第3章導(dǎo)數(shù)應(yīng)用文答案版

3.2導(dǎo)數(shù)的運(yùn)用(文1、★（2013海淀一模文）已知曲線f(x)lnx在點(diǎn)(x0,f(x0處的切線經(jīng)過點(diǎn)(01) A.e

【答案】【答案】3、★★(2011x[02]ysinxxcosx【答案 （開閉均可4、★★(2013f(xe|x||x|xf(x)k有兩個(gè)不同k的取值范圍是 【答案】5、★★(2014海淀期中，文)若函數(shù)f(x)sinxkxk A.

B.

D.(,【答案】 【答案】7、★★(2014石景一模)若存在實(shí)常數(shù)和,使得函數(shù)和對(duì)其定義域內(nèi)的任意實(shí)數(shù)分別滿足：8、★★★（2013房山二模文）f(xax3bx2cxd(a0,給出定義f'(xyf(x的導(dǎo)數(shù),f''(xf'(x的導(dǎo)數(shù),f''(x0x0,稱點(diǎn)(x0,f(x0yf(x的“拐點(diǎn)”.某同學(xué)經(jīng)過探究發(fā)現(xiàn):若f(x)1x31x21x1,則該函數(shù)的對(duì)稱中心 計(jì)算f(1)f(2)f(3) f(2012)

( 29、★★（2013延慶一模文）f(x)2a2lnx1x2ax(aR2(Ⅰ)a1時(shí),yf(x在點(diǎn)(1,f(1的切線方程(Ⅱ)f(x的單調(diào)性【答案】:f(x的定義域?yàn)?0,,

(x)

2ax

xa1時(shí),f(13,f(1)21102yf(x在點(diǎn)(1,f(1y2

(x)

x2axx

(x2a)(xa)xa0時(shí),f(x)x0,f(x在定義域?yàn)?0,上單調(diào)遞增a0時(shí),f(x0,x12a(舍去x2ax變化時(shí),f(x,f(x的變化情況如下此時(shí),f(x在區(qū)間(0a單調(diào)遞減,在區(qū)間(a,)上單調(diào)遞增a0時(shí),f(x0,x12ax2a(舍去x變化時(shí),f(x,f(x的變化情況如下此時(shí),f(x)在區(qū)間(0,2a)單調(diào)遞減,在區(qū)間(2a,10、★★（2013東城一模文）f(x)mlnxmm2時(shí),yf(x在點(diǎn)(1,f(1處的切線方程f(x的單調(diào)性f(xM,M0,求m的取值范圍【答案】(Ⅰ)當(dāng)m2時(shí),f(x)2lnxxf(x)21x2 f(13

(mR).yf(x在點(diǎn)(1,f(1y13(x1即3xy20(Ⅱ)f(x的定義域?yàn)?0f(x)mm1(m1)xm m≤0時(shí),x0f(x)mm10恒成立xf(x在區(qū)間(0上單調(diào)遞減m≥1時(shí),x0f(x)mm10恒成立xf(x在區(qū)間(0上單調(diào)遞增當(dāng)0m1時(shí),f(x)0,x

1

,f(x)0,x

1mf(x在區(qū)間

m內(nèi)單調(diào)遞增,在區(qū)間(

)內(nèi)單調(diào)遞減1 1(III)由(Ⅱ)f(x的定義域?yàn)?0m≤0m≥1時(shí),f(x在區(qū)間(0上單調(diào),f(x無最大值當(dāng)0m1時(shí),f(x在區(qū)間(0,m內(nèi)單調(diào)遞增,在區(qū)間(

)內(nèi)單調(diào)遞減1所以當(dāng)0m1f(x有最大值

1Mf(mm

m1M0,m

1 m0 m 1 1所以m的取值范圍是(

111、★★（2013豐臺(tái)一模文）f(x)

x

,g(x)bx23xh(x)f(xg(x,且h(1h(10a,b的值 g( 當(dāng)a=2且b=4時(shí),求函數(shù) 的單調(diào)區(qū)間,并求該函數(shù)在區(qū)間(-f2m1)上的最大值4h(x)f(xg(x)

(x

2bx3bb3aa4解得 2b3b或 因?yàn)閔(1

1 (1(Ⅱ)記

g(x),則(x)=(x+a)(bx2+3x)(x≠-a)fa=2,b=4,所以(x)(x2)(4x23x(x≠-(x)12x222x62(2x3)(3x1)令(x)0,x3,x1 x3,x1時(shí),(x0,當(dāng)3x1時(shí),(x0 函數(shù)(x)的單調(diào)遞增區(qū)間為(2),(23),(1) 單調(diào)遞減區(qū)間為(31 ①當(dāng)-2<m<3時(shí)(x)在(-2,m)上單調(diào)遞增2其最大值為(m)=4m311m26m②當(dāng)3≤m≤1時(shí)(x)在(-2,3)上單調(diào)遞增,在31)上單調(diào)遞減,在1,m) 調(diào)遞增,而319 (x)412、★★★（2013門頭溝一模文）f(x)

x2b

,其中bRf(xx1x軸平行,求b的值f(x的單調(diào)區(qū)間 b【答案】解:(Ⅰ)

(x)(x2依題意,f(1)0,得b經(jīng)檢驗(yàn)b

(Ⅱ)①當(dāng)b0時(shí),f(x)1xf(x的單調(diào)減區(qū)間為(0(0;②當(dāng)b0時(shí),

(x)

b.(x2bb令f(x)0,得x1 ,x2bbf(xf(x的情況如下x(, (x(, ( (b,f 00f↘↗↘f(x)的單調(diào)減區(qū)間為(

b),

b;單調(diào)增區(qū)間為(

b)b0時(shí),f(xDxR|x b

(x) 0D上恒成立(x2f(x的單調(diào)減區(qū)間為(b(

b),

b,);13、★★★（2013西城一模文）f(xexax,g(x)axlnx,其中a0(Ⅰ)f(x的極值(Ⅱ)M,f(xg(xM上具有相同的單調(diào)性,求a的取值范圍(Ⅰ)解:f(x的定義域?yàn)镽,

f(x)ex①當(dāng)a0時(shí),f(xex,f(x在R上單調(diào)遞增.從而f(x)沒有極大值,也沒有極小值②當(dāng)a0時(shí),f(x)0,xln(af(xf(x的情況如下x(x(, f 0f↘↗f(x的單調(diào)減區(qū)間為(ln(a;單調(diào)增區(qū)間為(ln(a),f(xf(ln(a))aaln(a);(Ⅱ)解:g(x的定義域?yàn)?0,

g(x)a1ax ③當(dāng)a0時(shí),f(x在R上單調(diào)遞增,g(x在(0,上單調(diào)遞減,不合題意④當(dāng)a0時(shí),g(x)0,g(x)在(0,上單調(diào)遞減1a0時(shí),ln(a)0此時(shí)f(x)(ln(a),)上單調(diào)遞增由于g(x)(0,上單調(diào)遞減,a1時(shí)ln(a)0,此時(shí)f(x)在(ln(a上單調(diào)遞減,由于f(x)在(0,上單調(diào)遞減,符合題意綜上a的取值范圍是(14、★★（2013東城期末文）f(x）1x3mx23m2x1mR3m1y

f(x在點(diǎn)(2,f(2f(x在區(qū)間(23m的取值范圍（Ⅰ）m1f(x1x3x23x13f'(xx22x3f'(2)5f(2)53

y55(x2)，即15x3y2503所以曲線yf(x)在點(diǎn)(2,f(2))處的切線方程為15x3y25 6（Ⅱ）f'(x）x22mx3m2 8當(dāng)m0時(shí)，f'(x）x20恒成立，不符合題 9m0時(shí)，f(x的單調(diào)遞減區(qū)間是(3mm，若f(x)在區(qū)間(233m則m

解得m 11m0時(shí)，f(x的單調(diào)遞減區(qū)間是(m3m，若f(x在區(qū)間(23m則3m3.m2綜上所述，實(shí)數(shù)m的取值范圍是m3或m 1315、★★（2013大興一模文）f(x)=(ax1)ex.(I)求函數(shù)f(x)的單調(diào)區(qū)間;(Ⅱ)a>0時(shí),f(x在區(qū)間[-20上的最小值【答案】解:f'(x)(ax1)'ex(ax1)(ex)'ex(axa(Ⅰ)①當(dāng)a0時(shí),fx)ex0,fx的單調(diào)增區(qū)間為②當(dāng)a0時(shí),fx0得

xa1,fx0得a

xa1afx的單調(diào)增區(qū)間為(a1,,fx的單調(diào)減區(qū)間為(,aa③當(dāng)a0時(shí),fx)0得

xa1,fx0得a

xa1afx的單調(diào)增區(qū)間為(,a1,fx的單調(diào)減區(qū)間為(aaa(a

時(shí) 即當(dāng)a1時(shí)

fx)在

aa

上是減函數(shù), (a

f(

f

a

)

a1上是增函數(shù),a

a②當(dāng)a1

時(shí) 即當(dāng)0a1時(shí)

fx在[2,0上是增函數(shù) fx在區(qū)間[-2,0]f(2)1綜上:當(dāng)a1時(shí)

fx在區(qū)間[-2,0]上最小值為

a1當(dāng)0a1時(shí)

fx在區(qū)間[-2,0]上最小值為116★★★（2013豐臺(tái)期末文f(xax2bxc)ex(a0yf的兩個(gè)零點(diǎn)為-3f(xf(x的極小值為-1f(x的極大值（Ⅰ）f(x2axb)exax2bxc)exax22ab)xbc]ex分g(x)ax22ab)xbc∵ex0yf'(x)g(x)ax22ab)xbcf(xgx符號(hào)a0x3,或x0gx>0,f(x)0當(dāng)3x0時(shí)，g(x)<0,即f(x)0 63（0，+∞3，0．……7（Ⅱ）由（Ⅰ）x=0f(xcbc9a3(2ab)bc解得a1,b1,c 11f(xx2x1)ex3）,（0，+∞f(xf(3)931)e317、★★（2014西城期末文

14xx(,a a(a1,f0f↘↗f(x)xa)ex，其中eaRf(xx[04]時(shí)，求函數(shù)f(x)的最小值（Ⅰ）f(x)xa)exxRf(xxa1)ex2分f(x)0xa3分xf(xf(x 分故f(x)的單調(diào)減區(qū)間為(,a1)；單調(diào)增區(qū)間為(a1,) 分（Ⅰ所以當(dāng)a1≤0a≥1時(shí)，f(x)在[04上單調(diào)遞增，故f(x)在[0,4]上的最小值為f(x)minf(0)a 8當(dāng)0a14，即5a1f(x在(0,a1上單調(diào)遞減，f(x在(a14)故f(x)在[0,4]上的最小值為f f(a1)ea1 分當(dāng)a1≥4a≤5f(x在[04故f(x)在[0,4]上的最小值為f f(4)(a 分所以函數(shù)f(x)在[04上的最小值為f分

e(a4)e4

5a

★★★（2013海淀期末文）f(x1x21g(xalnx在點(diǎn)(1,0 F(x)f(xmg(x)(m0求aFx在區(qū)間[1,e上的最小值（I）f(1g(10所以(1,0f(xg(x的圖象上f'(x)xg'(x)af'(1)1g'(1)ax所以a（Ⅱ）F(x)1x21mlnx，其定義域?yàn)閧x|x x2

3Fxx 5 x2當(dāng)m0時(shí)，F(xiàn)'(x)x 0 Fx在(0Fx在[1,eF(1

7 x2mm當(dāng)m0時(shí)，令F'(x)x mmx

0，得到x1

x2 0(舍m當(dāng)m

1時(shí)，即0m1F'(x0對(duì)(1,eFx在[1,e上單調(diào)遞增,F(1

9m當(dāng)m

e時(shí)，即me2時(shí)

F'(x0對(duì)(1,eFx在[1,eF(e)1e21

11m當(dāng)1 e,即1me2時(shí)m

F'(x)0對(duì)

m成立

F'(x)0對(duì)

m,e)Fx在(1,m單調(diào)遞減,在F(m1m1

m,e)m1m1mlnm………13m綜上，當(dāng)m1

Fx在[1,eF(1當(dāng)1me2Fx在[1,e

1m1mln 當(dāng)me2時(shí)

Fx在[1,eF(e)1e21m 19、★★★（2013通州期末文）fxx3ax2bxa2a,bfxx110b若對(duì)于任意的a4fxx02b（Ⅰ） 分f132abf11aba2a解 b

ab

3a當(dāng)b

時(shí)，fx3x28x1164132

，所以函數(shù)有極值點(diǎn) 4b當(dāng)a3時(shí)，fx3x120，所以函數(shù)無極值點(diǎn)b所以b11． 6Fa2xa3x2b0a4,x02都成立．8因?yàn)閤0所以Fa在a4,上為單調(diào)遞增函數(shù)或?yàn)槌?shù)函數(shù) ………9所以Fa F48x3x2b0對(duì)任意x0,2都成立 即b3x2 11 又3x28x3x

433

1616 3所 當(dāng)x4時(shí)，3x23

16 123所以b163所以b3

13法二：fx3x22axb0對(duì)任意a4,，x0,2都成立 分即b3x22axa4,x02 即b3x2 8 Fx3x22ax3x

a 33

9 F00，于是b0 10 a 當(dāng)4a0FxmaxF3

b

．……11a2 又3

，所以b3

123 b的最小值為3

1320、★★★（2013石景山期末文）f(x)=lnxax+1aRy=f(xP(1,f(1處的切線ly=f(x)(x1的圖象在直線ly=f(x有零點(diǎn)，求實(shí)數(shù)a（Ⅰ）f(x)=1xf(1a+1kl=f(1)=1al

2yf(1)=kl(x1)，即y=(1a)x 4（Ⅱ）Fx)=f(x1-a)x=lnxx+1，x>0F(x)=11=1(1x)，解F(x)=0得x=1. x(0,1(1,F(0F↗↘F(1)<0，所以x>0x1Fx)<0fx)<(1a)x即函數(shù)y=f(x)(x1)的圖像在直線l的下方 9y=fxf(x)=lnxax+10alnx+1xg(xlnx+1g(xlnx+1)1lnx+1lnx 解g(x)=0得 11g(x在(0,1)上單調(diào)遞增，在(1,+上單調(diào)遞減，x=1g(x)g(1)=1，所以a 1321、★★（2013昌平期末文）f(x)1x3a2x1a(aR) a1f(x)在[02x(0,+f(x)0恒成立，求a的取值范圍（I）a1f(x1x3x1f'(xx2 令f'(x)0，得x11,x2 2

1x012f-0+3fx12↘6↗76∴當(dāng)x[0,2]時(shí)，fx最大值為f2 76（Ⅱ）f'(xx2a2xa)(xaf'(x0,得xa

fxxafa1a3a3aa2a21 a

02a210所以faa2a210 9若 10fxxafa1a3a3aa2a21 f

a(2a21)0,由a0,得2a210,0a 3 3所以當(dāng)0a

3時(shí)對(duì)任意x0,f(x0都成立2綜上，a的取值范圍是[0，3 13222、★★（2013海淀二模文f(x)lnxg(x)a(a0x（I）a1時(shí)，若曲線yf(x)在點(diǎn)Mx0fx0處的切線與曲線yg(x)在點(diǎn)Px0gx0))x0（II）若x(0,e]f(x)g(x)3，求實(shí)數(shù)a的取值范圍2（I）a1,f'(x1g(x

分f(xMx0,fx0gxPx0gx0 11x f(xM(1,0yxgxP(11)yxx0（II）若x(0,e]，都有f(x)g(x)2F(x)f(xg(x3lnxa3

4 Fx在(0,e]F'(x)1

xa

6F'(xF(x)xx(0,a(a,F'(0F( 8當(dāng)aeFx在(0,eF(e)F(e)1a30，得a 所以a

10當(dāng)aeFx在(0a上單調(diào)遞減，在(a,e)上單調(diào)遞增，eF(aF(a)lnaa30，得ae ee

a

12e綜上 e

1323、★★（2014海淀期末文fx)xa)ex，其中a為常數(shù)fx是區(qū)間[3上的增函數(shù)，求實(shí)數(shù)afx)e2x[02時(shí)恒成立，求實(shí)數(shù)a的取值范圍 ：

f'(x)(xa

x 2f(x是區(qū)間[3所以f'(x)0，即xa10在[3,)上恒成 3yxa1所以滿足題意只需3a10，即a2 5（Ⅱ）f'(x)0xa

6f(x),f'(xx(,aaf0f↘↗10①當(dāng)a10，即a1f(x在[0,2]f(0)，f(0e2，解得ae2，所以此時(shí)，ae2 11②當(dāng)0a12，即3a1f(x在[0,2]f(a1)，f(a1e2，求解可得此不等式無解，所以a不存在 12③當(dāng)a12，即a3f(x在[0,2]f(2)f(2)e2，解得a 13綜上討論，所求實(shí)數(shù)a的取值范圍為[e224（2014石景山期末文f(xex2x（e為自然對(duì)數(shù)的底數(shù)f(x在點(diǎn)(0，f(0f(x．x1，2f(xmx成立，求實(shí)數(shù)m的取值范圍 （Ⅰ）f 分f(x)ex2得f(0)1 2所以曲線f(x)在點(diǎn)(0，f(0))處的切線方程為yx1 3（Ⅱ）f(x)ex2令f(x)0，即ex2=0，解得xln 5x(ln2)時(shí)，f(x)0，x(ln2)時(shí)，f(x)0，此時(shí)f(x)的單調(diào)遞減區(qū)間為(ln2)，單調(diào)遞增區(qū)間為(ln2) 7分（）x2

2]

f(xmxx2

ex2] x立 8m

ex2x

min

9

(x令g(x) 2，g(x) 1]122則g(x)ming(1)e2 12所以m(e2 1325、★★★（2013房山二模f(xax2)exx1處取得極值求af(x在mm1x1x2[02]，都有|f(x1f(x2|e（Ⅰ）f'(xaexax2)exaxa

1f'(10即(2a2)ex

2a

3a1x1f(xx2)exa（Ⅱ）f(x)x2ex f'(x)ex+x2exx1exx1f-0+f減增所以函數(shù)f(x)在,1遞減，在1,遞 4當(dāng)m1時(shí)，f(x)在m,m1單調(diào)遞增， (x)f(m)(m2)em當(dāng)0m1m1m

5f(x在m,1單調(diào)遞減，在1m1fmin(xf(1em0

6f(x)在m,m1單調(diào)遞減， (x)f(m1)(m 7(m2)em m綜 f(x)在m,m1上的最小值

0m(m1)em1 m由（Ⅰ）知f(x)x2ex

8f'(x)ex+x2exx1exf'(x0x因?yàn)閒(0

fmax(x

fmin(x)

11x1x2[02]，都有|f(x1f(x2|fmax(xfmin(x) 1326、★★★（2014東城二模文f(xlnxa(a0)x(Ⅰ)f(x(Ⅱ)P(x0y0yf

x(03

P(x0y0k1恒成立，求實(shí)數(shù)a的最小值2

f(x)lnxa，定義域?yàn)?0,) 1x則f|(x)1ax 3 a0,f(x)0x(a,f(x)0x(0a所以f(x)的單調(diào)遞增區(qū)間為(a,),單調(diào)遞減區(qū)間為(0, 7(Ⅱ)P(x0y0為切點(diǎn)的切線的斜率kkf(x)x0a (3

0 9x2 x20a1x2x對(duì)3

0恒成 112 又當(dāng)x0時(shí),31x2x1 13 2 所以a2

……1427、★★★（2013朝陽二模文f(x)

x21

a，g(x)alnxx（a0f(xa0x1x20eg(x1f(x2成立（Ⅰ） a(1

a(1x)(1f(x)(x21)2

(x2a0xf(xf(xx(,1f00f↘↗↘a0xf(xf(xx(,1f00f↗↘↗a0f(x的單調(diào)遞增區(qū)間為(1,1，單調(diào)遞減區(qū)間為(1(1；a0時(shí)，f(x)的單調(diào)遞增區(qū)間為(1(1)，單調(diào)遞減區(qū)間為(1,1. 5（Ⅱ）由（Ⅰ）a0f

(0,1)

f(x)f(0)

f

(1e]f(e)

e21

aax0e]f(x)ag(xalnxxg(xa1xg(x)0xa①當(dāng)0aeg(x0，得0xag(x0xagx在(0a上單調(diào)遞增，在(ae上單調(diào)遞減g(x)maxg(aalnaaaalnaaa(2lnaa(2lne)a0，x1x20eg(x1f(x2.aeg(x)0在(0egx在(0eg(x)maxg(eaeax1x20eg(x1f(x2綜上所述，對(duì)于任意x1,x20,e，總有g(shù)(x1)f(x2 1328、★★★（2014東城二模文已知aRf(x1x31(a2)x2bg(x)2alnx yf(xyg(x)在它們的交點(diǎn)(1ca（）F(x)f'(xg(x)，若對(duì)任意的x1x2(0)，且x1x2，都有F(x2F(x1)a，求ax2（Ⅰ）2g'(x)2a，g'(1)2axf'(1)g'(1可得2a(a3)1a1a

（Ⅱ）F(x)1x2(a2)x2alnx2x1x2F(x2F(x1)aF(xF(x

x)x

F(x2ax2F(x1ax1．設(shè)G(x)F(x)ax，xx(0

xF(x2F(x1)a

x 等價(jià)于G(x)F(xax在(0G(x)1x22alnx2x2 x22x可得G'(x)x 2 x0x22x2a0由2ax22x(x1)21，可得a1 13229、★★★（2014東城零模文nf(xxnbxn

(nN,b,c（Ⅰ）n2b

c1

(x在區(qū)間1,1 （Ⅱ）n2x1x2[1,1，有|f2(x1f2(x2|4，求b（Ⅰ） fn2)fn(1)2n21fn(x)在（2內(nèi)存在零點(diǎn)x

(2,1)時(shí)，fn(x)

10f

11）內(nèi)存在唯一零點(diǎn)在（，

在（， 2（Ⅱ）當(dāng)n22

f(x)

6bxcM4,b2b2

1b2時(shí)2M

(b)(b1)24恒成立 2M

(b)2

(-2

24恒成立綜上可知，-2b2 ．1430、★★★（2014朝陽期末文f(x)lnxg(x)ax1aRF(x)

f(x)g(x)yf(xxeF(xa0F(x沒有零點(diǎn)，求a的取值范圍【答案】解：(I)f(x)1f(xxek1 f(e1f(xxey11(xe),y1

4F(x)lnxax1

F(x)1a1ax

x0①當(dāng)a≤0F(x)0F(x在區(qū)間(0②當(dāng)a0F(x0x1F(x)0，解得0x1 綜上所述，當(dāng)a≤0F(x的增區(qū)間是(0當(dāng)a0時(shí)，函數(shù)F(x)的增區(qū)間是(0,1)，減區(qū)間是(1, 9 F(xF(x)lnxax10無解由（Ⅱ）知，當(dāng)a0F(x在區(qū)間(01上為增函數(shù),區(qū)間1 F(1a10F1ln1a11lna20 ae2所以實(shí)數(shù)a的取值范圍為(1, 1331、★★★（2014大興一模文）f(xx3x2xaa2yf(x)在點(diǎn)(0,f(0yf(xa的范圍【答案】解:(Ⅰ)f(xx3x2xa得xR,f(x)3x22x 2當(dāng)a2時(shí)，f(0)2，kf(0) 3切線方程:yx 4

f(x)3x22x1(3x1)(x1 f(x)0得x1x

……2 x,1 3 31 1f+0-0+f↗↘↗3f(xf(1)3

5

…………6yf(x有且僅有一個(gè)零點(diǎn)，須

a0，或a1 8a

或a1時(shí)，函數(shù)有且僅有一個(gè)零點(diǎn) 932、★★★（2014石景山一模文f(xx22a2lnx(a0f(xx1af(x[1e（Ⅰ）f 分f(x)2x

x

2x2x

2(xa)(xx

2f(x)x1處取得極值， 3所以a的值為 4（Ⅱ）令f(x)0，解得xa或xa（舍 5x(0a)af0f↘↗由上表知f(x)的單調(diào)遞增區(qū)間為(a，)，單調(diào)遞減區(qū)間為(0a) 8[e][e]f(110，只須在區(qū)間[1e]f(x)min0.[1ef f(e)e22a20解得0a

2ea2

10[1a)(aeef(x) f(a)a2(12lna)0，解得0a ee所以1a 12ee當(dāng)0a1時(shí)，f(x)在區(qū)間[1e]上單調(diào)遞增，f(x)minf(1)0，滿足題意.綜上，a的取值范圍為0a 13分e33、★★★（2014延慶一模f(x)x33ax2a(aR(Ⅰ)f(xyf(xx軸有且只有一個(gè)公共點(diǎn)，求a的取值范圍【答案】(Ⅰ)f(x)3x23a 1(1)當(dāng)a0f(x)0f(x在(,)上是增函數(shù)，…2a（2）當(dāng)a0時(shí)，令f(x)0，得x aaa令f(x)0，得x 或xaaaa令f(x)0，得 xaaf(x在

a和

a,)在[a

上是減函數(shù) ………5當(dāng)a0f(x在區(qū)間(,)單調(diào)遞增，所以題設(shè)成立………6aa當(dāng)a0時(shí)，f(x)在x 處達(dá)到極大值，在x 處達(dá)到極小值aaf(a0f(a0

a)3

a2a0或

3

2a

aa

a2a0a

aa

a2aa

………110a

………12（1（2） 1334、★★★（2014海淀二模文f(x)1x3ax24xba,bR且a03f(x)在點(diǎn)(0,f(0f(x)f(x在區(qū)間(1,1上有且僅有一個(gè)極值點(diǎn)，求實(shí)數(shù)a的取值范圍 】：

f'(x)x22ax

1f'(04 2f(0f(x)在x