年高考數(shù)學(xué)（文）課時(shí)作業(yè)（十四）第1課時(shí)導(dǎo)數(shù)與函數(shù)的單調(diào)性

課時(shí)作業(yè)(十四)第1課時(shí)導(dǎo)數(shù)與函數(shù)的單調(diào)性時(shí)間/45分鐘分值/100分基礎(chǔ)熱身1.函數(shù)f(x)=2xlnx的單調(diào)遞減區(qū)間為 ()A.0,1B.(0,+∞)C.(1,+∞) D.(∞,0)∪(1,+∞)2.函數(shù)f(x)的導(dǎo)函數(shù)f'(x)有下列信息:①f'(x)>0時(shí),1<x<2;②f'(x)<0時(shí),x<1或x>2;③f'(x)=0時(shí),x=1或x=2.則函數(shù)f(x)的大致圖像是 ()ABCD圖K1413.函數(shù)f(x)=x2+ax在(2,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為 (A.(∞,8) B.(∞,8]C.(∞,16) D.(∞,16]4.函數(shù)f(x)=x315x233x+6的單調(diào)遞減區(qū)間為.

5.函數(shù)f(x)=1+12x+cosx在0,π2能力提升6.[2017·湖北襄陽(yáng)聯(lián)考]已知函數(shù)f(x)=xln|x|,則f(x)的圖像大致為 ()ABCD圖K1427.[2017·長(zhǎng)沙長(zhǎng)郡中學(xué)月考]若函數(shù)f(x)=x+bx(b∈R)的導(dǎo)函數(shù)在區(qū)間(1,2)上有零點(diǎn),則f(x)在下列區(qū)間上單調(diào)遞增的是 (A.(∞,1] B.(1,0)C.(0,1) D.(2,+∞)8.[2017·天津耀華中學(xué)月考]已知函數(shù)f(x)=x+blnx在區(qū)間(0,2)上不是單調(diào)函數(shù),則b的取值范圍是 ()A.(∞,0) B.(∞,2)C.(2,0) D.(2,+∞)9.[2017·山東實(shí)驗(yàn)中學(xué)一診]已知函數(shù)f(x)=lnx+(x-b)2x(b∈R),若存在x∈12,2,使得f(x)>x·f'A.(∞,2) B.-∞,C.-∞,9D.(∞,3)10.函數(shù)f(x)=(x2)ex的單調(diào)遞減區(qū)間是.

11.若函數(shù)f(x)=x3ax2x+6在(0,1)內(nèi)單調(diào)遞減,則實(shí)數(shù)a的取值范圍是.

12.若函數(shù)f(x)=x2exax在R上存在單調(diào)遞增區(qū)間,則實(shí)數(shù)a的取值范圍是.

13.(15分)已知函數(shù)f(x)=x4+axlnx32,其中a∈R,且曲線y=f(x)在點(diǎn)(1,f(1))處的切線垂直于直線(1)求a的值;(2)求函數(shù)f(x)的單調(diào)區(qū)間.14.(15分)[2017·佛山一模]已知函數(shù)f(x)=eax+λlnx,其中a<0,e是自然對(duì)數(shù)的底數(shù).若f(x)在(0,+∞)上是單調(diào)函數(shù),求λ的取值范圍.難點(diǎn)突破15.(5分)已知定義在R上的函數(shù)y=f(x)滿足函數(shù)y=f(x1)的圖像關(guān)于直線x=1對(duì)稱(chēng),且當(dāng)x∈(∞,0)時(shí),f(x)+xf'(x)<0(f'(x)是函數(shù)f(x)的導(dǎo)函數(shù))成立.若a=sin12·fsin12,b=ln2·f(ln2),c=log1214·flog1214A.a>b>c B.b>a>cC.c>a>b D.a>c>b16.(5分)已知函數(shù)f(x)(x∈R)滿足f(1)=1,且f(x)的導(dǎo)函數(shù)f'(x)<12,則不等式f(x2)<x22+1課時(shí)作業(yè)(十四)第1課時(shí)導(dǎo)數(shù)與函數(shù)的單調(diào)性1.A[解析]函數(shù)f(x)的定義域是(0,+∞),且f'(x)=21x=2x-1x,令f'(x)<0,得2.C[解析]根據(jù)題意知,函數(shù)f(x)在(1,2)上是增函數(shù),在(∞,1),(2,+∞)上是減函數(shù).故選C.3.D[解析]由f(x)=x2+ax,得f'(x)=2xax2=2x3-ax2.因?yàn)閒(x)在(2,+∞)上單調(diào)遞增,所以2x3-ax2≥0,即a≤2x3,在(2,+∞)上恒成立,因?yàn)?x3>4.(1,11)[解析]由f(x)=x315x233x+6得f'(x)=3x230x33.令f'(x)<0,即3(x11)(x+1)<0,解得1<x<11,所以函數(shù)f(x)的單調(diào)遞減區(qū)間為(1,11).5.0,π6[解析]f'(x)=12sinx,令f'(x)>0,得sinx<12,因?yàn)閤∈0,π26.A[解析]當(dāng)x<0時(shí),f(x)=xln(x),f'(x)=11x=x-1x,函數(shù)f(x)在(∞,0)上為增函數(shù),所以排除選項(xiàng)B,C,D.當(dāng)x>0時(shí),f(x)=xlnx,f'(x)=11x=x-1x,當(dāng)0<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí),f'(x)>0,所以f(x)在(0,1)上單調(diào)遞減,在(1,+∞7.D[解析]f'(x)=1bx2=x2-bx2,令f'(x)=0得x2=b,則x=±b.因?yàn)楹瘮?shù)f(x)=x+bx(b∈R)的導(dǎo)函數(shù)在區(qū)間(1,2)上有零點(diǎn),所以1<b<2.由f'(x)>0,得x<b或x>b,則函數(shù)f(x)的單調(diào)遞增區(qū)間為(∞,b),(b,+∞),結(jié)合選項(xiàng)知,函數(shù)f(x)8.C[解析]f'(x)=1+bx=x+bx.設(shè)g(x)=x+b(x>0),則g(x)是增函數(shù),故需滿足g(0)=b<0,g(2)=b+2>0,即b<0,b>2,所以b∈(29.C[解析]由f(x)>x·f'(x),得f(x)+x·f'(x)>0,即[xf(x)]'>0,所以1x+2(xb)>0,即b<12x+x,當(dāng)x∈12,2時(shí)有解,易知當(dāng)x∈12,2時(shí),12x+x的最大值為14+210.(∞,1)[解析]因?yàn)閒(x)=(x2)ex,所以f'(x)=(x1)ex.令f'(x)<0,得x<1,所以f(x)的單調(diào)遞減區(qū)間為(∞,1).11.a≥1[解析]因?yàn)閒'(x)=3x22ax1,f(x)在(0,1)內(nèi)單調(diào)遞減,所以不等式3x22ax1<0在(0,1)內(nèi)恒成立,所以f'(0)≤0且f'(1)≤0,解得a≥1.12.(∞,2ln22)[解析]因?yàn)閒(x)=x2exax,所以f'(x)=2xexa.因?yàn)楹瘮?shù)f(x)=x2exax在R上存在單調(diào)遞增區(qū)間,所以f'(x)=2xexa>0有解,即a<2xex有解.設(shè)g(x)=2xex,則g'(x)=2ex,令g'(x)=0,解得x=ln2,則當(dāng)x<ln2時(shí),g'(x)>0,g(x)單調(diào)遞增,當(dāng)x>ln2時(shí),g'(x)<0,g(x)單調(diào)遞減,所以當(dāng)x=ln2時(shí),g(x)取得最大值,且g(x)max=g(ln2)=2ln22,所以a<2ln22.13.解:(1)f'(x)=14ax21x(由曲線y=f(x)在點(diǎn)(1,f(1))處的切線垂直于直線y=12x,f'(1)=34a=2,解得a=5(2)由(1)知f(x)=x4+54xln則f'(x)=x2令f'(x)=0,解得x=1或x=5.因?yàn)閒(x)的定義域?yàn)?0,+∞),所以當(dāng)x∈(0,5)時(shí),f'(x)<0,故f(x)在(0,5)上單調(diào)遞減;當(dāng)x∈(5,+∞)時(shí),f'(x)>0,故f(x)在(5,+∞)上單調(diào)遞增.14.解:f'(x)=aeax+λx=axeax+λ①若λ≤0,則f'(x)<0,則f(x)在(0,+∞)上單調(diào)遞減.②若λ>0,令g(x)=axeax+λ,其中a<0,x>0,則g'(x)=aeax(1+ax).令g'(x)=0,得x=1a當(dāng)x∈0,-1a時(shí),g'(x)<0,g(x)單調(diào)遞減,當(dāng)x∈-1a,+∞時(shí),g'(x)>0,g故當(dāng)x=1a時(shí),g(x)取極小值也是最小值,函數(shù)g(x)的最小值為g-1a因此當(dāng)λ1e≥0,即λ≥1e時(shí),g(x)≥0,此時(shí)f'(x)≥0,f(x)在(0,+∞)綜上所述,所求λ的取值范圍是(∞,0]∪1e15.A[解析]因?yàn)楹瘮?shù)y=f(x1)的圖像關(guān)于直線x=1對(duì)稱(chēng),所以y=f(x)的圖像關(guān)于y軸對(duì)稱(chēng),所以函數(shù)y=xf(x)為奇函數(shù).因?yàn)閇xf(x)]'=f(x)+xf'(x),所以當(dāng)x∈(∞,0)時(shí),[xf(x)]'=f(x)+xf'(x)<0,函數(shù)y=xf(x)單調(diào)遞減,當(dāng)x∈(0,+∞)時(shí),函數(shù)y=xf(x)單調(diào)遞減.因?yàn)?<sin12<12,1>ln2>lne=12,log1214=2,所以0<sin12<ln1