PAGEPAGE1第3講導(dǎo)數(shù)的簡潔應(yīng)用1.定積分01(2x+exA.e+2 B.e+1 C.e D.e-12.曲線y=ex在點(diǎn)(2,e2)處的切線與坐標(biāo)軸所圍成的三角形的面積為()A.94e2 B.2e2 C.e2 D.3.若函數(shù)f(x)=ax3+ln(x+2)在點(diǎn)(-1,-a)處取得極值,則a=()A.-1 B.1 C.-13 D.4.設(shè)函數(shù)f(x)=2(x2-x)lnx-x2+2x,則函數(shù)f(x)的單調(diào)遞減區(qū)間為()A.0,12 B.5.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)y=(1-x)f'(x)的圖象如圖所示,則下列結(jié)論肯定成立的是()A.函數(shù)f(x)有極大值f(2)和微小值f(1)B.函數(shù)f(x)有極大值f(-2)和微小值f(1)C.函數(shù)f(x)有極大值f(2)和微小值f(-2)D.函數(shù)f(x)有極大值f(-2)和微小值f(2)6.若函數(shù)f(x)=cosx+2xf'π6,則f-π3A.f-π3=fπ6 B.fC.f-π3<f7.若函數(shù)f(x)=2cosωx+π48.已知函數(shù)f(x)=x3+3ax2+3(a+2)x+1既有極大值,又有微小值,則實(shí)數(shù)a的取值范圍是.

9.設(shè)f(x)=1-x2,x10.已知函數(shù)f(x)=12x2+2ax-lnx,若f(x)在區(qū)間13,11.已知f(x)=ex-ax2,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=bx+1.(1)求a,b的值;(2)求f(x)在[0,1]上的最大值;12.已知a>0,函數(shù)f(x)=a2x3-3ax2+2,g(x)=-3ax+3.(1)若a=1,求函數(shù)f(x)的圖象在點(diǎn)(1,f(1))處的切線方程;(2)求函數(shù)f(x)在區(qū)間(-1,1)上的極值.13.(2024石家莊模擬)已知函數(shù)f(x)=x2-alnx.(1)當(dāng)a=2時,求函數(shù)f(x)的最小值;(2)若函數(shù)g(x)=f(x)+2x14.設(shè)f(x)=lnx-2ax+2a,g(x)=xlnx-ax2+(2a-1)x,a∈R.(1)求f(x)的單調(diào)區(qū)間;(2)f(x)=g'(x),已知g(x)在x=1處取得極大值,求實(shí)數(shù)a的取值范圍.

答案全解全析1.C01(2x+ex)dx=(x2+ex)

01=1+e2.D由題意,可得y'=ex,則所求切線的斜率k=e2,所求切線方程為y-e2=e2(x-2),即y=e2x-e2.∴S=12×1×e2=e3.Cf'(x)=3ax2+1x+2.由題意,得f'(-1)=3a+1=0.解得a=-4.B由題意可得,f(x)的定義域?yàn)?0,+∞),f'(x)=2(2x-1)lnx+2(x2-x)·1x-2x+2=(4x-2)lnx.由f'(x)<0,可得(4x-2)lnx<0,所以4x-2>0,ln5.D當(dāng)x<-2時,1-x>0,由(1-x)f'(x)>0?f'(x)>0,函數(shù)f(x)為增函數(shù);當(dāng)-2<x<1時,1-x>0,由(1-x)f'(x)<0?f'(x)<0,函數(shù)f(x)為減函數(shù);當(dāng)1<x<2時,1-x<0,由(1-x)·f'(x)>0?f'(x)<0,函數(shù)f(x)為減函數(shù);當(dāng)x>2時,1-x<0,由(1-x)·f'(x)<0?f'(x)>0,函數(shù)f(x)為增函數(shù).所以函數(shù)f(x)有極大值f(-2)和微小值f(2).故選D.6.C依題意,得f'(x)=-sinx+2f'π6∴f'π6=-sinπ6+2f'∴f'π6=1易知f'(x)=-sinx+1≥0,∴f(x)=cosx+x是R上的增函數(shù).又-π3<π6,∴f-π7.答案3解析由題意,得f'(x)=-2ωsinωx+π4,所以f'(0)=-2ωsinπ4=-3.所以ω=3.8.答案(-∞,-1)∪(2,+∞)解析f'(x)=3x2+6ax+3(a+2).因?yàn)楹瘮?shù)f(x)既有極大值,又有微小值,所以f'(x)=0有兩個不等實(shí)根.所以36a2-36(a+2)>0,解得a>2或a<-1.9.答案π2+4解析∫

-12f(x)dx=∫

-111-x2dx+∫

12(x10.答案43解析由題意知,f'(x)=x+2a-1x≥0在13,2上恒成立,即2a≥-x+1x在13,2上恒成立,∵當(dāng)x∈13,11.解析(1)f'(x)=ex-2ax,由已知,可得f'(1)=e-2a=b,f(1)=e-a=b+1.解得a=1,b=e-2.(2)令g(x)=f'(x)=ex-2x,則g'(x)=ex-2.令g'(x)=0,得x=ln2.故當(dāng)0≤x<ln2時,g'(x)<0,g(x)在[0,ln2)上單調(diào)遞減;當(dāng)ln2<x≤1時,g'(x)>0,g(x)在(ln2,1]上單調(diào)遞增.所以g(x)min=g(ln2)=2-2ln2>0.所以f(x)在[0,1]上單調(diào)遞增.所以f(x)max=f(1)=e-1.12.解析(1)由f(x)=a2x3-3ax2+2,得f'(x)=3a2x2-6ax.當(dāng)a=1時,f'(1)=-3,f(1)=0,所以f(x)的圖象在點(diǎn)(1,f(1))處的切線方程是y=-3x+3.(2)令f'(x)=0,得x1=0,x2=2a①當(dāng)0<2a<1,即a>2時,x改變時,f'(x),f(x)的改變狀況x(-1,0)0022f'(x)+0-0+f(x)↗極大值↘微小值↗故f(x)的極大值是f(0)=2;微小值是f2a=2-4②當(dāng)2a所以f(x)的極大值為f(0)=2,無微小值.13.解析(1)由題意知,函數(shù)f(x)的定義域?yàn)?0,+∞),當(dāng)a=2時,f'(x)=2x-2x=2由f'(x)<0,得0<x<1,故f(x)在(0,1)上單調(diào)遞減;由f'(x)>0,得x>1,故f(x)在(1,+∞)上單調(diào)遞增.所以函數(shù)f(x)的最小值為f(1)=1.(2)由題意,得g'(x)=2x-ax-2①若g(x)為[1,+∞)上的單調(diào)遞增函數(shù),則g'(x)≥0在[1,+∞)上恒成立,即a≤-2x+2x2設(shè)φ(x)=-2x+2x2∵φ(x)在[1,+∞)上單調(diào)遞增,∴φ(x)min=φ(1)=0.∴a≤0.②若g(x)為[1,+∞)上的單調(diào)遞減函數(shù),則g'(x)≤0在[1,+∞)上恒成立,即a≥-2x+2x2由①知φ(x)=-2x+2x2在[1,+∞)上單調(diào)遞增,且當(dāng)x趨向于無窮大時,φ(x)趨向于無窮大,∴φ(x)無最大值.故g(x)在[1,+∞)上不行綜上所述,a的取值范圍為(-∞,0].14.解析(1)由f(x)=lnx-2ax+2a,得f'(x)=1x-2a=1當(dāng)a≤0,x∈(0,+∞)時,f'(x)>0,f(x)單調(diào)遞增.當(dāng)a>0,x∈0,x∈12所以,當(dāng)a≤0時,f(x)的單調(diào)增區(qū)間為(0,+∞);當(dāng)a>0時,f(x)的單調(diào)增區(qū)間為0,12(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,g(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時,f(x)>0,g(x)單調(diào)遞增.所以g(x)在x=1處取得微小值,不合題意.②當(dāng)0<a<12時,1由(1)知,f(x)在0,可得當(dāng)x∈(0,1)時,f(x)<0,當(dāng)x∈1,所以g(x)在(0,1)內(nèi)單調(diào)遞減,在1,所以g(x)在x=1處