全國(guó)統(tǒng)考高考數(shù)學(xué)大一輪復(fù)習(xí)解題思維2高考中函數(shù)與導(dǎo)數(shù)解答題的提分策略備考試題文含解析

一輪復(fù)習(xí)精品資料(高中)PAGEPAGE1解題思維2高考中函數(shù)與導(dǎo)數(shù)解答題的提分策略1.〖2021貴陽(yáng)市四校聯(lián)考,12分〗已知函數(shù)f(x)=xex,f'(x)是f(x(1)求f(x)的極值;(2)當(dāng)x0<1時(shí),證明:f(x)≤f'(x0)(x-x0)+f(x0).2.〖2021黑龍江省六校聯(lián)考,12分〗設(shè)函數(shù)f(x)=xsinx+cosx-12ax2(1)當(dāng)a=12時(shí),討論f(x(2)當(dāng)a>13時(shí),證明:f(x)有且僅有兩個(gè)零點(diǎn)3.〖2021江西紅色七校聯(lián)考,12分〗已知函數(shù)f(x)=lnx-12ax2-(1)若函數(shù)f(x)在〖1,+∞)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.(2)若函數(shù)f(x)的圖象在x=1處的切線平行于x軸,則是否存在整數(shù)k,使不等式x〖f(x)+x-1〗>k(x-2)在x>e時(shí)恒成立?若存在,求出k的最大值;若不存在,請(qǐng)說(shuō)明理由.4.〖2020江西紅色七校聯(lián)考,12分〗已知f(x)=12x(x+2)-a(x+lnx)(a∈R)(1)討論f(x)的單調(diào)性;(2)若f(x1)=f(x2)(x1≠x2),證明:x1+x2>2a.答案解題思維2高考函數(shù)與導(dǎo)數(shù)解答題的提分策略1.(1)因?yàn)閒(x)=xex,所以f'(x)=令f'(x)=0,得x=1,當(dāng)x∈(-∞,1)時(shí),f'(x)>0,當(dāng)x∈(1,+∞)時(shí),f'(x)<0,所以f(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)有極大值,極大值為f(1)=1e,無(wú)極小值.(4分)(2)令g(x)=f(x)-f'(x0)(x-x0)-f(x0),則g'(x)=f'(x)-f'(x0)=1-設(shè)φ(x)=ex0(1-x)-ex(1-x0),則φ'(x)=-ex0-因?yàn)閤0<1,所以φ'(x)<0,所以φ(x)在R上單調(diào)遞減,又φ(x0)=0,所以當(dāng)x<x0時(shí),φ(x)>0,當(dāng)x>x0時(shí),φ(x)<0,即當(dāng)x<x0時(shí),g'(x)>0,當(dāng)x>x0時(shí),g'(x)<0,所以g(x)在區(qū)間(-∞,x0)上單調(diào)遞增,在區(qū)間(x0,+∞)上單調(diào)遞減,所以g(x)≤g(x0)=0,即f(x)≤f'(x0)(x-x0)+f(x0).(12分)2.(1)當(dāng)a=12時(shí),f(x)=xsinx+cosx-14則f'(x)=x(cosx-12),令f'(x)=0,得x=-π3或x=0或x=當(dāng)-π<x<-π3時(shí),f'(x)>0,當(dāng)-π3<x<0時(shí),f'(x)<0,當(dāng)0<x<π3時(shí),f'(x)>0,當(dāng)π3<所以f(x)在(-π,-π3),(0,π3)上單調(diào)遞增,在(-π3,0),((2)f(x)的定義域?yàn)?-∞,+∞),因?yàn)閒(-x)=-xsin(-x)+cos(-x)-12a(-x)2=xsinx+cosx-12ax2=f(x),所以f因?yàn)閒(0)=1>0,所以當(dāng)a>13時(shí),f(x)有且僅有兩個(gè)零點(diǎn)等價(jià)于當(dāng)a>13時(shí),f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn).f'(x)=x(cosx-a),當(dāng)a≥1時(shí),若x>0,則f'(x)<0,所以f(x)在(0,+∞)上單調(diào)遞減,因?yàn)閒(π)=-1-12aπ2<0,所以f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn);當(dāng)13<a<1時(shí),存在θ∈(0,π2),使得cosθ=當(dāng)0<x<θ時(shí),f'(x)>0,當(dāng)2kπ+θ<x<2kπ+2π-θ,k∈N時(shí),f'(x)<0,當(dāng)2kπ+2π-θ<x<2kπ+2π+θ,k∈N時(shí),f'(x)>0,所以f(x)在(0,θ)上單調(diào)遞增,在(2kπ+θ,2kπ+2π-θ)(k∈N)上單調(diào)遞減,在(2kπ+2π-θ,2kπ+2π+θ)(k∈N)上單調(diào)遞增,由tanθ=1a2-1,13<a<1,可得0<tanθ<22,當(dāng)k∈N時(shí),2kπ+2π+θ所以f(2kπ+2π+θ)=-12a〖(2kπ+2π+θ-tanθ)2-1〗<-16〖(2kπ+2π+θ-tanθ)2-1〗=-<0,因此f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn).綜上,當(dāng)a>13時(shí),f(x)有且僅有兩個(gè)零點(diǎn).(12分)3.(1)依題意,f'(x)=1x-ax-1=1-ax2-xx≥0在〖1,+∞)上恒成立,即1-ax2-x≥0在〖令φ(x)=1-xx2=(1x)2-1x=(1x-12)2-14(x≥1),則當(dāng)x=2時(shí),φ∴a≤-14,即實(shí)數(shù)a的取值范圍是(-∞,-14(2)依題意,f'(1)=1-a-1=0,∵a=0,∴f(x)=lnx-x,不等式x〖f(x)+x-1〗>k(x-2)在x>e時(shí)恒成立,即k<xlnx-x令g(x)=xlnx-xx-2(x>e),則g'令h(x)=x-2lnx(x>e),則h'(x)=1-2x>0,∴h(x)在(e,+∞)上單調(diào)遞增,∴h(x)>h(e)=e-2lne=e-2>0,即g'(x)>0在(e,+∞)上恒成立,∴g(x)在(e,+∞)上單調(diào)遞增,∴g(x)>elne-∴k≤0,(11分)∴整數(shù)k的最大值為0.(12分)4.(1)f(x)的定義域?yàn)?0,+∞),由f(x)=12x(x+2)-a(x+lnx得f'(x)=x+1-a-a若a≤0,則f'(x)>0恒成立,f(x)在(0,+∞)上單調(diào)遞增;若a>0,則當(dāng)0<x<a時(shí),f'(x)<0,當(dāng)x>a時(shí),f'(x)>0,則f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.綜上可得,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(6分)(2)由(1)知,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增,不存在f(x1)=f(x2)(x1≠x2),所以a>0.由(1)知當(dāng)a>0時(shí),f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增,存在f(x1)=f(x2).不妨設(shè)0<x1<a<x2,設(shè)g(x)=f(a+x)-f(a-x),x∈(0,a),則g'(x)=f'(a+x)+f'(a-x),又由(1)知f'(x)=(x+1)(x-a)x,可得g'(x)=f'(a+x)+因?yàn)閤∈(0,a),所以g'(x)=-2所以g(x)在(0,a)上單調(diào)遞減,所以g(x)<0,即當(dāng)x∈(0,a)時(shí),f(a+x)<f(a-x),由于0<a-x1<a,則f(a-(a-x1))>f(a+(a-x1)),即f(x1)=f(a-(a-x1))>f(a+(a-x1))=f(2a-x1).又f(x2)=f(x1),則有f(x2)>f(2a-x1).又x2>a,2a-x1>a,f(x)在(a,+∞)上單調(diào)遞增,所以x2>2a-x1,即x1+x2>2a.(12分)解題思維2高考中函數(shù)與導(dǎo)數(shù)解答題的提分策略1.〖2021貴陽(yáng)市四校聯(lián)考,12分〗已知函數(shù)f(x)=xex,f'(x)是f(x(1)求f(x)的極值;(2)當(dāng)x0<1時(shí),證明:f(x)≤f'(x0)(x-x0)+f(x0).2.〖2021黑龍江省六校聯(lián)考,12分〗設(shè)函數(shù)f(x)=xsinx+cosx-12ax2(1)當(dāng)a=12時(shí),討論f(x(2)當(dāng)a>13時(shí),證明:f(x)有且僅有兩個(gè)零點(diǎn)3.〖2021江西紅色七校聯(lián)考,12分〗已知函數(shù)f(x)=lnx-12ax2-(1)若函數(shù)f(x)在〖1,+∞)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.(2)若函數(shù)f(x)的圖象在x=1處的切線平行于x軸,則是否存在整數(shù)k,使不等式x〖f(x)+x-1〗>k(x-2)在x>e時(shí)恒成立?若存在,求出k的最大值;若不存在,請(qǐng)說(shuō)明理由.4.〖2020江西紅色七校聯(lián)考,12分〗已知f(x)=12x(x+2)-a(x+lnx)(a∈R)(1)討論f(x)的單調(diào)性;(2)若f(x1)=f(x2)(x1≠x2),證明:x1+x2>2a.答案解題思維2高考函數(shù)與導(dǎo)數(shù)解答題的提分策略1.(1)因?yàn)閒(x)=xex,所以f'(x)=令f'(x)=0,得x=1,當(dāng)x∈(-∞,1)時(shí),f'(x)>0,當(dāng)x∈(1,+∞)時(shí),f'(x)<0,所以f(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)有極大值,極大值為f(1)=1e,無(wú)極小值.(4分)(2)令g(x)=f(x)-f'(x0)(x-x0)-f(x0),則g'(x)=f'(x)-f'(x0)=1-設(shè)φ(x)=ex0(1-x)-ex(1-x0),則φ'(x)=-ex0-因?yàn)閤0<1,所以φ'(x)<0,所以φ(x)在R上單調(diào)遞減,又φ(x0)=0,所以當(dāng)x<x0時(shí),φ(x)>0,當(dāng)x>x0時(shí),φ(x)<0,即當(dāng)x<x0時(shí),g'(x)>0,當(dāng)x>x0時(shí),g'(x)<0,所以g(x)在區(qū)間(-∞,x0)上單調(diào)遞增,在區(qū)間(x0,+∞)上單調(diào)遞減,所以g(x)≤g(x0)=0,即f(x)≤f'(x0)(x-x0)+f(x0).(12分)2.(1)當(dāng)a=12時(shí),f(x)=xsinx+cosx-14則f'(x)=x(cosx-12),令f'(x)=0,得x=-π3或x=0或x=當(dāng)-π<x<-π3時(shí),f'(x)>0,當(dāng)-π3<x<0時(shí),f'(x)<0,當(dāng)0<x<π3時(shí),f'(x)>0,當(dāng)π3<所以f(x)在(-π,-π3),(0,π3)上單調(diào)遞增,在(-π3,0),((2)f(x)的定義域?yàn)?-∞,+∞),因?yàn)閒(-x)=-xsin(-x)+cos(-x)-12a(-x)2=xsinx+cosx-12ax2=f(x),所以f因?yàn)閒(0)=1>0,所以當(dāng)a>13時(shí),f(x)有且僅有兩個(gè)零點(diǎn)等價(jià)于當(dāng)a>13時(shí),f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn).f'(x)=x(cosx-a),當(dāng)a≥1時(shí),若x>0,則f'(x)<0,所以f(x)在(0,+∞)上單調(diào)遞減,因?yàn)閒(π)=-1-12aπ2<0,所以f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn);當(dāng)13<a<1時(shí),存在θ∈(0,π2),使得cosθ=當(dāng)0<x<θ時(shí),f'(x)>0,當(dāng)2kπ+θ<x<2kπ+2π-θ,k∈N時(shí),f'(x)<0,當(dāng)2kπ+2π-θ<x<2kπ+2π+θ,k∈N時(shí),f'(x)>0,所以f(x)在(0,θ)上單調(diào)遞增,在(2kπ+θ,2kπ+2π-θ)(k∈N)上單調(diào)遞減,在(2kπ+2π-θ,2kπ+2π+θ)(k∈N)上單調(diào)遞增,由tanθ=1a2-1,13<a<1,可得0<tanθ<22,當(dāng)k∈N時(shí),2kπ+2π+θ所以f(2kπ+2π+θ)=-12a〖(2kπ+2π+θ-tanθ)2-1〗<-16〖(2kπ+2π+θ-tanθ)2-1〗=-<0,因此f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn).綜上,當(dāng)a>13時(shí),f(x)有且僅有兩個(gè)零點(diǎn).(12分)3.(1)依題意,f'(x)=1x-ax-1=1-ax2-xx≥0在〖1,+∞)上恒成立,即1-ax2-x≥0在〖令φ(x)=1-xx2=(1x)2-1x=(1x-12)2-14(x≥1),則當(dāng)x=2時(shí),φ∴a≤-14,即實(shí)數(shù)a的取值范圍是(-∞,-14(2)依題意,f'(1)=1-a-1=0,∵a=0,∴f(x)=lnx-x,不等式x〖f(x)+x-1〗>k(x-2)在x>e時(shí)恒成立,即k<xlnx-x令g(x)=xlnx-xx-2(x>e),則g'令h(x)=x-2lnx(x>e),則h'(x)=1-2x>0,∴h(x)在(e,+∞)上單調(diào)遞增,∴h(x)>h(e)=e-2lne=e-2>0,即g'(x)>0在(e,+∞)上恒成立,∴g(x)在(e,+∞)上單調(diào)遞增,∴g(x)>elne-∴k≤0,(11分)∴整數(shù)k的最大值為0.(12分)4.(1)f(x)的定義域?yàn)?0,+∞),由f(x)=12x(x+2)-a(x+lnx得f'(x)=x+1-a-a若a≤0,則f'(x)>0恒成立,f(x)在(0,+∞)上單調(diào)遞增;若a>0,則當(dāng)0<x<a時(shí),f'(x)<0,當(dāng)x>a時(shí),f'(x)>0,則f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.綜上可得,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增.(6分)(2)由(1)知,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增,不存在f(x1)=f(x2)(x1≠x2),所以a>0.由(1)知當(dāng)a>0時(shí),f(x)在(0,a)上單調(diào)遞減,在(a,+∞)上單調(diào)遞增,存在f(x1)=f(x2).不妨設(shè)0<x1<a<x2,設(shè)g(x)=f(a+x)-f(a-x),x∈(0,a),則g'(x)=f'(a+x)+f'(a-x),又由(