考點規(guī)范練13導數(shù)與函數(shù)的單調(diào)性基礎(chǔ)鞏固組1.函數(shù)f(x)=(x3)ex的單調(diào)遞增區(qū)間是()A.(∞,2) B.(0,3)C.(1,4) D.(2,+∞)2.(2017浙江嘉興調(diào)研)已知函數(shù)f(x)=12x3+ax+4,則“a>0”是“f(x)在R上單調(diào)遞增”的(A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件3.設(shè)f'(x)是函數(shù)f(x)的導函數(shù),y=f'(x)的圖象如圖所示,則y=f(x)的圖象最有可能是()4.設(shè)函數(shù)f(x)=12x29lnx在區(qū)間[a1,a+1]上單調(diào)遞減,則實數(shù)a的取值范圍是(A.1<a≤2B.aB≥4C.a≤2 D.0<a≤35.函數(shù)f(x)在定義域R內(nèi)可導,若f(x)=f(2x),且當x∈(∞,1)時,(x1)f'(x)<0,設(shè)a=f(0),b=f12,c=f(3),則(A.a<b<cB.cB<b<aC.c<a<bD.bD<c<a6.函數(shù)f(x)=lnxx的單調(diào)遞增區(qū)間是7.(2017浙江麗水模擬)已知函數(shù)f(x)=lnx+2x,若f(x2+2)<f(3x),則實數(shù)x的取值范圍是.

8.(2017安徽合肥模擬)若函數(shù)f(x)=13x3+12x2+2ax在23,+∞上存在單調(diào)遞增區(qū)間,能力提升組9.(2017浙江湖州測試)已知f(x)是定義在R上的減函數(shù),其導函數(shù)f'(x)滿足f(x)f'(x)A.對于任意x∈R,f(x)<0B.對于任意x∈R,f(x)>0C.當且僅當x∈(∞,1),f(x)<0D.當且僅當x∈(1,+∞),f(x)>010.(2017浙江舟山模擬改編)若f(x)=12x2+bln(x+2)在[1,+∞)上是減函數(shù),則實數(shù)b的取值范圍是(A.[1,+∞) B.[1,+∞)C.(∞,1] D.(∞,1]11.(2017河北承德調(diào)考)已知f(x)是可導的函數(shù),且f'(x)<f(x)對于x∈R恒成立,則()A.f(1)<ef(0),f(2017)>e2017f(0)B.f(1)>ef(0),f(2017)>e2017f(0)C.f(1)>ef(0),f(2017)<e2017f(0)D.f(1)<ef(0),f(2017)<e2017f(0)12.(2017安徽馬鞍山二模)已知函數(shù)f(x)=x2lnx+1,g(x)=kx,若存在x0使得f(x0)=g(x0),則k的取值范圍是()A.(∞,1] B.[1,+∞)C.(∞,e] D.[e,+∞)13.(2017浙江杭州高級中學模擬)已知方程ln|x|ax2+32=0有4個不同的實數(shù)根,則實數(shù)a的取值范圍是()A.0,e2C.0,e214.(2017山東高考)若函數(shù)exf(x)(e=2.71828…是自然對數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)f(x)具有M性質(zhì).下列函數(shù)中所有具有M性質(zhì)的函數(shù)的序號為.

①f(x)=2x②f(x)=3x③f(x)=x3④f(x)=x2+215.(2017浙江名校協(xié)作體上學期聯(lián)考)已知函數(shù)f(x)=x2ex,若f(x)在t,t+1上不單調(diào),則實數(shù)t的取值范圍是16.(2017江蘇高考)已知函數(shù)f(x)=x32x+ex1ex,其中e是自然對數(shù)的底數(shù).若f(a1)+f(2a2)≤0,則實數(shù)a的取值范圍是17.(2017浙江麗水檢測)設(shè)f(x)=ex1+ax2(1)當a=43時,求f(x)單調(diào)區(qū)間(2)若f(x)為R上的單調(diào)函數(shù),求實數(shù)a的取值范圍.18.(2017浙江湖州調(diào)研)設(shè)函數(shù)f(x)=alnx+x-1x+1,(1)若a=0,求曲線y=f(x)在點(1,f(1))處的切線方程;(2)討論函數(shù)f(x)的單調(diào)性.答案：1.D因為f(x)=(x3)ex,則f'(x)=ex(x2),令f'(x)>0,得x>2,所以f(x)的單調(diào)遞增區(qū)間為(2,+∞).2.Af'(x)=32x2+a,當a≥0時,f'(x)≥0恒成立,故“a>0”是“f(x)在R上單調(diào)遞增”的充分不必要條件3.C由y=f'(x)的圖象易知當x<0或x>2時,f'(x)>0,故函數(shù)y=f(x)在區(qū)間(∞,0)和(2,+∞)上單調(diào)遞增;當0<x<2時,f'(x)<0,故函數(shù)y=f(x)在區(qū)間(0,2)上單調(diào)遞減.4.A∵f(x)=12x29lnx,∴f'(x)=x9x(當x9x≤0,即0<x≤3時,函數(shù)f(x)∴a1>0,且a+1≤3,解得1<a≤2.5.C因為當x∈(∞,1)時,(x1)f'(x)<0,所以f'(x)>0,所以函數(shù)f(x)在(∞,1)上是單調(diào)遞增函數(shù).所以a=f(0)<f12=b.又f(x)=f(2x),所以c=f(3)=f(1),所以c=f(1)<f(0)=a,所以c<a<b,故選C6.(0,e)由f'(x)=lnxx'=1-lnxx2>0(x>0),7.(1,2)由題可得函數(shù)定義域為(0,+∞),f'(x)=1x+2xln2,所以在定義域內(nèi)f'(x)>0,函數(shù)單調(diào)遞增,所以由f(x2+2)<f(3x)得x2+2<3x,所以1<x<28.-19,+∞對得f'(x)=x2+x+2a=x-122當x∈23,+∞時,f'(x)的最大值為f'令29+2a>0,解得a>所以實數(shù)a的取值范圍是-9.B∵f(x)f'(x)+x<1,f(x)是定義在R∴f(x)+f'(x)x>f'(x),∴f(x)+f'(x)(x1)>0,∴[(x1)f(x)]'>0,∴函數(shù)y=(x1)f(x)在R上單調(diào)遞增,而x=1時,y=0,則x<1時,y<0,當x∈(1,+∞)時,x1>0,故f(x)>0,又f(x)是定義在R上的減函數(shù),∴x≤1時,f(x)>0也成立,∴f(x)>0對任意x∈R成立,故選B.10.C由已知得f'(x)=x+bx+2≤0在[1,+∞)上恒成立,∴b≤(x+1)21在[1,+∞)上恒成立,∴b11.D令g(x)=f(x)ex,則g'(x)=f(x)ex'=f'(x)ex-f(x)(ex)'e2x=f'(x)-f(x)ex<0,所以函數(shù)g(x)=f12.B函數(shù)f(x)=x2lnx+1,g(x)=kx,若存在x0使得f(x0)=g(x0),等價于方程x2lnx+1=kx有正根,即方程k=xlnx+1x有正根,設(shè)h(x)=xlnx+1x,可得h'(x)=lnx+11x2,當x>1時,h'(x)>0,h(x)在(1,+∞)上遞增,當0<x<1時,h'(x)<0,h(x)在(0,1)上遞減,所以h(x)在(0,+∞)上有最小值h(1)=1,k的取值范圍是[1,+∞13.A由ln|x|ax2+32=0得ax2=ln|x|+3∵x≠0,∴方程等價于a=ln|x|+32x2,設(shè)f(x)=ln|x|+32x2,則函數(shù)f(則f'(x)=1x由f'(x)>0得2x(1+lnx)>0,得1+lnx<0,即lnx<1,得0<x<1e,此時函數(shù)單調(diào)遞增由f'(x)<0得2x(1+lnx)<0,得1+lnx>0,即lnx>1,得x>1e,此時函數(shù)單調(diào)遞減即當x>0時,x=1e時,函數(shù)f(x)取得極大值f1e=ln1e+321e2=-1+32e2=12e2,作出函數(shù)f(x)的圖象如圖14.①④①exf(x)=ex·2x=e2x在R上單調(diào)遞增,故f(x)=2x具有M性質(zhì);②exf(x)=ex·3x=e3x在R上單調(diào)遞減,故f(x)=3x不具有M性質(zhì);③exf(x)=ex·x3,令g(x)=ex·x3,則g'(x)=ex·x3+ex·3x2=x2e∴當x>3時,g'(x)>0,當x<3時,g(x)<0,∴exf(x)=ex·x3在(∞,3)上單調(diào)遞減,在(3,+∞)上單調(diào)遞增,故f(x)=x3不具有M性質(zhì);④exf(x)=ex(x2+2),令g(x)=ex(x2+2),則g'(x)=ex(x2+2)+ex·2x=ex(x+1∴exf(x)=ex(x2+2)在R上單調(diào)遞增,故f(x)=x2+2具有M性質(zhì).15.(3,2)∪(1,0)由題意,得f'(x)=ex(x2+2x),∴f(x)在(∞,2),(0,+∞)上單調(diào)遞增,(2,0)上單調(diào)遞減,又f(x)在[t,t+1]上不單調(diào),∴t<-2,t+1>-2或t<0,t+1>016.-1,12因為f(x)=x3+2x+1exex=f(x),所以函數(shù)f(x)是奇函數(shù),因為f'(x)=3x22+ex+ex≥3x22+2ex·e-x≥0,所以f(x)在R上單調(diào)遞增,又f(a1)+f(2a2)≤0,即f(2a2)≤f(1a),所以2a2≤1a,即2a2+a17.解對f(x)求導得f'(x)=ex·(1)當a=43時,若f'(x)=0,則4x28x+3=解得x1=32,x2=12.結(jié)合x-1133f'(x)+00+f(x)↗極大值↘極小值↗所以增區(qū)間為-∞,1(2)若f(x)為R上的單調(diào)函數(shù),則f'(x)在R上不變號,結(jié)合①與條件a>0,知ax22ax+1≥0在R上恒成立,即Δ=4a24a=4a(a1)≤0,又a>0,得0<a≤1.所以實數(shù)a的取值范圍為{a|0<a≤1}.18.解(1)由題意知a=0時,f(x)=x-1x+1,x∈(0,此時f'(x)=2(x+1)2.可得f'(1)=12,又f(1)=0,所以曲線y=f(x)在(1,f(1))處的切線方程為(2)函數(shù)f(x)的定義域為(0,+∞).f'(x)=a當a≥0時,f'(x)>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增.當a<0時,令g(x)=ax2+(2a+2)x+a,由于Δ=(2a+2)24a2=4(2a+1).①當a=12時,Δ=0,f'(x)=-12(x-1)2x(x+1②當a<12時,Δ<0,g(x)<f'(x)<0,函數(shù)f(x)在(0,+∞)上單調(diào)遞減.③當12<a<0時,Δ>0設(shè)x1,x2(x1<x2)是函數(shù)g(x)的兩