課時(shí)規(guī)范練14函數(shù)與方程高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2026基礎(chǔ)鞏固練123456789101112131415161.方程lnx=4-2x的根所在的區(qū)間是(

)A.(0,1) B.(1,2)C.(2,3) D.(3,4)B解析

令f(x)=ln

x+2x-4,顯然f(x)=ln

x+2x-4在(0,+∞)內(nèi)單調(diào)遞增.又因?yàn)閒(1)=2-4=-2<0,f(2)=ln

2+4-4=ln

2>0,f(x)的圖象在區(qū)間[1,2]上是連續(xù)不斷的,由函數(shù)零點(diǎn)存在定理可知f(x)=ln

x+2x-4的零點(diǎn)所在的區(qū)間為(1,2),所以ln

x=4-2x的根所在的區(qū)間為(1,2).故選B.12345678910111213141516

C12345678910111213141516

123456789101112131415163.(2024·北京西城模擬)函數(shù)f(x)=ex|lnx|-1的零點(diǎn)個(gè)數(shù)是(

)A.0 B.1C.2 D.3C解析

令f(x)=0,可得|ln

x|=e-x,作出函數(shù)y=|ln

x|與y=e-x的圖象(如圖所示),由圖可知,函數(shù)y=|ln

x|與y=e-x的圖象的交點(diǎn)個(gè)數(shù)為2,故f(x)的零點(diǎn)個(gè)數(shù)為2,故選C.12345678910111213141516

C

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D12345678910111213141516

123456789101112131415166.(多選題)(2024·河南信陽模擬)已知函數(shù)f(x)=|1-2x|,實(shí)數(shù)a,b(a<b)是函數(shù)y=f(x)-m的兩個(gè)零點(diǎn),則下列結(jié)論正確的有(

)A.m>1B.0<m<1C.2a+2b=2D.a+b<0BCD12345678910111213141516

123456789101112131415167.(2024·云南昆明期末)設(shè)函數(shù)f(x)=ln(|x|+1),g(x)=-x2+a,若曲線f(x)與曲線g(x)有兩個(gè)交點(diǎn),則實(shí)數(shù)a的取值范圍是

.(0,+∞)解析

當(dāng)x≥0時(shí),f(x)=ln(x+1),當(dāng)x<0時(shí),f(x)=ln(-x+1),函數(shù)圖象示意圖為由g(x)=-x2+a與f(x)=ln(|x|+1)的圖象有兩個(gè)交點(diǎn)知a的取值范圍是(0,+∞).12345678910111213141516

2

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綜合提升練B

1234567891011121314151611.(2024·江蘇南通模擬)f(x)是定義在R上的奇函數(shù),當(dāng)x∈[-1,1]時(shí),f(x)=x,f(1+x)=f(1-x),令g(x)=f(x)-lgx,則函數(shù)g(x)的零點(diǎn)個(gè)數(shù)為(

)A.4 B.5C.6 D.7B12345678910111213141516解析

由f(1+x)=f(1-x)可知f(x)的圖象關(guān)于直線x=1對稱,又由f(1+x)=f(1-x)可得f(2+x)=f(-x)=-f(x),所以f(4+x)=-f(2+x)=f(x),因此f(x)的周期為4.作出f(x)的圖象(如圖所示),g(x)=f(x)-lg

x的零點(diǎn)個(gè)數(shù)即為f(x)的圖象與y=lg

x圖象的交點(diǎn)個(gè)數(shù),因?yàn)閘g

9<1,lg

10=1,由圖象可得f(x)的圖象與y=lg

x圖象的交點(diǎn)個(gè)數(shù)為5,故選B.12345678910111213141516

BCD12345678910111213141516

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A解析

由f(4-x)=(4-x)2-4(4-x)+a[e2(4-x)-4+e4-2(4-x)]=x2-4x+a(e4-2x+e2x-4)=f(x),可得函數(shù)f(x)的圖象關(guān)于直線x=2對稱,要使函數(shù)f(x)=x2-4x+a(e2x-4+e4-2x)有唯一的零點(diǎn),則f(2)=0,即4-8+2a=0,得a=2,故選A.1234567891011121314151614.(13分)(2024·北京模擬)已知關(guān)于x的不等式ax2+bx+c>0的解集為(1,t),記函數(shù)f(x)=ax2+(a-b)x-c.(1)求證:函數(shù)y=f(x)必有兩個(gè)不同的零點(diǎn);(2)若函數(shù)y=f(x)的兩個(gè)零點(diǎn)分別為m,n,求|m-n|的取值范圍.12345678910111213141516

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