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)4.5.1函數(shù)的零點(diǎn)與方程的解分層演練含解析新人教A版必修第一冊(cè)

4.5.1函數(shù)的零點(diǎn)與方程的解分層演練綜合提升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≤1,1+loA.12,0 B.-2,0 C.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.若f(x)=x2-x-1,x≥2或x≤-1,5.推斷函數(shù)f(x)=x12-(12)解:方法1(單調(diào)性法)因?yàn)閥=x12在區(qū)間[0,+∞)上單調(diào)遞增,y=(12)x在R上單調(diào)遞減,所以f(x)=x12-(12)x在區(qū)間[0,+∞)上單調(diào)遞增.又因?yàn)閒(0)=-1<0,f(1)=12>0,所以f(x)=x方法2(數(shù)形結(jié)合法)令f(x)=x12-(12x12=(12)x,在同一平面直角坐標(biāo)系中作出y=x12和y=(由圖象可知兩個(gè)函數(shù)的圖象只有一個(gè)交點(diǎn),故函數(shù)f(x)=x12-(12)B級(jí)實(shí)力提升6.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=2.7.已知a∈R,當(dāng)x>0時(shí),f(x)=log21x+a.(1)若函數(shù)f(x)的圖象過點(diǎn)(1,1),求此時(shí)函數(shù)f(x)的解析式;(2)若函數(shù)g(x)=f(x)+2log2x只有一個(gè)零點(diǎn),求實(shí)數(shù)a的值.解:(1)因?yàn)閍∈R,當(dāng)x>0時(shí),f(x)=log2(1x+a)函數(shù)f(x)的圖象過點(diǎn)(1,1),所以f(1)=log2(1+a)=1,解得a=1,此時(shí)函數(shù)f(x)=log2(1x+1)(2)g(x)=f(x)+2log2x=log2(1x+a)+2log2xlog2(x+ax2).因?yàn)楹瘮?shù)g(x)=f(x)+2log2x只有一個(gè)零點(diǎn),所以ax2+x=1只有一個(gè)解,所以當(dāng)a=0時(shí),x=1,滿意題意;當(dāng)a≠0時(shí),ax2+x-1=0只有一個(gè)根,則Δ=12-4a×(-1)=0,解得a=-14綜上所述,a=0或a=-148.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+2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-x1+-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)于a=21+2x.因?yàn)楹瘮?shù)y=2x在R上單調(diào)遞增且值域?yàn)?0,+∞),所以y=21+2x在R上單調(diào)遞減且值域?yàn)?0,2),所以當(dāng)a≤0或a≥2時(shí),函數(shù)f(C級(jí)挑戰(zhàn)創(chuàng)新9.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:AC10.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;解析: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)有三個(gè)零點(diǎn).(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[