（新教材適用）2023-2024學(xué)年高中數(shù)學(xué)第二章導(dǎo)數(shù)及其應(yīng)用6用導(dǎo)數(shù)研究函數(shù)的性質(zhì)61函數(shù)的單調(diào)性課后訓(xùn)練北師大版選擇性

6.1函數(shù)的單調(diào)性課后訓(xùn)練鞏固提升A組1.函數(shù)y=f(x)的導(dǎo)函數(shù)f'(x)的圖象如圖所示,則下面結(jié)論正確的是().(第1題)A.在區(qū)間(2,1)上,函數(shù)f(x)單調(diào)遞增B.在區(qū)間(1,3)上,函數(shù)f(x)單調(diào)遞減C.在區(qū)間(4,5)上,函數(shù)f(x)單調(diào)遞增D.在區(qū)間(2,3)上,函數(shù)f(x)不是單調(diào)函數(shù)解析:由f'(x)的圖象知,在區(qū)間(3,2),(2,4)上,f'(x)<0,函數(shù)f(x)單調(diào)遞減,在區(qū)間(1,2),(4,5)上,f'(x)>0,函數(shù)f(x)單調(diào)遞增,故選C.答案:C2.函數(shù)y=x·ex的單調(diào)遞減區(qū)間為().A.(∞,0) B.(∞,1)C.(0,+∞) D.(1,+∞)解析:y'=exx·ex=ex(1x),由y'<0,得x>1.因此,函數(shù)y=x·ex的單調(diào)遞減區(qū)間為(1,+∞).答案:D3.已知函數(shù)f(x)=(1+x)22ln(1+x),則f(x)的單調(diào)遞減區(qū)間為().A.(2,0) B.(1,0)C.(0,+∞) D.(0,1)解析:函數(shù)f(x)的定義域?yàn)?1,+∞),f'(x)=2x2+4x令f'(x)<0,得1<x<0,所以函數(shù)f(x)的單調(diào)遞減區(qū)間為(1,0).答案:B4.函數(shù)y=13x3+ax2+bx3(a,b∈R)在區(qū)間(∞,1),(3,+∞)上單調(diào)遞增,在區(qū)間(1,3)上單調(diào)遞減,則a+b的值為()A.1 B.2 C.1 D.2解析:∵y'=x2+2ax+b,且原函數(shù)在區(qū)間(∞,1),(3,+∞)上單調(diào)遞增,在區(qū)間(1,3)上單調(diào)遞減,∴1與3是關(guān)于x的方程x2+2ax+b=0的兩根.∴1+3=2a,1×3=b,可得a=2,b=3.∴a+b=1.答案:C5.已知函數(shù)f(x),g(x)滿足當(dāng)x∈R時(shí),f'(x)g(x)+f(x)g'(x)>0,若a>b,則有().A.f(a)g(a)=f(b)g(b)B.f(a)g(a)>f(b)g(b)C.f(a)g(a)<f(b)g(b)D.f(a)g(a)與f(b)g(b)的大小關(guān)系不定解析:由題意知[f(x)g(x)]'>0,從而函數(shù)y=f(x)g(x)在R上是增函數(shù),又a>b,所以f(a)g(a)>f(b)g(b).答案:B6.函數(shù)f(x)=xln(x)的單調(diào)遞減區(qū)間為;單調(diào)遞增區(qū)間為.

解析:函數(shù)f(x)的定義域?yàn)?∞,0),f'(x)=ln(x)+x·1-x·(1)=ln(x)+令f'(x)<0,得x>1e令f'(x)>0,得x<1e故函數(shù)f(x)的單調(diào)遞減區(qū)間為-1e,答案:-7.若函數(shù)f(x)=exax1在區(qū)間(2,3)上為減函數(shù),則a的取值范圍為.

解析:由題意知,f'(x)=exa≤0在(2,3)上恒成立.即a≥ex在x∈(2,3)上恒成立.∵2<x<3,∴e2<ex<e3.只需a≥e3.當(dāng)a=e3時(shí),f'(x)=exe3在x∈(2,3)上,f'(x)<0,即f(x)在(2,3)上為減函數(shù),∴a≥e3.答案:[e3,+∞)8.設(shè)p:函數(shù)f(x)=x3+2x2+mx+1在區(qū)間(∞,+∞)上單調(diào)遞增,q:m≥43,則p是q的條件.(填“充分不必要”“必要不充分”“充要”或“既不充分也不必要”解析:f'(x)=3x2+4x+m.若函數(shù)f(x)在區(qū)間(∞,+∞)上單調(diào)遞增,則f'(x)≥0在區(qū)間(∞,+∞)上恒成立,由Δ≤0,得m≥43,故p?q反之,若m≥43,則f'(x)≥0,且不恒等于0,即f(x)在區(qū)間(∞,+∞)上單調(diào)遞增,故q?p因此,p是q的充要條件.答案:充要9.丹麥數(shù)學(xué)家琴生(Jensen)是19世紀(jì)對(duì)數(shù)學(xué)分析做出卓越貢獻(xiàn)的人,特別是在函數(shù)的凹凸性與不等式方面留下了很多寶貴的成果.設(shè)函數(shù)f(x)在區(qū)間(a,b)上的導(dǎo)函數(shù)為f'(x),f'(x)在區(qū)間(a,b)上的導(dǎo)函數(shù)為f″(x),若在區(qū)間(a,b)上,f″(x)<0恒成立,則稱函數(shù)f(x)在區(qū)間(a,b)上為“凸函數(shù)”.已知f(x)=x44-t3x3+32x2在區(qū)間(1,4)上為“凸函數(shù)”解析:由f(x)=x44-t3x3+32x2可得f'(x)=x3tx2+3x,f″(x)=3因?yàn)閒(x)=x44-t3x3+32x2在區(qū)間(1,4)上為“凸函數(shù)”,所以當(dāng)x∈(1,4)時(shí),f″(x)=3x22tx+3令g(x)=32x+1x.因?yàn)間'(x)=321-1x2>0在區(qū)間(1,4)上恒成立,所以函數(shù)g(x)在區(qū)間(1,4)上單調(diào)遞增,所以答案:5110.求函數(shù)f(x)=sinx2+cos解:∵2+cosx≠0,∴函數(shù)f(x)的定義域?yàn)镽.f'(x)=(2+cos令f'(x)=0,得cosx=12從而x=2kπ±2π3,k∈當(dāng)x∈2kπ-2π3,2kπ+2π3當(dāng)x∈2kπ+2π3,2kπ+4π3故函數(shù)f(x)的單調(diào)遞增區(qū)間為2kπ2π3,2kπ+2π3(k∈Z),單調(diào)遞減區(qū)間為2kπ+2π3,2kπ+4π3(kB組1.下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞增的是().A.y=sinx B.y=xe2C.y=x3x D.y=lnxx解析:顯然函數(shù)y=sinx在區(qū)間(0,+∞)上不單調(diào),故排除A;對(duì)于函數(shù)y=xe2,因?yàn)閑2為大于零的常數(shù),所以不用求導(dǎo)就知函數(shù)y=xe2在區(qū)間(0,+∞)上單調(diào)遞增;對(duì)于C,y'=3x21=3x+故函數(shù)在區(qū)間-∞在區(qū)間-3對(duì)于D,y'=1x1(x>0),故函數(shù)在區(qū)間(1,+∞)上單調(diào)遞減,在區(qū)間(0,1)上單調(diào)遞增.故選B答案:B2.(多選題)若函數(shù)y=exf(x)(e=2.71828…是自然對(duì)數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)y=f(x)具有性質(zhì)M.下列函數(shù)中,不具有性質(zhì)M的是().A.f(x)=2x B.f(x)=x2C.f(x)=3x D.f(x)=cosx解析:設(shè)函數(shù)g(x)=exf(x),對(duì)于A,g(x)=ex·2x=e2x,在定義域R上為增函數(shù),具有性質(zhì)M.對(duì)于B,g(x)=ex·x2,則g'(x)=x(x+2)ex,由g'(x)>0得x<2或x>0,g(x)=ex·x2在定義域R上不是增函數(shù),不具有性質(zhì)M.對(duì)于C,g(x)=ex·3x=e3x在定義域R上是減函數(shù),不具有性質(zhì)M.對(duì)于D,g(x)=excosx,則g'(x)=2excosx+π4,g'(x)>0在定義域R上不恒成立,不具有性質(zhì)M.答案:BCD3.設(shè)函數(shù)f(x),g(x)分別是定義在R上的奇函數(shù)和偶函數(shù),g(x)恒不為0,當(dāng)x<0時(shí),f'(x)g(x)f(x)g'(x)>0,且f(3)=0,則不等式f(x)g(x)<0的解集是().A.(3,0)∪(3,+∞)B.(3,0)∪(0,3)C.(∞,3)∪(3,+∞)D.(∞,3)∪(0,3)解析:令F(x)=f(x)g(x).∵f(x),g(x)分別為定義在RF'(x)=f'(∵當(dāng)x<0時(shí),f'(x)g(x)f(x)g'(x)>0,∴F'(x)>0,∴函數(shù)F(x)在區(qū)間(∞,0)上單調(diào)遞增.又F(3)=f(3)g(3)=∴當(dāng)x<3時(shí),F(x)<0;當(dāng)3<x<0時(shí),F(x)>0.又F(x)為奇函數(shù),∴當(dāng)0<x<3時(shí),F(x)<0;當(dāng)x>3時(shí),F(x)>0.而不等式f(x)g(x)<0和f(x)g(x∴不等式f(x)g(x)<0的解集為(∞,3)∪(0,3).答案:D4.已知函數(shù)y=f(x)對(duì)任意的x∈π2,π2,滿足f'(x)cosx+f(x)sinx>0(其中f'(x)是函數(shù)f(x)的導(dǎo)函數(shù)),則下列不等式成立的是(A.2f-B.2fπC.f(0)>2fπD.f(0)>2解析:因?yàn)楹瘮?shù)y=f(x)對(duì)任意的x∈-π2,π2,滿足f'(x)cosx+f所以f(x)cos所以函數(shù)f(x)cos因?yàn)棣?<π3<所以f-即2f-π故A正確.答案:A5.已知在R上的可導(dǎo)函數(shù)f(x)的圖象如圖所示,則不等式(x22x3)·f'(x)>0的解集為.

(第5題)解析:由函數(shù)f(x)的圖象可知,在區(qū)間(∞,1)上,f'(x)>0;在區(qū)間(1,1)上,f'(x)<0;在區(qū)間(1,+∞)上,f'(x)>0.由(x22x3)f'(x)>0,得x解得x<1或1<x<1或x>3.故不等式的解集為(∞,1)∪(1,1)∪(3,+∞).答案:(∞,1)∪(1,1)∪(3,+∞)6.若函數(shù)y=43x3+bx有三個(gè)單調(diào)區(qū)間,則實(shí)數(shù)b的取值范圍是.解析:若函數(shù)y=43x3+bx有三個(gè)單調(diào)區(qū)間,則關(guān)于x的方程y'=4x2+b=0有兩個(gè)不相等的實(shí)數(shù)根.由Δ>0,得b>0答案:(0,+∞)7.定義在R上的函數(shù)f(x)=ax3+bx2+cx+3同時(shí)滿足以下條件:①f'(1)=0;②f(x)的導(dǎo)函數(shù)是偶函數(shù);③f(x)在x=0處的切線與第一象限、第三象限的平分線垂直.求函數(shù)y=f(x)的解析式和單調(diào)區(qū)間.解:函數(shù)f(x)的導(dǎo)數(shù)f'(x)=3ax2+2bx+c.已知f