高考數(shù)學(xué)二輪復(fù)習(xí)專題二第三講導(dǎo)數(shù)簡(jiǎn)單應(yīng)用習(xí)題文

第三講導(dǎo)數(shù)的簡(jiǎn)單應(yīng)用1.已知函數(shù)f(x)=xsinx+ax,且f'(π2)=1,則a=()ln??2.(2018河北石家莊模擬)已知f(x)=??,此中e為自然對(duì)數(shù)的底數(shù),則( )A.f(2)>f(e)>f(3)B.f(3)>f(e)>f(2)C.f(e)>f(2)>f(3)D.f(e)>f(3)>f(2)3.(2018安徽合肥質(zhì)量檢測(cè))已知直線2x-y+1=0與曲線y=aex+x相切,此中e為自然對(duì)數(shù)的底數(shù),則實(shí)數(shù)a的值是( )4.(2018遼寧沈陽(yáng)質(zhì)量檢測(cè))設(shè)函數(shù)f(x)=xex+1,則()A.x=1為f(x)的極大值點(diǎn)B.x=1為f(x)的極小值點(diǎn)C.x=-1為f(x)的極大值點(diǎn)D.x=-1為f(x)的極小值點(diǎn)5.已知定義在R上的函數(shù)f(x)知足f(-3)=f(5)=1,f'(x)為f(x)的導(dǎo)函數(shù),且導(dǎo)函數(shù)y=f'(x)的圖象如圖所示,則不等式f(x)<1的解集是( )A.(-3,0)B.(-3,5)C.(0,5)D.(-∞,-3)∪(5,+∞)6.(2018江西南昌模擬)已知函數(shù)f(x)是定義在R上的偶函數(shù),設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),若對(duì)隨意x>0都有2f(x)+xf'(x)>0建立,則( )A.4f(-2)<9f(3)B.4f(-2)>9f(3)C.2f(3)>3f(-2)1D.3f(-3)<2f(-2)7.(2018河北“五個(gè)一名校結(jié)盟”模擬2的單一遞減區(qū)間是.)函數(shù)f(x)=x-2lnx8.(2018廣西桂林、百色、梧州、崇左、北海五市聯(lián)考)直線x=a(a>0)分別與曲線y=2x+1,y=x+lnx交于A,B兩點(diǎn),則|AB|的最小值為.9.已知函數(shù)f(x)=lnx,g(x)=x2+mx(m∈R),若函數(shù)f(x)的圖象在點(diǎn)(1,f(1))處的切線與函數(shù)g(x)的圖象相切,則m的值為.110.已知函數(shù)f(x)=x3-2x+ex-e??,此中e是自然對(duì)數(shù)的底數(shù).若f(a-1)+f(2a2)≤0,則實(shí)數(shù)a的取值范圍是.11.已知函數(shù)f(x)=excosx-x.(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)求函數(shù)f(x)在區(qū)間[0,π2]上的最大值和最小值.12.(2018江西南昌模擬)設(shè)函數(shù)f(x)=2lnx-mx2+1.議論函數(shù)f(x)的單一性;當(dāng)f(x)有極值時(shí),若存在x0,使得f(x0)>m-1建立,務(wù)實(shí)數(shù)m的取值范圍.2答案精解精析π1.A∵f'(x)=sinx+xcosx+a,且f'(2)=1,sinπ+πcosπ+a=1,即a=0.2222.D∵f(x)=ln??1-ln??函數(shù)f(x)單一遞加;當(dāng)??,∴f'(x)=2.令f'(x)=0,解得x=e.當(dāng)x∈(0,e)時(shí),f'(x)>0,??x∈(e,+∞)時(shí),f'(x)<0,函數(shù)f(x)單一遞減,故f(x)在x=e處獲得最大值f(e).∵f(2)-f(3)=ln2-ln3=3ln2-2ln3=ln8-ln9<0,∴f(2)<f(3),則f(e)>f(3)>f(2).應(yīng)選D.23663.C∵y=aex+x,∴y'=aexx+x相切的切點(diǎn)坐標(biāo)為(m,n),則y'|x=mm+1.設(shè)直線2x-y+1=0與曲線y=ae=ae+1=2,得aem=1.又n=aem+m=2m+1,∴m=0,n=1,∴a=1.應(yīng)選C.4.D由題意得f'(x)=(x+1)ex,令f'(x)=0,得x=-1.當(dāng)x∈(-∞,-1)時(shí),f'(x)<0;當(dāng)x∈(-1,+∞)時(shí),f'(x)>0,則f(x)在(-∞,-1)上單一遞減,在(-1,+∞)上單一遞加,所以x=-1為f(x)的極小值點(diǎn).應(yīng)選D.5.B由題圖可知,當(dāng)x>0時(shí),f'(x)>0,f(x)是增函數(shù);當(dāng)x<0時(shí),f'(x)<0,f(x)是減函數(shù).又f(-3)=f(5)=1,所以不等式f(x)<1的解集是(-3,5).應(yīng)選B.6.A依據(jù)題意,令g(x)=x2f(x),則g'(x)=2xf(x)+x2f'(x).又對(duì)隨意x>0都有2f(x)+xf'(x)>0建立,則當(dāng)x>0時(shí),有g(shù)'(x)=x[2f(x)+xf'(x)]>0恒建立,即函數(shù)g(x)在(0,+∞)上為增函數(shù).又由函數(shù)f(x)是定義在R上的偶函數(shù),則f(-x)=f(x),則有g(shù)(-x)=(-x)2f(-x)=x2f(x)=g(x),即函數(shù)g(x)也為偶函數(shù),3則有g(shù)(-2)=g(2),且g(2)<g(3),則有g(shù)(-2)<g(3),即有4f(-2)<9f(3).應(yīng)選A.7.答案(0,1)222(??+1)(??-1)分析函數(shù)f(x)=x-2lnx的定義域?yàn)?0,+∞),f'(x)=2x-??=??,令f'(x)<0,得0<x<1,∴函數(shù)f(x)的單一遞減區(qū)間是(0,1).答案2分析設(shè)A(a,yA),B(a,yB),則yA=2a+1,yB=a+lna.所以|AB|=|yA-yB|=|a+1-lna|.令f(a)=a+1-lna,則f'(a)=1-1,所以函數(shù)f(a)在(0,1)上單一遞減,在(1,+∞)上單一遞加,于是當(dāng)a=1時(shí),f(a)獲得最小值2,??即|AB|min=2.9.答案-1或31分析易知f(1)=0,f'(x)=??,進(jìn)而獲得f'(1)=1,函數(shù)f(x)的圖象在點(diǎn)(1,f(1))處的切線方程為y=x-1.解法一:設(shè)直線y=x-1與函數(shù)g(x)=x2+mx(m∈R)的圖象相切于點(diǎn)P(x0,y0),進(jìn)而可得g'(x0)=1,g(x0)=x0-1.又g'(x)=2x+m,所以有{??'(??0)=2??0+m=1,得??2=1,解得{??0=1,或{??0=-1,??20+m??0=??0-1,0??=-1??=3.解法二:聯(lián)立{??=??-1,得x2+(m-1)x+1=0,所以=(m-1)2-4=0,解得m=-1或m=3.??=??2+mx,10.答案[-1,1]2分析易知函數(shù)f(x)的定義域?yàn)镽,對(duì)于原點(diǎn)對(duì)稱.3x1∵f(x)=x-2x+e-e??,3-x131x∴f(-x)=(-x)-2(-x)+e-e-??=-x+2x+e??-e=-f(x),f(x)為奇函數(shù).2x122又f'(x)=3x-2+e+e??≥3x-2+2=3x≥0(當(dāng)且僅當(dāng)x=0時(shí),取“=”),進(jìn)而f(x)在R上單一遞加,2221∴f(a-1)+f(2a)≤0?f(a-1)≤f(-2a)?-2a≥a-1,解得-1≤a≤2.11.分析(1)由于f(x)=excosx-x,所以f'(x)=ex(cosx-sinx)-1,所以f'(0)=0.又由于f(0)=1,所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=1.(2)設(shè)h(x)=ex(cosx-sinx)-1,4則h'(x)=ex(cosx-sinx-sinx-cosx)x=-2esinx.π當(dāng)x∈(0,2)時(shí),h'(x)<0,所以h(x)在區(qū)間[0,π2]上單一遞減.x∈(0,π所以對(duì)隨意2],有h(x)<h(0)=0,即f'(x)<0.π所以函數(shù)f(x)在區(qū)間[0,2]上單一遞減.所以f(x)在區(qū)間[0,πππ.2]上的最大值為f(0)=1,最小值為f(2)=-2212.分析(1)函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=??-2mx-2(????2-1),=??當(dāng)m≤0時(shí),f'(x)>0,∴f(x)在(0,+∞)上單一遞加;當(dāng)m>0時(shí),令f'(x)>0,則0<x<1,√??令f'(x)<0,則x>1,∴f(x)在(0,√????)上單一遞加,在(√????,+∞)上單一遞減.??(2)由(1)知,當(dāng)f(x)有極值時(shí),m>0,且f(x)在(0,√??√??,+∞)上單一遞減,??)上單一遞加,在(??∴f(x)max=f(√