導(dǎo)數(shù)大題練習(xí)題

1、導(dǎo)數(shù)練習(xí)題(B)答案1.(本題滿(mǎn)分12分)“已知函數(shù)f(x)ax3bx2(c3a2b)xd的圖象如圖所/示.(I)求c,d的值；/(II)若函數(shù)f(x)在x2處的切線(xiàn)方程為3xy110,Jyy求函數(shù)f(x)的解析式；I(III)在(II)的條件下，函數(shù)yf(x)與y1f(x)5xm的圖象有3三個(gè)不同的交點(diǎn)，求m的取值范圍.解：函數(shù)f(x)的導(dǎo)函數(shù)為f1(x)3ax22bxc3a2b(2分)由圖可知函數(shù)f(x)的圖象過(guò)點(diǎn)(0,3),且f1(1)0得d3d3(4分)3a2bc3a2boe0(II)依題意f'(2)3且f(2)512a4b3a2b38a4b6a4b35解得a1,b6所以f(x

2、)x36x29x3(8分)(III)f(x)3x212x9.可轉(zhuǎn)化為：x36x29x3x24x35xm有三個(gè)不等實(shí)根，即：gxx37x28xm與x軸有三個(gè)交點(diǎn)；2gx3x14x83x2x4,x2,323乙344,gx+0一0+gx增極大值減極小值增g8m,g416m.(10分)327當(dāng)且僅當(dāng)g28m。且g416m。時(shí)，有三個(gè)交點(diǎn),327(12分)故而，16m竺為所求.-272.(本小題滿(mǎn)分12分)已知函數(shù)f(x)aInxax3(aR).今若函數(shù)g(x)(2分)x2f1(x)£在區(qū)間(I)求函數(shù)f(x)的單調(diào)區(qū)間；(II)函數(shù)f(x)的圖象的在x4處切線(xiàn)的斜率為(1,3)上不是單調(diào)函數(shù)

3、，求m的取值范圍.解：f1(x)a(1x)(x0)x當(dāng)a01,f(x)的單調(diào)增區(qū)間為0,1，減區(qū)間為1,當(dāng)a0t,f(x)的單調(diào)增區(qū)間為1,，減區(qū)間為0,1;當(dāng)a=1時(shí)，f(x)不是單調(diào)函數(shù)(5分)(II)f(4)3aI得a2,f(x)2lnx2x3g(x)1x3(m2)x22x,g'(x)x2(m4)x2(6分)g(x)在區(qū)間(1,3)上不是單調(diào)函數(shù)，且g'(0)2g'(i)Qg'(3)0.(8分)m3,19(10分)m(-,3)(12分)m,333 .(本小題滿(mǎn)分14分)c的圖象經(jīng)過(guò)坐標(biāo)原點(diǎn)，且在x1處取得極大值.已知函數(shù)f(x)x3ax2bx(I)求實(shí)數(shù)a

4、的取值范圍；2(II)若萬(wàn)程f(x)組包恰好有兩個(gè)不同的根，求f(x)的解析式;9)|81.(III)對(duì)于(II)中的函數(shù)f(x),對(duì)任意、R,求證：|f(2sin)f(2sin解：f(0)0c0,f(x)3x22axb,f(1)0b2a32f(x)3x22ax(2a3)(x1)(3x2a3),由f(x)0x1或x空因?yàn)楫?dāng)x1時(shí)取得極大值,3所以2a_f1a3,所以a的取值范圍是:(,3);3(4分)(10分)(14分)(II)由下表:x(,1)12a3(1,3)2a332a3(3,)f(x)+0-0-f(x)遞增極大值a2遞減極小值-a6(2a3)227遞增2依題意得：i(2a3)2也坦，解

5、得：a9279所以函數(shù)f(x)的解析式是：f(x)x39x215x(III)對(duì)任意的實(shí)數(shù)，都有22sin2,22sin2,在區(qū)間-2,2有：f(2)8363074,f(1)7,f(2)836302f(x)的最大值是f(1)7,f(x)的最小值是f(2)8363074函數(shù)f(x)在區(qū)間2,2上的最大值與最小值的差等于81,所以|f(2sin)f(2sin)|81.(4分)號(hào)，列表4 .(本小題滿(mǎn)分12分)已知常數(shù)a0,e為自然對(duì)數(shù)的底數(shù)，函數(shù)f(x)exx,g(x)x2alnx.(I)寫(xiě)出f(x)的單調(diào)遞增區(qū)間，并證明eaa;(II)討論函數(shù)yg(x)在區(qū)間(1,ea)上零點(diǎn)的個(gè)數(shù).解：f(x)

6、ex10,得f(x)的單調(diào)遞增區(qū)間是(0,),(2分).a0,/.f(a)f(0)1,/.eaa1a,即eaa.2a2a2(x)(x)(II)g(x)2x-22-,由g(x)0,得xg(x)g(x)(0,2)烏22-0單調(diào)遞減極小值(等，)2+單調(diào)遞增等時(shí),函數(shù)yg(x)取極小值g(墳)a(iin-,無(wú)極大值.222(6分)2aaee由(I)eaa,a，二e2aa，.二e'a2'22g(1)10,g(ea)e2aa2(eaa)(eaa)0(8分)(i)當(dāng)2a1,即0a2時(shí)，函數(shù)yg(x)在區(qū)間(1,ea)不存在零點(diǎn)(ii)當(dāng)三1,即a2時(shí)2若a(1lna)0,即2a2e時(shí)，函數(shù)

7、yg(x)在區(qū)間(1,ea)不存在零點(diǎn)e；若a(1ln$0,即a2e時(shí)，函數(shù)yg(x)在區(qū)間(1,ea)存在一個(gè)零點(diǎn)x若2(1ina)0,即a2e時(shí)，函數(shù)yg(x)在區(qū)間(1,ea)存在兩個(gè)零點(diǎn)；綜上所述，yg(x)在(1,ea)上，我們有結(jié)論：當(dāng)0a2e時(shí)，函數(shù)f(x)無(wú)零點(diǎn)；當(dāng)a2e時(shí)，函數(shù)f(x)有一個(gè)零點(diǎn)；當(dāng)a2e時(shí)，函數(shù)f(x)有兩個(gè)零點(diǎn).(12分)5.(本小題滿(mǎn)分14分)已知函數(shù)f(x)ln(x1)k(x1)1.(I)當(dāng)k1時(shí)，求函數(shù)f(x)的最大值；(II)若函數(shù)f(x)沒(méi)有零點(diǎn)，求實(shí)數(shù)k的取值范圍;解：(I)當(dāng)k1時(shí),f(x)f(x)定義域?yàn)?1,+),令f(x)0,得x2,(

8、2分).當(dāng)x(1,2)日1f(x)0,當(dāng)x(2,)時(shí)，f(x)0,f(x)在(1,2)內(nèi)是增函數(shù)，在(2,)上是減函數(shù).當(dāng)x2時(shí)，f(x)取最大值f(2)0(4分)(II)當(dāng)k0時(shí)，函數(shù)yln(x1)圖象與函數(shù)yk(x1)1圖象有公共點(diǎn)，函數(shù)f(x)有零點(diǎn)，不合要求；(8分)11kkxk(x)八當(dāng)k0時(shí)，f(x)kk-(6分)x1x1x1k1k1.1.令f(x)0,得xJ,.x(1,2)時(shí)，f(x)0,x(1-,)時(shí)，f(x)0,kkkf(x)在(1,1，內(nèi)是增函數(shù)，在1；)上是減函數(shù)，f(x)的最大值是f(1)lnk,k:函數(shù)f(x)沒(méi)有零點(diǎn)，lnk0,k1,因此，若函數(shù)f(x)沒(méi)有零點(diǎn)，則

9、實(shí)數(shù)k的取值范圍k(1,).(10分)6.(本小題滿(mǎn)分12分)已知x2是函數(shù)f(x)(x2ax2a3)ex的一個(gè)極值點(diǎn)(e2.718).(I)求實(shí)數(shù)a的值；(II)求函數(shù)f(x)在x|,3的最大值和最小值.解：(I)由f(x)(x2ax2a3)ex可得f(x)(2xa)ex(x2ax2a3)exx2x2是函數(shù)f(x)的一個(gè)極值點(diǎn)，.二(a5)e20,解得a5(II)由f(x)(x2)(x1)ex0,得£但在(由f(x)0,得f(x)在在(1,2)遞減f(2)e2是f(x)在x|,3的最小值；(2a)xa3exf(2)0,1)遞增，在(2,(4分)(6分)遞增,372f(2)4e，f(

10、x)在x7.(本小題滿(mǎn)分3.3f(3)e3f(3)f(-)23,3的最大值是f(3)214分)已知函數(shù)f(x)x24x(2a)lnx,(a73-e24R,a11一e(4e.e40)(8分)37)0,f(3)f(-)2(12分)(II)求函數(shù)解：(I)f(x)f(x)在區(qū)間e,e2上的最小值.(I)當(dāng)a=18時(shí)，求函數(shù)f(x)的單調(diào)區(qū)間;f'(x)2x4由f'(x)0得(x24x16lnx,162(x2)(x4)xxx2)(x4)0,解得x4或x2注意到x0,所以函數(shù)f(x)的單調(diào)遞增區(qū)間是(4,+8)由f'(x)0得(x2)(x4)0,解得-2<x<4,注意

11、到x0,所以函數(shù)f(x)的單調(diào)遞減區(qū)間是(0,4.綜上所述，函數(shù)f(x)的單調(diào)增區(qū)間是(4,+s),單調(diào)減區(qū)間是(0,46分(H)在xe,e2日寸，f(x)x24x(2a)lnx22a2x4x2af'(x)2x4,xx設(shè)g(x)2x24x2a當(dāng)a0時(shí)，有=16+4>2(2a)8a0,此時(shí)g(x)0,所以f'(x)0,所以f(x)minf(e)e24e當(dāng)a0時(shí)，A=16令f'(x)0,即2x242(24x2令f'(x)0,即2x24x22a)f(x)在e,e2上單調(diào)遞增,a0,12a或x2解得12a二；.2a/-2ax1-若1浮而2,即a支(e21)2時(shí)，f

12、(x)在區(qū)間e,e2單調(diào)遞減，所以f(x)minf(e2)e44e242a.若e1尊e2,即2(e1)2a2(e21)2時(shí)間，f(x)在區(qū)間e,1盤(pán)上單調(diào)遞減，在區(qū)間1出,e2上單調(diào)遞增,22所以f(x)minf(1必)a2a3(2a)ln(13).222若1=aW,即0a<e1)2時(shí)，f(x)在區(qū)間e,e2單調(diào)遞增，2所以f(x)minf(e)e24e2a綜上所述，當(dāng)a>2e21)2日寸，f(x)mina44e242a；jl當(dāng)2(e1)2a2(e21)2時(shí)，f(x)mina后3(2a)ln(1卓)；22當(dāng)aW2(e1)2時(shí)，f(x)mine24e2a14分8.(本小題滿(mǎn)分12分)

13、已知函數(shù)f(x)x(x6)alnx在x(2,)上不具有單調(diào)性.(I)求實(shí)數(shù)a的取值范圍；(II)若f(x)是f(x)的導(dǎo)函數(shù)，設(shè)g(x)f(x)6X、x2,不等式解：(I)f(x)f(x)在x|g(xOg(x2)|7區(qū)x2|恒成立.2aca2x6xa2x6,xx(2,)上不具有單調(diào)性，.,.在x(2,與，試證明：對(duì)任意兩個(gè)不相等正數(shù)x(2分)上f(x)有正也有負(fù)也有0,(4分)即二次函數(shù)y2x26xa在x(2,)上有零點(diǎn)y2x26xa是對(duì)稱(chēng)軸是x的實(shí)數(shù)a的取值范圍(,4)(II)由(I)g(x)2x三馬,xx方法1：g(x)f(x)E62xaxxa4a4,g(x)22xx3,開(kāi)口向上的拋物線(xiàn),

14、2，、T(x0),x3442x34x42y22262a0(8分)(6分)8124(2x3)x"寸，h(x)取最小值Hg(x)|8x是增函數(shù)，設(shè)h(x)223,h(x)xxh(x)在(0。是減函數(shù)，在(|,)增函數(shù)，當(dāng)從而g(x)|7,/.(g(x)37x)0,函數(shù)yXi、x2是兩個(gè)不相等正數(shù)，不妨設(shè)XiX2,則g(X2)38一X22738g(x1)于Xg(X2)g(X1)38(X2xj,X2Xi0,.g(x1)g(x2)38x1x227-g(xi)g(x2)X1X2383827，即|g(x1)g(x2)|萬(wàn)以|(12分)方法2：M(x1,g(x1)、N(x2,g(x2)是曲線(xiàn)g(xi

15、)g(x2)2(xX2)aX1X22(x1X2)22xX222X1X2a2X1X24(_：X瓦)3aX1X2x2設(shè)tL,tX1X20,令kMNu(t)24t3由u(t)0,得t2,由u(t)0得0t3u(t)在(0,|)上是減函數(shù)，在(|,u(t)在t2處取極小值38,u(t)327即|g(xjg(七)|3y|x19.(本小題滿(mǎn)分12分)x2I已知函數(shù)f(x)(I)討論函數(shù)(II)證明：若g(x)上任意兩相異點(diǎn),2X1X2,a424t,u(t)(8分)4t(3t2),23,)上是增函數(shù),37,所以ax(a1)lnx,a1.f(x)的單調(diào)性；a5,則對(duì)任意x，X2(0,),x1(1)f(x)的定

16、義域?yàn)?0,),f'(x)xg(x1)g(x2)X1X2*2,有3827f(xi)f(x2)1.(12分)X12xaxa1xX2(x1)(x1a)x2分若a11,即a2,則f'(x)UL.故f(x)在(0,)單調(diào)增加.(ii)若a11,而a當(dāng)x(0,a1)及x(1,)單調(diào)增加.(iii)若a11,即a單調(diào)增加.(II)考慮函數(shù)g(x)x1,故1a2,則當(dāng)x(a1,1)時(shí)，f'(x)0.(1,)時(shí)，f'(x)0,故f(x)在(a1,1)單調(diào)減少,在(0,a-1),2,同理可得f(x)在(1,a1)單調(diào)減少，在(0,1),(a1,f(x)x12-xax2(a1)ln

17、xx.由g'(x)x(a(a1)1(.a11)2.由于aa5,故g'(x)0,即g(x)在(0,)單調(diào)增加，從而當(dāng)g(x1)g(x2)0,即f(x1)f(X2)X1X20,x1x20時(shí)有故f(X1)_f(X2)X1X210.(本小題滿(mǎn)分14分)已知函數(shù)f(x)L22f(X1)f(X2)X1X2時(shí)，有X1X2alnx,g(x)(a1)x,a1.f卜)f(x1)X2X1(I)若函數(shù)f(x),g(x)在區(qū)間1,3上者B是單調(diào)函數(shù)且它們的單調(diào)性相同，求實(shí)數(shù)值范圍；a的取(II)若a(1,e(e2.71828L),設(shè)F(x)f(x)g(x),求證：當(dāng)xx21,a時(shí)，不等式|F(x)F(x

18、2)|1成立.解：f(x)xa,g(x)a1,x函數(shù)f(x),g(x)在區(qū)間1,3上都是單調(diào)函數(shù)且它們的單調(diào)性相同,(2分)二當(dāng)X1,3時(shí),2f(x)g(x)()0恒成立,(4分)即(aa1)(x2a)1+2在*x(II)F(x)1,12x20恒成立，1,3時(shí)恒成立,或alnx,F(x)定義域是(0,1+2在xx1,3時(shí)恒成立,(a1)x,F(x)在(0,1)是增函數(shù),在F(x)(1,e,即(a1)(Xa)(x1)(1,a)實(shí)際減函數(shù),在(a,)是增函數(shù)當(dāng)x1時(shí)，F(xiàn)(x)取極大值MF(1)a當(dāng)xa日寸，F(xiàn)(x)取極小值mF(a)alna121-a21,a,/.|F(x1)F(x2)|Mm|Mm

19、設(shè)G(a)M12-a2：G(a)1G(a)G(a)al12-a21alna1alna一，貝UG(a)alna2(1,e,G(a)0i(1,e是增函數(shù)，G(a)1.alnaa(1,e也小臺(tái)函數(shù)/.G(a)而L2二當(dāng)G(e),即G(a)21 (e1)de-12 212-ee2-2(31)21,(6分)(8分)(10分)(12分)1,/.G(a)Mm1x,x21,a時(shí),不等式|F(xJFd)|1成立.(14分)11.(本小題滿(mǎn)分12分)設(shè)曲線(xiàn)C:f(x)lnxex(e2.71828(I)求函數(shù)f(x)的極值；(II)對(duì)于曲線(xiàn)C上的不同兩點(diǎn)"七,3)使直線(xiàn)AB的斜率等于f(%).),f(x)

20、表示f(x)導(dǎo)函數(shù).,BNyz),xx2,求證：存在唯一的x0(為公)，解：f(x)當(dāng)x變化時(shí),11ex彳日1一e0,倚x-xxef(x)與f(x)變化情況如下表:x(0,1)e1e(，)ef(x)十0一f(x)單調(diào)遞增極大值單調(diào)遞減f(x)取得極大值f(1)2,沒(méi)有極小值;e當(dāng)x1時(shí),e(4分)(II)(方法i)f(X0)kAB,ieX0Inx2Inxie(x2xi).x2xi,X2XiX0In*0Xi(X2X1)g(Xi)Xi0,設(shè)g(x),X2,、xIn一(x2xi)XiXiIn(X2Xix3g(Xi)X1In2i0,g(xi)是Xi的增函數(shù),Xi*XiX2,一g(Xi)X2g(x2)X

21、2InX2(X2X2)0;g(X2)x2In*(x2XiXi),g(X2)/X2In也i0,gd)是X2的增函數(shù),XiXiX2,二g(X2)Xig(x)XiInXi(XiXi)0,函數(shù)g(x)又x2i,Xi函數(shù)g(x)xIn上(X2Xi)在(為岱)內(nèi)有零點(diǎn),XIn0,函數(shù)g(x)xIn2(X2x1)在保田)是增函數(shù),(i0分)Xix2xiXXi此上在(為公)內(nèi)有唯一零點(diǎn)X。，命題成立Xi(i2分)(方法2)f(X0)iInx2Inxie(x2xi)一e,即x0Inx2x0Inxixi設(shè)g(x)xInx2xInxiX0X2XiX0(x,X2),且X0唯一貝Ug(x1)xiInx2xiInxixi

22、x2,再設(shè)h(x)xInx2xInxxx2,0xx2,.=h(x)Inx2Inx0h(x)xInx2xInxxx2在0xX2是增函數(shù)g(Xi)h(Xi)h(x2)0,同理gd)0方程xInx2xInXixX20在區(qū)區(qū))有解(10分):一次函數(shù)在(Xi,X2)g(x)(InX2Inxjx為x2是增函數(shù)方程xInx2xInXixX20在小保區(qū))有唯一解，命題成立(12分)注：僅用函數(shù)單調(diào)性說(shuō)明，沒(méi)有去證明曲線(xiàn)C不存在拐點(diǎn)，不給分.12.(本小題滿(mǎn)分14分)定義F(x,y)(1x)y,x,y(0,),令函數(shù)f(x)F(3,Iog2(2xx24),寫(xiě)出函數(shù)f(x)的定義域;(II)令函數(shù)g(x)F(1,Io