2025高考數(shù)學(xué)一輪復(fù)習(xí)-3.3導(dǎo)數(shù)與函數(shù)的極值、最值-專項訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-3.3導(dǎo)數(shù)與函數(shù)的極值、最值-專項訓(xùn)練基礎(chǔ)鞏固練1.如圖是f(x)的導(dǎo)函數(shù)f'(x)的圖象,則f(x)的極小值點的個數(shù)為()A.1 B.2 C.3 D.42.設(shè)函數(shù)f(x)=2x+lnx,則(A.x=12為f(x)B.x=12為f(x)C.x=2為f(x)的極大值點D.x=2為f(x)的極小值點3.若x=1是函數(shù)f(x)=(x2+ax-1)ex-1的極值點,則f(x)的極大值為()A.-1 B.-2e-3 C.5e-3 D.14.當(dāng)x=1時,函數(shù)f(x)=alnx+b+1x取得極小值4,則a+b=(A.7 B.8 C.9 D.105.函數(shù)f(x)=x2ex+1在x∈[-1,3]A.1 B.9e-4 C.0 D.4e-36.已知函數(shù)f(x)=xex-a和g(x)=lnxx+b有相同的極大值,則A.2 B.0 C.-3 D.-17.已知一正四棱柱(底面為正方形的直四棱柱)內(nèi)切于底面半徑為1,高為2的圓錐,當(dāng)正四棱柱的體積最大時,該正四棱柱的底面邊長為()A.223 B.23 C.2 D8.(多選題)已知函數(shù)f(x)=xlnx+x2,x0是函數(shù)f(x)的極值點,下列結(jié)論中正確的有()A.0<x0<1e B.x0>C.f(x0)+2x0<0 D.f(x0)+2x0>09.已知函數(shù)f(x)=lnx-ax存在最大值0,則a=.10.已知函數(shù)f(x)=2xlnx,g(x)=-x2+ax-3.對一切x∈(0,+∞),f(x)≥g(x)恒成立,求實數(shù)a的取值范圍.綜合提升練11.已知f(x)是定義域為R的偶函數(shù),若當(dāng)x≤0時,f(x)=(x+1)3ex+1,則函數(shù)f(x)的極值點的個數(shù)是()A.5 B.4 C.3 D.212.(多選題)已知函數(shù)f(x)=(x-2)ex-ax2+2ax-2a,若f(x)有兩個不同的極值點x1,x2(x1<x2),且當(dāng)0<x<x2時,恒有f(x)<-2a,則a的可能取值有()A.a=e2 B.a=eC.a=e2 D.a=13.用數(shù)學(xué)的眼光看世界就能發(fā)現(xiàn)很多數(shù)學(xué)之“美”.現(xiàn)代建筑講究線條感,曲線之美讓人稱奇.衡量曲線彎曲程度的重要指標(biāo)是曲率,曲線的曲率定義如下:若f'(x)是f(x)的導(dǎo)函數(shù),f″(x)是f'(x)的導(dǎo)函數(shù),則曲線y=f(x)在點(x,f(x))處的曲率K=|f″(x)|{1+[f'(x)]2}32,則曲線f(x)=x在(1,1)處的曲率為;正弦曲線g(14.(2023南通檢測)已知函數(shù)f(x)=ax-lnx-ax(1)若x>1,f(x)>0,求實數(shù)a的取值范圍.(2)設(shè)x1,x2是函數(shù)f(x)的兩個極值點,證明:|f(x1)-f(x2)|<1-創(chuàng)新應(yīng)用練15．(1)證明:當(dāng)0<x<1時,x-x2<sinx<x.(2)已知函數(shù)f(x)=cosax-ln(1-x2),若x=0是f(x)的極大值點,求a的取值范圍.參考答案1.A2.D3.C4.A5.C6.B7.A8.AD9.10.解由題可知,當(dāng)x∈(0,+∞)時,f(x)≥g(x)恒成立,即2xlnx≥-x2+ax-3在x∈(0,+∞)上恒成立,得a≤2lnx+x+3x在x∈(0,+∞)上恒成立.設(shè)h(x)=2lnx+x+3x(x>0),則h'(x)=2x?3x2+1=(x+3)(x-1)x2.當(dāng)x∈(0,1)時,h'(x)<0,h(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時,h'(x)>0,h(x)單調(diào)遞增,所以h(x)min=h(1)=2ln1+1+31=4,11.C12.BD13.2514.(1)解依題意得,f'(x)=a-1x+ax①當(dāng)a≤0時,在x∈(1,+∞)上,f'(x)<0,所以f(x)在(1,+∞)上單調(diào)遞減,所以f(x)<f(1)=0,所以a≤0不符合題設(shè).②當(dāng)0<a<12時,令f'(x)=0,得ax2-x+a=0,解得x1=1-1-4a22a所以當(dāng)x∈(1,x2)時,f'(x)<0,所以f(x)在(1,x2)上單調(diào)遞減,此時f(x)<f(1)=0,所以0<a<12不符合題設(shè)③當(dāng)a≥12時,判別式Δ=1-4a2≤0,所以f'(x)≥0,所以f(x)在(1,+∞)上單調(diào)遞增,所以f(x)>f(1)=綜上,實數(shù)a的取值范圍是1(2)證明由(1)知,當(dāng)0<a<12時,f(x)在(0,x1),(x2,+∞)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,所以x1是f(x)的極大值點,x2是f(x)的極小值點.由(1)知,x1x2=1,x1+x2=1a,則x2-x1=(x1+x2)2-4x1x2=1-4a2a.綜上,要證|f(x1)-f因為x2-x1-f(x1)+f(x2)=(a+1)(x2-x1)-lnx2x1+a(x2-x1)x1x2=2a=2(x2=2x2x1-11+x2x1+x2x1?x1x2-lnx2x1,設(shè)t=x2x1>1,則g(t)=2(t-1)t+1+所以x2-x1-f(x1)+f(x2)>0,即f(x1)-f(x2)<x2-x1成立,所以原不等式成立.15.(1)證明構(gòu)建F(x)=x-sinx,x∈(0,1),則F'(x)=1-cosx>0對任意的x∈(0,1)恒成立,則F(x)在(0,1)上單調(diào)遞增,可得F(x)>F(0)=0,所以x>sinx,x∈(0,1).構(gòu)建G(x)=sinx-(x-x2)=x2-x+sinx,x∈(0,1),則G'(x)=2x-1+cosx,x∈(0,1).構(gòu)建g(x)=G'(x),x∈(0,1),則g'(x)=2-sinx>0對任意的x∈(0,1)恒成立,則g(x)在(0,1)上單調(diào)遞增,可得g(x)>g(0)=0,即G'(x)>0對任意的x∈(0,1)恒成立,則G(x)在(0,1)上單調(diào)遞增,可得G(x)>G(0)=0,所以sinx>x-x2,x∈(0,1).綜上所述,x-x2<sinx<x.(2)解令1-x2>0,解得-1<x<1,即函數(shù)f(x)的定義域為(-1,1),若a=0,則f(x)=-ln(1-x2),x∈(-1,1).因為y=-lnu在定義域內(nèi)單調(diào)遞減,y=1-x2在(-1,0)上單調(diào)遞增,在(0,1)上單調(diào)遞減,則f(x)=-ln(1-x2)在(-1,0)上單調(diào)遞減,在(0,1)上單調(diào)遞增,故x=0是f(x)的極小值點,不合題意,所以a≠0.當(dāng)a≠0時,令b=|a|>0,因為f(x)=cosax-ln(1-x2)=cos(|a|x)-ln(1-x2)=cosbx-ln(1-x2),且f(-x)=cos(-bx)-ln[1-(-x)2]=cosbx-ln(1-x2)=f(x),所以函數(shù)f(x)在定義域內(nèi)為偶函數(shù),由題意可得,f'(x)=-bsinbx-2xx2-1,x(ⅰ)當(dāng)0<b2≤2時,取m=min1b,1,x∈(0,m),則bx由(1)可得f'(x)=-bsinbx-2xx2-1>-b2x-2xx2-1=x(b2x2+2-b2)1-x即當(dāng)x∈(0,m)?(0,1)時,f'(x)>0,則f(x)在(0,m)上單調(diào)遞增,結(jié)合偶函數(shù)的對稱性可知,f(x)在(-m,0)上單調(diào)遞減,所以x=0是f(x)的極小值點,不合題意.(ⅱ)當(dāng)b2>2時,取x∈0,1b?(0,1),由(1)可得f'(x)=-bsinbx-2xx2-1<-b(bx-b2x2)-2xx2-1=x1-x2構(gòu)建h(x)=-b3x3+b2x2+b3x+2-b2,x∈0,1b,則h'(x)=-3b3x2+2b2x+b3,x∈0,1b,且h'(0)=b3>0,h'1b=b3-b>0,則h'(x)>0對任意的x∈0,1b恒成立,可知h(x)在0,1b上單調(diào)遞增,且h(0)=2-b2<0,h1b=2>0,所