2024-2025學(xué)年高中數(shù)學(xué)第一章導(dǎo)數(shù)及其應(yīng)用1.3.1函數(shù)的單調(diào)性與導(dǎo)數(shù)課時素養(yǎng)評價含解析新人教A版選修2-2

PAGE課時素養(yǎng)評價五函數(shù)的單調(diào)性與導(dǎo)數(shù)(15分鐘30分)1.若在區(qū)間(a,b)內(nèi),f′(x)>0,且f(a)≥0,則在(a,b)內(nèi)有 ()A.f(x)>0 B.f(x)<0 C.f(x)=0 D.不能確定【解析】選A.因為f(x)在(a,b)上為增函數(shù),所以f(x)>f(a)≥0.2.函數(shù)y=fQUOTE的圖象如圖所示,則導(dǎo)函數(shù)y=f′QUOTE的圖象的大致形態(tài)是()【解析】選D.函數(shù)y=fQUOTE的單調(diào)性是先減,再增,最終變?yōu)槌?shù)函數(shù),那么,導(dǎo)函數(shù)y=f′QUOTE的符號為:先負(fù),后正,最終變?yōu)?,選項D符合題意.3.函數(shù)f(x)=lnx-x的單調(diào)遞減區(qū)間為 ()A.QUOTE,QUOTE B.QUOTE∪QUOTEC.QUOTE D.QUOTE【解析】選C.由題可得f′(x)=QUOTE-1=QUOTE,令f′(x)<0,即QUOTE<0,解得x>1或x<0,又因為x>0,故x>1.4.函數(shù)f(x)=ax3-x2+x-5在(-∞,+∞)上單調(diào)遞增,則實數(shù)a的取值范圍是________.

【解析】f′(x)=3ax2-2x+1.由題意知3ax2-2x+1≥0在(-∞,+∞)上恒成立,所以QUOTE解得a≥QUOTE.答案:QUOTE5.推斷函數(shù)f(x)=QUOTE-1在(0,e)及(e,+∞)上的單調(diào)性.【解析】f′(x)=QUOTE=QUOTE.當(dāng)x∈(0,e)時,lnx<lne=1,1-lnx>0,x2>0,所以f′(x)>0,f(x)為增函數(shù).當(dāng)x∈(e,+∞)時,lnx>lne=1,1-lnx<0,x2>0,所以f′(x)<0,f(x)為減函數(shù).(30分鐘60分)一、選擇題(每小題5分,共25分)1.若f(x)=ax3+bx2+cx+d(a>0)為增函數(shù),則 ()A.b2-4ac>0 B.b>0,c>0C.b=0,c>0 D.b2-3ac≤0【解析】選D.由f(x)為增函數(shù),知f′(x)=3ax2+2bx+c≥0.所以Δ=4b2-12ac≤0.即b2-3ac≤0.2.(2024·洛陽高二檢測)函數(shù)fQUOTE在其定義域內(nèi)可導(dǎo),其圖象如圖所示,則導(dǎo)函數(shù)y=f′QUOTE的圖象可能為 ()【解析】選C.由函數(shù)fQUOTE的圖象可知,函數(shù)fQUOTE在自變量漸漸增大的過程中,函數(shù)先遞增,然后遞減,再遞增,當(dāng)x>0時,函數(shù)fQUOTE單調(diào)遞增,所以導(dǎo)數(shù)f′QUOTE的符號是正,負(fù),正,正,只有選項C符合題意.3.函數(shù)fQUOTE=sin2x+2cosxQUOTE,則fQUOTE ()A.在QUOTE上單調(diào)遞增 B.在QUOTE上單調(diào)遞減C.在QUOTE上單調(diào)遞減 D.在QUOTE上單調(diào)遞增【解析】選C.因為f(x)=sin2x+2cosx=2sinxcosx+2cosx=2cosx(sinx+1).所以f′(x)=2[(cosx)′(sinx+1)+cosx(sinx+1)′]=2[-sinx(sinx+1)+cos2x]=-2QUOTE,令f′(x)>0,即QUOTE<0,故-1<sinx<QUOTE(0≤x≤π)?x∈QUOTE∪QUOTE,故fQUOTE在QUOTE和QUOTE上單調(diào)遞增,可得在QUOTE上單調(diào)遞減.4.函數(shù)f(x)的導(dǎo)函數(shù)f′(x)滿意f′(x)>f(x)在R上恒成立,且f(1)=e,則下列推斷肯定正確的是 ()A.f(0)<1 B.f(-1)<f(0)C.f(0)>0 D.f(-1)>0【解析】選A.令函數(shù)F(x)=QUOTE,則F′(x)=QUOTE,因為f′(x)>f(x),所以F′(x)>0,故函數(shù)F(x)是定義在R上的增函數(shù),所以F(1)>F(0),即QUOTE>QUOTE,又f(1)=e,故有f(0)<1.【補償訓(xùn)練】已知函數(shù)f(x)=x2-cosx,則fQUOTE,f(0),fQUOTE的大小關(guān)系是 ()A.f(0)<fQUOTE<fQUOTEB.f(0)<fQUOTE<fQUOTEC.fQUOTE<fQUOTE<f(0)D.fQUOTE<f(0)<fQUOTE【解析】選B.因為函數(shù)f(-x)=(-x)2-cos(-x)=x2-cosx=f(x),所以f(x)為偶函數(shù),所以fQUOTE=fQUOTE,f′(x)=2x+sinx,當(dāng)0<x<QUOTE時,f′(x)=2x+sinx>0,所以函數(shù)在QUOTE上遞增,所以f(0)<fQUOTE<fQUOTE,即f(0)<fQUOTE<fQUOTE.5.已知函數(shù)f(x)=x2-alnx+1在(1,2)內(nèi)不是單調(diào)函數(shù),則實數(shù)a的取值范圍是 ()A.QUOTE B.QUOTEC.QUOTE∪QUOTE D.QUOTE【解析】選A.因為f′(x)=2x-QUOTE,fQUOTE=x2-alnx+1在QUOTE內(nèi)不是單調(diào)函數(shù),故2x-QUOTE=0在QUOTE內(nèi)有解,即a=2x2在QUOTE內(nèi)有解,所以2<a<8.二、填空題(每小題5分,共15分)6.設(shè)函數(shù)f(x)=x(ex-1)-QUOTEx2,則f(x)的單調(diào)遞增區(qū)間是________,單調(diào)遞減區(qū)間是________.

【解析】f′(x)=ex-1+xex-x=(ex-1)(x+1).當(dāng)x∈(-∞,-1)時,f′(x)>0;當(dāng)x∈(-1,0)時,f′(x)<0;當(dāng)x∈(0,+∞)時,f′(x)>0.故f(x)在(-∞,-1),(0,+∞)上單調(diào)遞增,在(-1,0)上單調(diào)遞減.答案:(-∞,-1)和(0,+∞)(-1,0)7.若函數(shù)fQUOTE=x3-ax2+3x+1在區(qū)間QUOTE上單調(diào)遞減,則實數(shù)a的取值范圍為________.

【解析】由題意得:f′(x)=3x2-2ax+3,因為f(x)在區(qū)間QUOTE上單調(diào)遞減,所以f′(x)≤0在區(qū)間QUOTE上恒成立,所以QUOTE?QUOTE解得a≥QUOTE.答案:QUOTE8.若函數(shù)f(x)=x-QUOTEsin2x+asinx在(-∞,+∞)上單調(diào)遞增,則a的取值范圍是________.

【解析】f′(x)=1-QUOTEcos2x+acosx≥0對x∈R恒成立,故1-QUOTE(2cos2x-1)+acosx≥0,即acosx-QUOTEcos2x+QUOTE≥0恒成立,設(shè)t=cosx,t∈[-1,1],則-QUOTEt2+at+QUOTE≥0對t∈[-1,1]恒成立,構(gòu)造f(t)=-QUOTEt2+at+QUOTE,開口向下的二次函數(shù)f(t)的最小值的可能值為端點值,故只需保證QUOTE解得-QUOTE≤a≤QUOTE.答案:QUOTE三、解答題(每小題10分,共20分)9.已知函數(shù)f(x)=x2-(a+2)x+alnx,試探討f(x)的單調(diào)性.【解析】f′(x)=2x-(a+2)+QUOTE=QUOTE,x>0.當(dāng)a≤0時,易知f(x)在(0,1)上為減函數(shù),在[1,+∞)上為增函數(shù);當(dāng)0<a<2時,f(x)在QUOTE上為增函數(shù),在QUOTE上為減函數(shù),在(1,+∞)上為增函數(shù);當(dāng)a=2時,f(x)在(0,+∞)上為增函數(shù);當(dāng)a>2時,f(x)在(0,1)上為增函數(shù),在QUOTE上為減函數(shù),在QUOTE上為增函數(shù).10.已知函數(shù)f(x)=x3+ax2-1(x∈R).(1)當(dāng)a=1時,求函數(shù)f(x)在點(1,1)處的切線方程.(2)若函數(shù)f(x)在x∈(1,2)上單調(diào)遞減,求實數(shù)a的取值范圍.【解析】(1)因為f′(x)=3x2+2ax,所以當(dāng)a=1時,f(x)=x3+x2-1,點(1,1)在f(x)上,f′(1)=3+2=5,所以y-1=5(x-1),即5x-y-4=0,所以函數(shù)f(x)在點(1,1)處的切線方程為5x-y-4=0.(2)因為函數(shù)f(x)在x∈(1,2)上單調(diào)遞減,所以f′(x)=3x2+2ax≤0對x∈(1,2)恒成立,所以a≤-QUOTEx,因為-3<-QUOTEx<-QUOTE,所以a≤-3,所以實數(shù)a的取值范圍為(-∞,-3].已知函數(shù)f(x)=(x-2)ex+a(x-1)2,探討f(x)的單調(diào)性.【解析】f′(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(1)設(shè)a≥0,則當(dāng)x∈(-∞,1)時,f′(x)<0;當(dāng)x∈(1,+∞)時,f′(x)>0.所以f(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.(2)設(shè)a<0,由f′(x)=0得x=1或x=ln(-2a).①若a=-QUOTE,則f′(x)=(x-1)(ex-e),所以f(x)在(-∞,+∞)單調(diào)遞增.②若a>-QUOTE,則ln(-2a)<1,故當(dāng)x∈(-∞,ln(-2a))∪(1,+∞)時,f′(x