4.5函數(shù)的應(yīng)用（二）4.5.1函數(shù)的零點(diǎn)與方程的解A級(jí)基礎(chǔ)鞏固1.函數(shù)f(x)=x+1x的零點(diǎn)的個(gè)數(shù)為 (A.0 B.1 C.2 D.3答案:A2.已知函數(shù)f(x)=2x-1,x≤1A.12C.12答案:D3.函數(shù)f(x)=lnx-3x的零點(diǎn)所在的區(qū)間是 (A.(1,2) B.(1,e)C.(e,3) D.(0,1)答案:C4.多選題已知函數(shù)f(x)的圖象是連綿不斷的曲線,有如下的x,f(x)的對(duì)應(yīng)值表:x123456f(x)-82-3568則函數(shù)f(x)肯定存在零點(diǎn)的區(qū)間有 ()A.(2,3)和(3,4) B.(3,4)和(5,6)C.(1,2) D.(5,6)解析:由題中表格可知f(1)f(2)<0,f(2)f(3)<0,f(3)f(4)<0,依據(jù)零點(diǎn)存在定理可知函數(shù)f(x)在區(qū)間(1,2),(2,3),(3,4)上肯定存在零點(diǎn).答案:AC5.多選題下列說(shuō)法錯(cuò)誤的是 ()A.冪函數(shù)的增長(zhǎng)速度比一次函數(shù)的增長(zhǎng)速度快B.對(duì)隨意的x>0,xn>logaxC.對(duì)隨意的x>0,ax>logaxD.不肯定存在x0,使得當(dāng)x>x0時(shí),總有ax>xn>logax答案:ABCB級(jí)實(shí)力提升6.若f(x)=x2-x-1,x≥2或x≤-1,解析:令f(x)-x=0,得x≥2解得x≥2所以x=1+2或1,所以函數(shù)g(x)的零點(diǎn)為1+2,1.7.若函數(shù)f(x)=lnx+x-4有零點(diǎn)的區(qū)間為(k,k+1),k∈Z,則k=2.解析:方法1(應(yīng)用零點(diǎn)存在定理求解)易知函數(shù)f(x)=lnx+x-4在定義域內(nèi)為增函數(shù),且f(2)=ln2-2<0,f(3)=ln3-1>0,所以f(2)f(3)<0,所以函數(shù)f(x)在區(qū)間(2,3)上有零點(diǎn).所以k=2.方法2(轉(zhuǎn)化為函數(shù)圖象交點(diǎn)問(wèn)題)令y1=4-x,y2=lnx.在同一平面直角坐標(biāo)系中作出兩個(gè)函數(shù)的圖象,如圖所示.當(dāng)x=2時(shí),y1=2,y2=ln2<2,當(dāng)x=3時(shí),y1=1,y2=ln3>1,所以函數(shù)f(x)的零點(diǎn)在區(qū)間(2,3)上.故k=2.8.多空題已知函數(shù)f(x)=-(1)當(dāng)m=0時(shí),函數(shù)f(x)的零點(diǎn)個(gè)數(shù)為3;(2)假如函數(shù)f(x)恰有兩個(gè)零點(diǎn),那么實(shí)數(shù)m的取值范圍為[-2,0)∪[4,+∞).解析:(1)當(dāng)m=0時(shí),函數(shù)f(x)=-當(dāng)x≤0時(shí),令-x2-2x=0,解得x=0或x=-2;當(dāng)x>0時(shí),令x-4=0,解得x=4.所以當(dāng)m=0時(shí),函數(shù)f(x)有3個(gè)零點(diǎn).(2)作出函數(shù)y=-x2-2x和y=x-4的圖象,如圖所示.要使函數(shù)f(x)恰有兩個(gè)零點(diǎn),則-2≤m<0或m≥4,即實(shí)數(shù)m的取值范圍是[-2,0)∪[4,+∞).C級(jí)挑戰(zhàn)創(chuàng)新9.已知函數(shù)f(x)=a-(1)當(dāng)a=1時(shí),推斷函數(shù)f(x)的奇偶性并證明;(2)探討f(x)的零點(diǎn)個(gè)數(shù).解:(1)當(dāng)a=1時(shí),函數(shù)f(x)=-1+證明如下:依題意,得函數(shù)f(x)的定義域?yàn)镽,關(guān)于原點(diǎn)對(duì)稱.又因?yàn)閒(-x)=-1+2-x-f(x),所以函數(shù)f(x)為奇函數(shù).(2)對(duì)f(x)進(jìn)行化簡(jiǎn),得f(x)=a-21+2x,所以f(x)=0等價(jià)