高考數(shù)學(xué)二輪復(fù)習(xí)專(zhuān)題1函數(shù)與導(dǎo)數(shù)專(zhuān)題檢測(cè)1課件

專(zhuān)題檢測(cè)一12345678910111213141516171819一、選擇題1.(2024北京平谷模擬)下列函數(shù)中,在區(qū)間(0,+∞)內(nèi)單調(diào)遞減的是(

)C.f(x)=2-x D.f(x)=-x2+xC12345678910111213141516171819A1234567891011121314151617181912345678910111213141516171819B123456789101112131415161718194.(2024重慶南開(kāi)中學(xué)模擬)已知函數(shù)f(x)的部分圖象如圖所示,則f(x)的解析式可能是(

)C12345678910111213141516171819123456789101112131415161718195.(2024廣東佛山二模)若函數(shù)f(x)=alnx+(a≠0)既有極大值也有極小值,則下列結(jié)論一定正確的是(

)A.a<0 B.b<0 C.ab>-1 D.a+b>0B1234567891011121314151617181912345678910111213141516171819A.a<b<c B.c<b<aC.b<c<a D.a<c<bB12345678910111213141516171819123456789101112131415161718197.(2024北京海淀一模)函數(shù)f(x)是定義在(-4,4)內(nèi)的偶函數(shù),其圖象如圖所示,f(3)=0.設(shè)f'(x)是f(x)的導(dǎo)函數(shù),則關(guān)于x的不等式f(x+1)·f'(x)≥0的解集是(

)A.[0,2] B.[-3,0]∪[3,4) C.(-5,0]∪[2,4) D.(-4,0]∪[2,3)D12345678910111213141516171819解析

由f(3)=0,且f(x)為偶函數(shù),故f(-3)=0,由導(dǎo)數(shù)性質(zhì)結(jié)合題圖可得當(dāng)x∈(-4,0)時(shí),f'(x)<0,當(dāng)x∈(0,4)時(shí),f'(x)>0,f'(0)=0,解得-4<x<3.12345678910111213141516171819即-4<x<0,由f(x+1)=0,可得x+1=±3,即x=2或x=-4(舍去,不在定義域內(nèi)),由f'(x)=0,可得x=0.綜上所述,關(guān)于x的不等式f(x+1)·f'(x)≥0的解集為(-4,0]∪[2,3).故選D.123456789101112131415161718198.(2024青海一模)我們把函數(shù)圖象上任一點(diǎn)的橫坐標(biāo)與縱坐標(biāo)之積稱(chēng)為該點(diǎn)的“積值”.設(shè)函數(shù)

圖象上存在不同的三點(diǎn)A,B,C,其橫坐標(biāo)從左到右依次為x1,x2,x3,且其縱坐標(biāo)均相等,則A,B,C三點(diǎn)“積值”之和的最大值為(

)A.5ln6-30 B.5ln6-60 C.6ln5-30 D.6ln5-60A12345678910111213141516171819解析

依題意,A,B,C三點(diǎn)“積值”之和為(x1+x2+x3)y,y=f(x1)=f(x2)=f(x3),因?yàn)?/p>

可得f(x)在(-∞,-3)和(0,+∞)內(nèi)單調(diào)遞增,在(-3,0)內(nèi)單調(diào)遞減,當(dāng)x→-∞時(shí),f(x)→-∞,f(-3)=14,f(-6)=f(0)=5;當(dāng)x→+∞時(shí),f(x)→+∞,可畫(huà)出函數(shù)f(x)的大致圖象如圖.且有x1<x2<x3,使得f(x1)=f(x2)=f(x3),那么必有x1∈[-6,-3),x2∈(-3,0],x3∈[ln

6,ln

15),且x1,x2關(guān)于x=-3對(duì)稱(chēng),即x1+x2=-6,y=f(x1)=f(x2)=f(x3),y∈[5,14),12345678910111213141516171819則A,B,C三點(diǎn)“積值”之和(x1+x2+x3)y=yln(1+y)-6y,h'(y)=ln(1+y)+1-5=ln(1+y)-4,顯然h'(y)在[5,14)內(nèi)單調(diào)遞增,h'(y)<ln

15-4<0,所以h(y)在[5,14)內(nèi)單調(diào)遞減,h(y)max=6ln

6-31<0,所以h(y)<0,所以φ'(y)<0,φ(y)在[5,14)內(nèi)單調(diào)遞減,當(dāng)y=5時(shí)取最大值,φ(5)=5ln

6-30,故選A.12345678910111213141516171819二、選擇題BCD12345678910111213141516171819解析

令x=y=1,則2f(1)f(0)=f(1)+f(1)=2f(1),所以f(0)=1,令x=-1,y=1,則2f(0)f(-1)=f(-1)+f(1)=2f(-1),所以f(-1)=f(1)=-1,故A錯(cuò)誤;12345678910111213141516171819令y=-x,則2f(0)f(x)=f(x)+f(-x)=2f(x),所以f(x)=f(-x),所以f(x)為偶函數(shù),由B可知,f(1-x)=-f(x),所以f(1-x)=-f(x)=-f(-x),則有f(2-x)=-f(1-x)=f(x),故C正確;由C可知f(2-x)=f(x),又f(x)為偶函數(shù),所以f(2-x)=f(-x),則f(x)的周期為2,f(1)=-1,f(2)=f(0)=1,所以f(1)+f(2)+f(3)+…+f(2

025)=1

012×0-1=-1,故D正確.故選BCD.1234567891011121314151617181910.(2024河南開(kāi)封二模)高斯是德國(guó)著名的數(shù)學(xué)家,近代數(shù)學(xué)奠基者之一.用其名字命名的高斯取整函數(shù)為f(x)=[x],[x]表示不超過(guò)x的最大整數(shù),例如[-3.5]=-4,[2.1]=2.下列命題中正確的有(

)A.?x∈R,f(x)=x-1B.?x∈R,n∈Z,f(x+n)=f(x)+nC.?x,y>0,f(lgx)+f(lgy)=f(lg(xy))D.?n∈N*,f(lg1)+f(lg2)+f(lg3)+…+f(lgn)=92BD12345678910111213141516171819解析

對(duì)于A,當(dāng)x∈Z時(shí),f(x)=x,當(dāng)x?Z時(shí),f(x)∈Z,而x-1?Z,因此f(x)≠x-1,A錯(cuò)誤;對(duì)于B,?x∈R,n∈Z,令f(x)=m,則m≤x<m+1,m+n≤x+n<m+n+1,因此f(x+n)=m+n=f(x)+n,B正確;對(duì)于D,n∈N*,當(dāng)1≤n≤9時(shí),f(lg

n)=0,當(dāng)10≤n≤99時(shí),f(lg

n)=1,而f(lg

100)=2,因此f(lg

1)+f(lg

2)+f(lg

3)+…+f(lg

99)+f(lg

100)=92,此時(shí)n=100,D正確.故選BD.1234567891011121314151617181911.(2024海南?？谀M)已知函數(shù)f(x)的定義域?yàn)镽,其導(dǎo)函數(shù)為f'(x),且2f(x)+f'(x)=x,f(0)=-,則(

)A.f(-1)>-2B.f(1)>-1C.f(x)在(-∞,0)內(nèi)單調(diào)遞減D.f(x)在(0,+∞)內(nèi)單調(diào)遞增ABD1234567891011121314151617181912345678910111213141516171819三、填空題01234567891011121314151617181913.(2024陜西安康模擬)已知函數(shù)f(x)=2x3-2mx+m(m∈R),g(x)=-3x2,若關(guān)于x的不等式f(x)≤g(x)在區(qū)間[1,+∞)上有解,則實(shí)數(shù)m的取值范圍是

.

[5,+∞)12345678910111213141516171819cos1-sin11234567891011121314151617181912345678910111213141516171819四、解答題15.(13分)(2024山東煙臺(tái)一模)已如曲線f(x)=ax2+x-2lnx+b(a,b∈R)在x=2處的切線與直線x+2y+1=0垂直.(1)求a的值;(2)若f(x)≥0恒成立,求b的取值范圍.123456789101112131415161718191234567891011121314151617181916.(15分)(2024福建漳州模擬)已知函數(shù)f(x)=aex+x+1.(1)討論f(x)的單調(diào)性;(2)當(dāng)x>1時(shí),f(x)>+x,求實(shí)數(shù)a的取值范圍.解

(1)函數(shù)定義域?yàn)镽,且f'(x)=aex+1.當(dāng)a≥0時(shí),f'(x)>0,所以f(x)在R上單調(diào)遞增.當(dāng)a<0時(shí),令f'(x)>0,可得x<-ln(-a),令f'(x)<0,可得x>-ln(-a),所以f(x)在(-∞,-ln(-a))上單調(diào)遞增,在(-ln(-a),+∞)上單調(diào)遞減.綜上所述,當(dāng)a≥0時(shí),f(x)在R上單調(diào)遞增;當(dāng)a<0時(shí),f(x)在(-∞,-ln(-a))上單調(diào)遞增,在(-ln(-a),+∞)上單調(diào)遞減.12345678910111213141516171819因此ex+ln

a+ln

a+x>ln(x-1)+x-1,即ex+ln

a+x+ln

a>eln(x-1)+ln(x-1).令h(x)=ex+x,則有h(x+ln

a)>h(ln(x-1))對(duì)于x∈(1,+∞)恒成立.因?yàn)閔'(x)=ex+1>0,所以h(x)在R上單調(diào)遞增,故只需x+ln

a>ln(x-1),即ln

a>ln(x-1)-x在(1,+∞)上恒成立.當(dāng)x∈(1,2)時(shí),F'(x)>0,當(dāng)x∈(2,+∞)時(shí),F'(x)<0,所以F(x)在(1,2)內(nèi)單調(diào)遞增,在(2,+∞)上單調(diào)遞減,所以F(x)≤F(2)=-2.1234567891011121314151617181917.(15分)(2024湖南益陽(yáng)模擬)已知函數(shù)f(x)=ax2-xlnx.(1)若函數(shù)f(x)在(0,+∞)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)若a=2e,證明:f(x)<xex+1.(1)解

由已知得f'(x)=ax-ln

x-1.因?yàn)閒(x)在(0,+∞)上單調(diào)遞增,所以當(dāng)x>0時(shí),f'(x)≥0,即

,所以當(dāng)0<x<1時(shí),h'(x)>0,當(dāng)x>1時(shí),h'(x)<0,即h(x)在(0,1)內(nèi)單調(diào)遞增,在(1,+∞)上單調(diào)遞減,因此h(x)max=h(1)=1,所以a≥1,故實(shí)數(shù)a的取值范圍是[1,+∞).123456789101112131415161718191234567891011121314151617181918.(17分)(2024安徽合肥模擬)已知函數(shù)f(x)=alnx+x2,其中a∈R.(1)討論f(x)的單調(diào)性;(2)當(dāng)a=1時(shí),證明:f(x)≤x2+x-1;1234567891011121314151617181912345678910111213141516171819(2)證明

當(dāng)a=1時(shí),f(x)=ln

x+x2,要證明f(x)≤x2+x-1,即證ln

x≤x-1,即證ln

x-x+1≤0.設(shè)g(x)=ln

x-x+1,則g'(x)=