2024-2025學(xué)年新教材高中數(shù)學(xué)第六章導(dǎo)數(shù)及其應(yīng)用6.2.2.1函數(shù)的導(dǎo)數(shù)與極值課時素養(yǎng)評價含解析新人教B版選擇性必修第三冊

PAGE課時素養(yǎng)評價十八函數(shù)的導(dǎo)數(shù)與極值(25分鐘·50分)一、選擇題(每小題5分,共20分,多選題全部選對的得5分,選對但不全的得3分,有選錯的得0分)1.如圖是函數(shù)y=f(x)的導(dǎo)函數(shù)y=f′(x)的圖象,則下列說法正確的是()A.x=a是函數(shù)y=f(x)的微小值點B.當(dāng)x=-a或x=b時,函數(shù)f(x)的值為0C.函數(shù)y=f(x)關(guān)于點(0,c)對稱D.函數(shù)y=f(x)在(b,+∞)上單調(diào)遞增【解析】選D.結(jié)合導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系可知,A中,在x=a旁邊,f′(x)<0,故x=a不是微小值點;B中,導(dǎo)數(shù)為0時,函數(shù)值不肯定為0;C中,導(dǎo)函數(shù)的對稱性與原函數(shù)的對稱性沒有關(guān)系;D中,當(dāng)x>b時,f′(x)>0,函數(shù)f(x)單調(diào)遞增.2.(2024·張家界高二檢測)函數(shù)f(x)=(x2-3x+1)ex的極大值是 ()A.-3e B.-e2 C.2e2 D.QUOTE【解析】選D.f(x)=(x2-3x+1)ex,x∈R.f′(x)=(2x-3)ex+(x2-3x+1)ex=(x2-x-2)ex=(x-2)(x+1)ex.令f′(x)=0,解得x=-1,2.令f′(x)>0,解得x>2,或x<-1.令f′(x)<0,解得-1<x<2.所以函數(shù)f(x)在(-∞,-1),(2,+∞)上單調(diào)遞增,在(-1,2)上單調(diào)遞減.所以x=-1時,函數(shù)f(x)取得極大值,f(-1)=QUOTE.3.(多選題)對于函數(shù)f(x)=x3-3x2,給出選項中正確的是 ()A.f(x)是增函數(shù),無極值B.f(x)是減函數(shù),無極值C.f(x)的單調(diào)遞增區(qū)間為(-∞,0),(2,+∞),單調(diào)遞減區(qū)間為(0,2)D.f(0)=0是極大值,f(2)=-4是微小值【解析】選CD.f′(x)=3x2-6x.令f′(x)=3x2-6x>0,得x>2或x<0;令f′(x)=3x2-6x<0,得0<x<2,所以函數(shù)f(x)在區(qū)間(-∞,0)和(2,+∞)上單調(diào)遞增,在區(qū)間(0,2)上單調(diào)遞減.當(dāng)x=0和x=2時,函數(shù)分別取得極大值0和微小值-4.所以CD正確.4.(2024·荊州高二檢測)若函數(shù)f(x)=(x-a)3-3x+b的極大值為M,微小值為N,則M-N ()A.與a有關(guān),且與b有關(guān) B.與a無關(guān),且與b有關(guān)C.與a無關(guān),且與b無關(guān) D.與a有關(guān),且與b無關(guān)【解析】選C.f′(x)=3(x-a)2-3,令f′(x)=0,得x=a-1,或x=a+1,當(dāng)x改變時,f′(x)的改變狀況如下表:x(-∞,a-1)a-1(a-1,a+1)a+1(1+a,+∞)f′(x)+0-0+因此,函數(shù)f(x)在x=a+1處取得微小值,函數(shù)f(x)在x=a-1處取得極大值,M-N=f(a-1)-f(a+1)=-1-3(a-1)+b-1+3(a+1)-b=4.二、填空題(每小題5分,共10分)5.已知函數(shù)f(x)=ax3+bx2+cx,其導(dǎo)函數(shù)y=f′(x)的圖象經(jīng)過點(1,0),(2,0),如圖所示,則下列說法中正確的是________(填寫序號).

①當(dāng)x=QUOTE時函數(shù)取得微小值;②f(x)有兩個極值點;③當(dāng)x=2時函數(shù)取得微小值;④當(dāng)x=1時函數(shù)取得極大值.【解析】從題中圖象上可以看到:當(dāng)x∈(-∞,1)時,f′(x)>0;當(dāng)x∈(1,2)時,f′(x)<0;當(dāng)x∈(2,+∞)時,f′(x)>0,所以f(x)有兩個極值點1和2,且當(dāng)x=2時函數(shù)取得微小值,當(dāng)x=1時函數(shù)取得極大值.只有①不正確.答案:②③④6.f(x)=x(x-c)2在x=2處有極大值,則常數(shù)c的值為________.

【解析】因為x=2是f(x)的極大值點,又f(x)=x(x2-2cx+c2),所以f′(x)=x(2x-2c)+x2-2cx+c2=3x2-4cx+c2.所以f′(2)=c2-8c+12=0.得c=2或c=6.當(dāng)c=2時,不能取極大值,故c=6.答案:6三、解答題(每小題10分,共20分)7.設(shè)a為實數(shù),函數(shù)f(x)=ex-2x+2a,x∈R,求f(x)的單調(diào)區(qū)間與極值.【解析】由f(x)=ex-2x+2a,x∈R知f′(x)=ex-2,x∈R.令f′(x)=0,得x=ln2.于是當(dāng)x改變時,f′(x),f(x)的改變狀況如下表:x(-∞,ln2)ln2(ln2,+∞)f′(x)-0+f(x)↘2(1-ln2+a)↗故f(x)的單調(diào)遞減區(qū)間是(-∞,ln2),單調(diào)遞增區(qū)間是(ln2,+∞);且f(x)在x=ln2處取得微小值.微小值為f(ln2)=eln2-2ln2+2a=2(1-ln2+a),無極大值.8.設(shè)函數(shù)y=x3+ax2+bx+c的圖象如圖所示,且與y=0在原點相切,若函數(shù)的微小值為-4.(1)求a,b,c的值.(2)求函數(shù)的遞減區(qū)間.【解析】(1)因為函數(shù)的圖象經(jīng)過點(0,0),易得c=0.又圖象與x軸相切于點(0,0),且y′=3x2+2ax+b,故0=3×02+2a×0+b,解得b=0.所以y=x3+ax2,則y′=3x2+2ax.令y′=0,解得x=0或x=-QUOTEa,即x=0和x=-QUOTEa是極值點.由圖象知函數(shù)在x=0處取極大值,故在x=-QUOTEa處取微小值.當(dāng)x=-QUOTEa時,函數(shù)有微小值-4,所以QUOTE+aQUOTE=-4,整理得a3=-27,解得a=-3.故a=-3,b=0,c=0.(2)由(1)得y=x3-3x2,則y′=3x2-6x,令y′<0,即3x2-6x<0,解得0<x<2,所以函數(shù)的遞減區(qū)間是(0,2).(15分鐘·30分)1.(5分)如圖是函數(shù)f(x)=x3+bx2+cx+d的大致圖象,則QUOTE+QUOTE等于 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選C.函數(shù)f(x)=x3+bx2+cx+d的圖象過點(0,0),(1,0),(2,0),得d=0,b+c+1=0,4b+2c+8=0,則b=-3,c=2,f′(x)=3x2+2bx+c=3x2-6x+2,且x1,x2是函數(shù)f(x)=x3+bx2+cx+d的兩個極值點,即x1,x2是方程3x2-6x+2=0的實根,所以QUOTE+QUOTE=(x1+x2)2-2x1x2=4-QUOTE=QUOTE.2.(5分)設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f′(x),且函數(shù)f(x)在x=-2處取得微小值,則函數(shù)y=xf′(x)的圖象可能是 ()【解析】選C.由題意可得f′(-2)=0,而且當(dāng)x∈(-∞,-2)時,f′(x)<0,此時xf′(x)>0;當(dāng)x∈(-2,+∞)時,f′(x)>0,此時若x∈(-2,0),xf′(x)<0,若x∈(0,+∞),xf′(x)>0,所以函數(shù)y=xf′(x)的圖象可能是C.3.(5分)(2024·福州高二檢測)設(shè)a<0,若函數(shù)y=ex+2ax,x∈R有小于零的極值點,則實數(shù)a的取值范圍是________.

【解析】因為y=ex+2ax,a<0,所以y′=ex+2a.由題意知ex+2a=0有小于0的實根,令y1=ex,y2=-2a,則兩曲線交點在其次象限,結(jié)合圖象:易得0<-2a<1?-QUOTE<a<0,故實數(shù)a的取值范圍是QUOTE.答案:QUOTE4.(5分)若函數(shù)y=x·2x在x=x0時取微小值,則x0=________.

【解析】令y′=2x+x·2xln2=2x(1+xln2)=0,得x=-QUOTE.所以當(dāng)x>-QUOTE時,y′>0,函數(shù)遞增;當(dāng)x<-QUOTE時,y′<0,函數(shù)遞減.所以當(dāng)x=-QUOTE時函數(shù)取微小值.答案:-QUOTE【加練·固】三次函數(shù)當(dāng)x=1時有極大值4,當(dāng)x=3時,有微小值0,且函數(shù)過原點,則此函數(shù)是 ()A.y=x3+6x2+9x B.y=x3-6x2+9xC.y=x3-6x2-9x D.y=x3+6x2-9x【解析】選B.三次函數(shù)過原點,結(jié)合選項,可設(shè)f(x)=x3+bx2+cx,f′(x)=3x2+2bx+c,由題設(shè)知,f′(1)=3+2b+c=0,f′(3)=27+6b+c=0,解得b=-6,c=9.所以f(x)=x3-6x2+9x;f′(x)=3x2-12x+9=3(x-1)(x-3).當(dāng)x=1時,f(x)極大值=4;當(dāng)x=3時,f(x)微小值=0,滿意條件.5.(10分)已知函數(shù)f(x)=x2-2lnx,h(x)=x2-x+a.(1)求函數(shù)f(x)的極值.(2)設(shè)函數(shù)k(x)=f(x)-h(x),若函數(shù)k(x)在[1,3]上恰有兩個不同零點,求實數(shù)a的取值范圍.【解析】(1)f(x)的定義域是(0,+∞).令f′(x)=2x-QUOTE=0,得x=1.當(dāng)x∈(0,1)時,f′(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時,f′(x)>0,f(x)單調(diào)遞增;所以f(x)在x=1處取得微小值,又f(1)=1,所以f(x)的微小值為1,無極大值.(2)k(x)=f(x)-h(x)=x-2lnx-a(x>0),所以k′(x)=1-QUOTE,令k′(x)>0,得x>2,令k′(x)<0,得0<x<2,所以k(x)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增.要使函數(shù)k(x)在[1,3]上恰有兩個不同零點,則需QUOTE解得2-2ln2<a≤3-2ln3.1.若函數(shù)f(x)=alnx-ex有極值點,則實數(shù)a的取值范圍是 ()A.(-e,+∞) B.(1,e)C.(1,+∞) D.(0,+∞)【解析】選D.因為函數(shù)f(x)=alnx-ex,x∈(0,+∞),所以f′(x)=QUOTE-ex,①當(dāng)a≤0時,f′(x)<0,函數(shù)f(x)在(0,+∞)上單調(diào)遞減,無極值點,②當(dāng)a>0時,依據(jù)y=QUOTE與y=ex的圖象,如圖所示,設(shè)兩個函數(shù)在第一象限的交點的橫坐標(biāo)為x0,當(dāng)x∈(0,x0)時,QUOTE>ex,f′(x)>0,函數(shù)f(x)在區(qū)間(0,x0)上單調(diào)遞增;當(dāng)x∈(x0,+∞)時,QUOTE<ex,f′(x)<0,所以函數(shù)f(x)在(x0,+∞)上單調(diào)遞減,所以當(dāng)a>0時,函數(shù)f(x)有一個極大值點.2.設(shè)f(x)=xlnx-ax2+(2a-1)x,a∈R.(1)令g(x)=f′(x),求g(x)的單調(diào)區(qū)間.(2)已知f(x)在x=1處取得極大值,求實數(shù)a的取值范圍.【解析】(1)由f′(x)=lnx-2ax+2a,可得g(x)=lnx-2ax+2a,x∈(0,+∞),則g′(x)=QUOTE-2a=QUOTE,當(dāng)a≤0,x∈(0,+∞)時,g′(x)>0,函數(shù)g(x)單調(diào)遞增;當(dāng)a>0,x∈QUOTE時,g′(x)>0,函數(shù)g(x)單調(diào)遞增,x∈QUOTE時,g′(x)<0,函數(shù)g(x)單調(diào)遞減.所以當(dāng)a≤0時,g(x)的單調(diào)遞增區(qū)間為(0,+∞);當(dāng)a>0時,g(x)的單調(diào)遞增區(qū)間為QUOTE,單調(diào)遞減區(qū)間為QUOTE.(2)由(1)知,f′(1)=0.①當(dāng)a≤0時,f′(x)單調(diào)遞增.所以當(dāng)x∈(0,1)時,f′(x)<0,f(x)單調(diào)遞減.當(dāng)x∈(1,+∞)時,f′(x)>0,f(x)單調(diào)遞增.所以f(x)在x=1處取得微小值,不合題意.②當(dāng)0<a<QUOTE時,QUOTE>1,由(1)知f′(x)在QUOTE內(nèi)單調(diào)遞增,可得當(dāng)x∈(0,1)時,f′(x)<0,x∈QUOTE時,f′(x)>0,所以f(x)在(0,1)內(nèi)單調(diào)遞減,