2025版高中數(shù)學(xué)一輪復(fù)習(xí)課時(shí)作業(yè)梯級練十八導(dǎo)數(shù)中的三大難點(diǎn)問題課時(shí)作業(yè)理含解析新人教A版

PAGE課時(shí)作業(yè)梯級練十八導(dǎo)數(shù)中的三大難點(diǎn)問題一、選擇題(每小題5分,共35分)1.已知函數(shù)f(x)的定義域?yàn)閇-1,4],部分對應(yīng)值如表:x-10234f(x)12020f(x)的導(dǎo)函數(shù)y=f′(x)的圖象如圖所示.當(dāng)1<a<2時(shí),函數(shù)y=f(x)-a的零點(diǎn)的個數(shù)為 ()A.1 B.2 C.3 【解析】選D.依據(jù)導(dǎo)函數(shù)圖象,知2是函數(shù)的微小值點(diǎn),函數(shù)y=f(x)的大致圖象如圖所示.由于f(0)=f(3)=2,1<a<2,所以y=f(x)-a的零點(diǎn)的個數(shù)為4.2.(2024·南充模擬)已知當(dāng)x≥1時(shí),關(guān)于x的不等式mx≥QUOTE+2lnx恒成立,則m的取值范圍為 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選A.依題意,mx≥QUOTE+2lnx,所以m≥QUOTE+QUOTE,令gQUOTE=QUOTE+QUOTE,故g′QUOTE=QUOTE;令hQUOTE=x-xlnx-1,則h′QUOTE=-lnx,故當(dāng)x∈QUOTE時(shí),h′QUOTE=-lnx≤0;h(1)=0,h(x)≤0,g′(x)≤0,故gQUOTE=QUOTE+QUOTE在QUOTE上單調(diào)遞減,故m≥QUOTE=gQUOTE=1,所以m的取值范圍為QUOTE.3.已知函數(shù)fQUOTE=x2-xsinx,若a=fQUOTE,b=fQUOTE,c=fQUOTE,則 ()A.a>b>c B.b>a>cC.c>b>a D.c>a>b【解析】選B.fQUOTE=QUOTE-QUOTEsinQUOTE=x2-xsinx=fQUOTE,故fQUOTE為偶函數(shù),故只需考慮x∈QUOTE的單調(diào)性即可.f′(x)=2x-sinx-xcosx=x-sinx+xQUOTE,當(dāng)x∈QUOTE時(shí),設(shè)hQUOTE=x-sinx,則h′QUOTE=1-cosx≥0,所以hQUOTE在QUOTE上單調(diào)遞增,即hQUOTE>h(0)=0,故x>sinx,而xQUOTE≥0明顯成立,故f′(x)>0,故fQUOTE在x∈QUOTE上單調(diào)遞增.a=fQUOTE=fQUOTE,b=fQUOTE=fQUOTE,0.23<0.2<log5QUOTE<log53<1<log35,由函數(shù)單調(diào)性可知fQUOTE<fQUOTE<fQUOTE,即c<a<b.4.(2024·渭南模擬)設(shè)函數(shù)f(x)=(x-a)2+(3lnx-3a)2,若存在x0,使f(x0)≤QUOTE,則實(shí)數(shù)a的值為 ()A.QUOTE B.QUOTE C.QUOTE D.1【解析】選A.f(x)表示點(diǎn)(x,3lnx)與點(diǎn)(a,3a)間的距離的平方,分別令g(x)=3lnx,h(x)=3x,設(shè)過點(diǎn)P(x0,3lnx0)的函數(shù)g(x)的切線l平行于直線y=3x.g′(x)=QUOTE,由QUOTE=3,解得x0=1,所以切點(diǎn)P(1,0).點(diǎn)P到直線y=3x的距離d=QUOTE.所以存在x0=1,使f(x0)≤QUOTE,過點(diǎn)P且與直線y=3x垂直的直線方程為y=-QUOTE(x-1).聯(lián)立QUOTE解得QUOTE則實(shí)數(shù)a=QUOTE.5.(2024·石家莊模擬)已知函數(shù)fQUOTE對于隨意x∈R,均滿意fQUOTE=fQUOTE,當(dāng)x≤1時(shí),f(x)=QUOTE(其中e為自然對數(shù)的底數(shù)),若存在實(shí)數(shù)a,b,c,dQUOTE滿意fQUOTE=fQUOTE=fQUOTE=fQUOTE,則(a+b+c+d)b-ea的取值范圍為 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選D.由f(x)=f(2-x)知f(x)關(guān)于x=1對稱,如圖,因此a+d=b+c=2,所以a+b+c+d=4,又因?yàn)閒(a)=f(b),所以ea=lnb+2,因此(a+b+c+d)b-ea=4b-lnb-2,由題意知QUOTE<b≤QUOTE,令g(b)=4b-lnb-2QUOTE,g′(b)=4-QUOTE=QUOTE,令g′(b)=0得b=QUOTE,故g(b)在QUOTE上單調(diào)遞減,在QUOTE上單調(diào)遞增,故g(b)min=gQUOTE=2ln2-1,由gQUOTE=QUOTE,gQUOTE=QUOTE-1,得gQUOTE-gQUOTE=QUOTE-QUOTE+1=QUOTE>0,故g(b)∈QUOTE.6.已知函數(shù)f(x)=QUOTE+(m+1)ex+2(m∈R)有兩個極值點(diǎn),則實(shí)數(shù)m的取值范圍為 ()A.QUOTE B.QUOTEC.QUOTE D.(0,+∞)【解析】選B.函數(shù)fQUOTE的定義域?yàn)镽,f′QUOTE=x+QUOTEex.因?yàn)楹瘮?shù)fQUOTE有兩個極值點(diǎn),所以f′QUOTE=x+QUOTEex有兩個不同的零點(diǎn),故關(guān)于x的方程-m-1=QUOTE有兩個不同的解,令gQUOTE=QUOTE,則g′QUOTE=QUOTE,當(dāng)x∈QUOTE時(shí),g′QUOTE>0,當(dāng)x∈QUOTE時(shí),g′QUOTE<0,所以函數(shù)gQUOTE=QUOTE在區(qū)間QUOTE上單調(diào)遞增,在區(qū)間QUOTE上單調(diào)遞減,又當(dāng)x→-∞時(shí),gQUOTE→-∞;當(dāng)x→+∞時(shí),gQUOTE→0,且gQUOTE=QUOTE,故0<-m-1<QUOTE,所以-1-QUOTE<m<-1.7.已知函數(shù)y=a+8lnxQUOTE的圖象上存在點(diǎn)P,函數(shù)y=-x2-2的圖象上存在點(diǎn)Q,且P,Q關(guān)于x軸對稱,則a的取值范圍是 ()A.[6-8ln2,e2-6]B.[e2-6,+∞)C.QUOTED.QUOTE【解析】選D.函數(shù)y=-x2-2的圖象與函數(shù)y=x2+2的圖象關(guān)于x軸對稱,依據(jù)已知得函數(shù)y=a+8lnxQUOTE的圖象與函數(shù)y=x2+2的圖象有交點(diǎn),即方程a+8lnx=x2+2在[QUOTE,e]上有解,即a=x2+2-8lnx在[QUOTE,e]上有解,令g(x)=x2+2-8lnx,x∈[QUOTE,e],則g′(x)=2x-QUOTE=QUOTE,當(dāng)x∈[QUOTE,2)時(shí),g′(x)<0;當(dāng)x∈(2,e]時(shí),g′(x)>0.故當(dāng)x=2時(shí),g(x)取最小值,g(2)=6-8ln2,由于g(QUOTE)=10+QUOTE,g(e)=e2-6.故當(dāng)x=QUOTE時(shí),g(x)取到最大值10+QUOTE.所以6-8ln2≤a≤10+QUOTE.二、填空題(每小題5分,共15分)8.(2024·北京模擬)能夠說明“在某個區(qū)間(a,b)內(nèi),假如函數(shù)y=f(x)在這個區(qū)間內(nèi)單調(diào)遞增,那么f′(x)>0恒成立”是假命題的一個函數(shù)是(寫出函數(shù)表達(dá)式和區(qū)間).

【解析】若f(x)=x3,x∈(-1,1),易知f(x)=x3在(-1,1)上單調(diào)遞增;但f′(x)=3x2,在x=0時(shí)f′(x)=0,不滿意f′(x)>0恒成立,是假命題.答案:f(x)=x3,x∈(-1,1)(答案不唯一)9.(2024·鄭州調(diào)研)已知函數(shù)f(x)=cosπx,g(x)=eax-a+QUOTE(a>0),若?x1,x2∈[0,1],使得f(x1)=g(x2),則實(shí)數(shù)a的取值范圍是.

【解析】設(shè)集合F,G分別為函數(shù)f(x)與g(x)在區(qū)間[0,1]上的值域,則F=[-1,1].由a>0,知ea>1,所以g(x)=(ea)x-a+QUOTE在[0,1]上單調(diào)遞增.所以G=[-a+QUOTE,ea-a+QUOTE].又?x1,x2∈[0,1],使得f(x1)=g(x2),所以F∩G≠?.因?yàn)閔(a)=ea-a+QUOTE在(0,+∞)上遞增,所以h(a)>h(0)=QUOTE≥1恒成立,所以只需-a+QUOTE≤1,即a≥QUOTE.答案:QUOTE10.(2024·曲靖模擬)已知函數(shù)f(x)=ax3-3x2+1,若f(x)存在唯一的零點(diǎn)x0,且x0>0,則實(shí)數(shù)a的取值范圍是.

【解析】因?yàn)閒(x)=ax3-3x2+1,所以f′(x)=3ax2-6x=3x(ax-2),f(0)=1;①當(dāng)a=0時(shí),f(x)=-3x2+1有兩個零點(diǎn),不成立;②當(dāng)a>0時(shí),f(x)=ax3-3x2+1在(-∞,0)上有零點(diǎn),故不成立;③當(dāng)a<0時(shí),f(x)=ax3-3x2+1在(0,+∞)上有且只有一個零點(diǎn);故f(x)=ax3-3x2+1在(-∞,0)上沒有零點(diǎn);而當(dāng)x=QUOTE時(shí),f(x)=ax3-3x2+1在(-∞,0)上取得最小值;故f(QUOTE)=QUOTE-3·QUOTE+1>0;故a<-2;綜上所述,實(shí)數(shù)a的取值范圍是(-∞,-2).答案:(-∞,-2)1.(5分)若函數(shù)fQUOTE與gQUOTE滿意:存在實(shí)數(shù)t,使得fQUOTE=g′QUOTE,則稱函數(shù)gQUOTE為fQUOTE的“友導(dǎo)”函數(shù).已知函數(shù)g(x)=QUOTEkx2-x+3為函數(shù)fQUOTE=x2lnx+x的“友導(dǎo)”函數(shù),則k的最小值為 ()A.QUOTE B.1 C.2 D.QUOTE【解析】選C.g′(x)=kx-1,由題意,g(x)為函數(shù)f(x)的“友導(dǎo)”函數(shù),即方程x2lnx+x=kx-1有解,故k=xlnx+QUOTE+1,記p(x)=xlnx+QUOTE+1,則p′(x)=1+lnx-QUOTE=QUOTE+lnx,當(dāng)x>1時(shí),QUOTE>0,lnx>0,故p′(x)>0,故p(x)遞增;當(dāng)0<x<1時(shí),QUOTE<0,lnx<0,故p′(x)<0,故p(x)遞減,故p(x)≥p(1)=2,故由方程k=xlnx+QUOTE+1有解,得k≥2,所以k的最小值為2.2.(5分)已知函數(shù)f(x)=x3-3x+1,若?x1∈[a,b],?x2∈[a,b],使得fQUOTE=fQUOTE,且x1≠x2,則b-a的最大值為 ()A.2 B.3 C.4 【解析】選C.因?yàn)閒(x)=x3-3x+1,所以f′QUOTE=3x2-3,令f′QUOTE=0,即3x2-3=0,解得x1=-1,x2=1,當(dāng)x<-1時(shí),f′QUOTE>0,所以fQUOTE在QUOTE上單調(diào)遞增;當(dāng)-1≤x≤1時(shí),f′QUOTE<0,所以fQUOTE在QUOTE上單調(diào)遞減;當(dāng)x>1時(shí),f′QUOTE>0,所以fQUOTE在QUOTE上單調(diào)遞增.所以fQUOTE在x=-1處取得極大值,極大值為f(-1)=-1+3+1=3;在x=1處取得微小值,微小值為f(1)=1-3+1=-1.令fQUOTE=3,即x3-3x+1=3,即QUOTE=0,解得x=-1(舍)或x=2.令fQUOTE=-1,即x3-3x+1=-1,即QUOTE=0,解得x=1(舍)或x=-2,所以b-a的最大值為2-QUOTE=4.3.(10分)已知函數(shù)f(x)=xlnx-x,兩相異正實(shí)數(shù)x1,x2滿意f(x1)=f(x2).求證:x1+x2>2.【證明】f′(x)=lnx,當(dāng)x∈(0,1)時(shí),f(x)單調(diào)遞減,當(dāng)x>1時(shí),f(x)單調(diào)遞增,且f(1)=-1,如圖所示,不妨設(shè)x1<1<x2,要證x1+x2>2,即證x2>2-x1,只須要證f(2-x1)<f(x2),又f(x1)=f(x2),所以只需證f(2-x1)<f(x1),設(shè)g(x)=f(x)-f(2-x)(x∈(0,1)),則g′(x)=f′(x)-[f(2-x)]′=lnx+ln(2-x),0<x<1,符號不易推斷,再設(shè)h(x)=lnx+ln(2-x),0<x<1,則h′(x)=QUOTE-QUOTE=QUOTE>0,所以h(x)在(0,1)上單調(diào)遞增,所以h(x)<h(1)=0,所以g(x)在(0,1)上單調(diào)遞減,所以g(x)>g(1)=0,所以f(x)-f(2-x)>0,0<x<1,所以f(x1)>f(2-x1)成立,所以x1+x2>2.4.(10分)(2024·全國Ⅰ卷)已知函數(shù)fQUOTE=QUOTE-x+alnx.(1)探討fQUOTE的單調(diào)性.(2)若fQUOTE存在兩個極值點(diǎn)x1,x2,證明:QUOTE<a-2.【解析】(1)f(x)的定義域?yàn)?0,+∞),f′(x)=-QUOTE-1+QUOTE=-QUOTE.(i)若a≤2,則f′(x)≤0,當(dāng)且僅當(dāng)a=2,x=1時(shí),f′(x)=0,所以f(x)在(0,+∞)上單調(diào)遞減.(ii)若a>2,令f′(x)=0得,x=QUOTE或x=QUOTE.當(dāng)x∈QUOTE∪QUOTE時(shí),f′(x)<0;當(dāng)x∈QUOTE時(shí),f′(x)>0.所以f(x)在QUOTE,QUOTE上單調(diào)遞減,在QUOTE上單調(diào)遞增.(2)由(1)知,f(x)存在兩個極值點(diǎn),當(dāng)且僅當(dāng)a>2.由于f(x)的兩個極值點(diǎn)x1,x2滿意x2-ax+1=0,所以x1x2=1,不妨設(shè)x1<x2,則x2>1.由于QUOTE=-QUOTE-1+aQUOTE=-2+aQUOTE=-2+aQUOTE,所以QUOTE<a-2等價(jià)于QUOTE-x2+2lnx2<0.設(shè)函數(shù)g(x)=QUOTE-x+2lnx,由(1)知,g(x)在(0,+∞)上單調(diào)遞減,又g(1)=0,從而當(dāng)x∈(1,+∞)時(shí),g(x)<0.所以QUOTE-x2+2lnx2<0,即QUOTE<a-2.【加練備選·拔高】設(shè)函數(shù)f(x)=lnx-a(x-1)ex,其中a∈R.(1)若a≤0,探討f(x)的單調(diào)性.(2)若0<a<QUOTE,①證明f(x)恰有兩個零點(diǎn);②設(shè)x0為f(x)的極值點(diǎn),x1為f(x)的零點(diǎn),且x1>x0,證明3x0-x1>2.【解析】(1)由已知,f(x)的定義域?yàn)?0,+∞),且f′(x)=QUOTE-[aex+a(x-1)ex]=QUOTE.因此當(dāng)a≤0時(shí),1-ax2ex>0,從而f′(x)>0,所以f(x)在(0,+∞)內(nèi)單調(diào)遞增.(2)①由(1)知f′(x)=QUOTE.令g(x)=1-ax