上的最小值是124.函數(shù)f(x)=(13)5.已知指數(shù)函數(shù)f(x)的圖象經(jīng)過點(diǎn)P(3,8),且函數(shù)g(x)的圖象與f(x)的圖象關(guān)于y軸對(duì)稱.(1)求函數(shù)g(x)的解析式;(2)若g(2x2-3x+1)>g(x2+2x-5),求x的取值范圍.解:(1)設(shè)f(x)=ax(a>0,且a≠1),因?yàn)楹瘮?shù)f(x)的圖象過點(diǎn)(3,8),所以a3=8,得a=2,所以f(x)=2x,所以f(x)關(guān)于y軸對(duì)稱的函數(shù)g(x)=12x.(2)因?yàn)間(x)在R上是減函數(shù),由g(2x2-3x+1)>g(x2+2x-5),可得2x2-3x+1<x2+2x-5,x2-5x+6<0,解得2<x<3,故x的取值范圍是(2,3).B級(jí)實(shí)力提升6.二次函數(shù)y=ax2+bx與指數(shù)函數(shù)y=bax的圖象可能為 ()ABCD解析:因?yàn)閥=(ba)x是指數(shù)函數(shù),所以ba>0,即a,b同號(hào).所以二次函數(shù)y=ax2+bx圖象的對(duì)稱軸x=-b2a<0,解除選項(xiàng)B,D;由A,C項(xiàng)中指數(shù)函數(shù)的圖象,得0<ba<1,則-12<-b2a<0,即二次函數(shù)的頂點(diǎn)的橫坐標(biāo)在區(qū)間(答案:A7.若f(x)=23x-1+m是奇函數(shù),則常數(shù)解析:f(x)=23x-1+因?yàn)閒(x)為奇函數(shù),所以對(duì)于定義域上隨意的實(shí)數(shù)x,都有f(x)+f(-x)=0,即(23x-1+m)+(2所以m=-13x-1-13-8.已知函數(shù)f(x)是定義在R上的奇函數(shù),且當(dāng)x>0時(shí),f(x)=(13)x-1(1)求函數(shù)f(x)的解析式,并作出函數(shù)f(x)的圖象;(2)當(dāng)x∈[2,4]時(shí),不等式f(2-5x)<f(2x2-mx+20)恒成立,求m的取值范圍.解:(1)當(dāng)x<0時(shí),-x>0,f(-x)=(13)-x-1=3x-1又因?yàn)閒(x)是奇函數(shù),所以f(-x)=-f(x),故當(dāng)x<0時(shí),f(x)=-3x+1.當(dāng)x=0時(shí),f(0)=0,滿意x>0時(shí)的解析式.故f(x)=(f(x)的圖象如下:(2)由(1)可知f(x)在R上單調(diào)遞減,故由f(2-5x)<f(2x2-mx+20)可得2-5x>2x2-mx+20,即m>2(x+9x)+5對(duì)x∈[2,4]恒成立,只須要m>[2(x+9x)+5]max即可.當(dāng)x∈[2,4]時(shí),[2(x+9x)+5]max=18,故C級(jí)挑戰(zhàn)創(chuàng)新9.多選題下列大小關(guān)系正確的是 ()A.0.65<π0<50.6B.40.9<12-1.5<80.48C.0.4<0.40.2<20D.37

37<37

47<47解析:選項(xiàng)A,0.65<1,50.6>1,故選項(xiàng)A正確;選項(xiàng)B,40.9=21.8,(12)-1.5=21.5,80.48=21.44,因此40.9>(12)-1.5>80.選項(xiàng)C,0.4=0.40.5<0.40.2<1,而20.選項(xiàng)D,(47)

37>(37)

37>答案:AC10.多空題已知函數(shù)f(x)=a·2x,x≥0,2-x,