2012年高 1（2012）

ln(x1)

yf(x（2012） 若2a2a2b3b,則 B．若2a2a2b3b,則C．若2a2a2b3b,則 D．若2a2a2b3b,則（2012（重慶理f(xRf(x,且y(1x)f(x的圖像如題(8)圖所示,則下列結(jié)論中一定成立的是 f(xf(2)ff(xf(2ff(xf(2)ff(xf(2f（2012） x1為f(x)的極大值 B．x1為f(x)的極小值C．x1為f(x)的極大值 D．x1為f(x)的極小值．（2012年高考（山東理））a0a1f(xaxR上是減函數(shù)g(x)(2a)x3在R上是增函數(shù)” ．（2012年高考（理））已知二次函數(shù)yf(x)的圖象如圖所示,則它與x軸所圍圖形的面積yyOx第35

3

2

π2（2012A．4

5

6

D．7（2012） 2或 B．9或 C．1或 D．3或22（2012）

f(xABCA(0,0),B1函數(shù)yxf(x)(0x1)的圖像與x軸圍成的圖形的面積 .x10（2012 與直線xa,y0所圍成封閉圖形的面積為a2,則ax 11（2012年高考（江西理）計(jì)算定積分1(x2sinx)dx 12（12 理）曲線yx3x3在點(diǎn)1,3處的切線方程 13（2012年高考（ 理）已知函數(shù)f(x)=xln(x+a)的最小值為0,其中a>0.(Ⅰ)求a的值;x[0,+f(xkx2成立,求實(shí)數(shù)kn(Ⅲ)證明2ln(2n+1)<2(nN*ni=12i14（2012（新課標(biāo)理）f(xf(xf(1)ex1f(0)x1x22f(xf(x1x2axb,求(a1)b215（2012年高考（浙江理））已知a>0,bR,函數(shù)fx4ax32bxab

fx+|2a-AEDBCAEDBC16（2012）f(xalnx求a

13x1其中aRyf(x在點(diǎn)(1,f(1y f(xn17（2012（陜西理）f(xxnbxn

(nN,b,c設(shè)n2b

c1

(x在區(qū)間1,1 設(shè)n2x1x2[1,1,有|f2(x1f2(x2|4,求b,在(1)xf(x在1,1x,

18（2012（山東理f(xlnxkke2.71828yf(x在點(diǎn)(1,f(1x求kf(x)g(x)x2xf'(xf'(xf(xx0g(x1e2x19（2012）x3

axb(abRa,b為常數(shù)yf(xy

x在(0,0)2求ab證明:當(dāng)0

f(x)

9xx20（2

f(xxx0x0y

f(x已知a，b是實(shí)數(shù),1和1是函數(shù)f(x)x3ax2bx的兩個(gè)極值點(diǎn).[來源:求a和bg(xg(x)f(x2g(x設(shè)h(xff(xc,其中c[2，2]yh(x21（2012）x∈R,f(x≥1af(xA(x1,f(x1B(x2,f(x2))(x1x2,記直線AB的斜率為K,問:是否存在x0∈(x1,x2),使f(x0kx0的取值范圍;若不存在,請(qǐng)說明理由.22（2012(Ⅰ)

f的設(shè)a10,

0,b1,

為正有理數(shù).若b1

1,則abab 題 注:當(dāng)為正有理數(shù)時(shí),有求導(dǎo)(x)x123（2012年高考（廣東理））(不等式、導(dǎo)數(shù))設(shè) Bfx2x331ax26axD24（2012）

a

yf(x在點(diǎn)(1,f(1xf(xayf(xP,曲線在該點(diǎn)處的切線與曲線只有一個(gè)P.25（12）f(xaxcosxx[0,].f(xf(x1sinxa26（12 理）已知函數(shù)f(x)ax21(a0),g(x)x3bxyf(xyg(x在它們的交點(diǎn)(1,c)處具有公共切線,求ab當(dāng)a24bf(xg(x的單調(diào)區(qū)間,并求其在區(qū)間(127（2012）((I)f(x在[0

b(a(II)yf(x在點(diǎn)(2,f(2y3x;求ab22012年高 【解析】選B[來源:g(x)ln(1x)xg(x) 1g(x)01x0,g(x)0x0g(x)g(0)x0或1x0f(x

AC2a2a2b3b,必有2a2a2b2bfx2x2xfx2xln220恒成fx2x2xx>0a>b成立.其余選項(xiàng)用同樣方法排除.x2,1x0,由(1xf(x0f(x0f(x2x1,1x0,由(1xf(x0f(x0f(x為減;1x2,1x0,由(1xf(x)0f(x)0f(x為減;x2,1x0,由(1x)f(x)0f(x)0f(x為增.00f(xx1)exf(x0x1x<-1f(x0f(xxexx>-f(x0f(xxexx1f(xf(x)ax在R0a1g(x)(2a)x32a0,所以a2f(x)axRg(x)2a)x3為增函數(shù)”A.

yf(xx21S1(x21)dx(1x3x)14

x 2 x

x)dx

x2

x2 0

6S

1,故P ,答案6x軸有兩個(gè)不同x軸恰有兩個(gè)公共點(diǎn),結(jié)合該函數(shù)的圖像,可得極大值或者極小值為零即可滿足要求.而f(x)3x233(x)(x1)x1時(shí)取得極值f(10f(10可得c20或c20,即c2y y 1xy5PMNODx 0x1,f(x)

221010x,1x2yxf(x)

0x2,210x210x,1x ,y=xf(x)的分段解析式中的兩部分拋物線形12方向及頂點(diǎn)位置不同,如圖2MNOOMPODMPS=155 [評(píng)注]對(duì)于曲邊圖形,現(xiàn)行中不出微積分,能用微積分求此面積的考生恐是極少的,而對(duì)于極大aa

2 2x x

S2

x2|a

a2a2,所以a2

,所以a 3

sinx)dx3cosx|13cos13cos1

3

解析:2xy10.y 31212,所以切線方程為y32x1,即2xy10f(x的定義域?yàn)?af(x)xln(xx

f(x)minf(1a1a0a（2）g(xkx2f(xkx2xln(x1)(xg(x)0x[0,+上恒成立g(x)min0g(1)k1ln20kg(x)2kx

x

x(2kx2kx①當(dāng)2k10(k1)g(x00x12k

g(xg(00與2②當(dāng)k1g(x02得：實(shí)數(shù)k12

g(0)0符合（3）由（2）xln(x11x2x02x

,n)： [ln(2i1)ln(2i1)] (2ii i當(dāng)n12ln3

當(dāng)i2時(shí)

(2i 2i1ln(2i1ln(2i12ln312n1 【解析】(1)f(xf(1)ex1f(0)x1x2f(xf(1)ex1f(02x1f(0f(x)f(1)ex1x1x2f(0)f(1)e112f(x)exx1x2g(x)f(x)ex12g(x)ex10yg(xxR

f(x)0f(0)x0,f(x)0f(0)xf(xf(xexx12且單調(diào)遞增區(qū)間為(0,單調(diào)遞減區(qū)間為((2)f(x1x2axbh(xexa1)xb0得h(x)exa2①當(dāng)a10h(x0yh(xxRxh(x)與h(x)0②當(dāng)a10h(x0xln(a1h(x0xln(axln(a1h(x)mina1a1ln(a1b0(a1)b(a1)2a1)2ln(a1)(a10)F(xx2x2lnx(x0F(xx(12ln

eF(x)00x e,F(x)0xeee當(dāng)x 時(shí),F(x)maxeee當(dāng)a 1,b 時(shí),(a1)b的最大值為ee2fx12ax22bfxf14a2bab3ab=|2a-b|﹢a;b>0fx12ax22b0≤x≤1此時(shí)fx的最大值為:

(0)，f}bfx+|2a-b|﹢a≥0,gx=﹣fx≤|2a-b|﹢a.gx0≤x≤1)|2a-b|﹢bgx4ax32bxabgx12ax22b0b≤0gx12ax22b<00≤x≤1gxg0ab3ab=|2a-

x bb<0gx12ax22b0≤xb

xmax{g( }3bbb3b

bbfx0≤x≤1﹣(|2a-b|﹢a)要大.baba

和b33ab

z=a+bP(1,2zmax3zmin解:(1)fxalnx13x1fxa1 從而a130a (2)由(1)知fxlnx x1x0 3x22xfx x

fx(3x1)(x 令fx0,解得x1,x (因x fxx1f13.n解析:(1b1c1n2f(xxnxn1∵f()

(1)10,∴

(x在1,1n

1 1

x1,1f(xnxn11 ∴f(x在1,1f(x在1,1 2當(dāng)n2f(xx2bx2x1x2[1,1都有|f2(x1f2(x2|4f2(x在[1,1M4此分類討論如下:(ⅰ)當(dāng)|b|1,即|b|22M|f2(1f2(1|2|b|4當(dāng)1b0,即0b22M

2(1)

(b)b1)24 當(dāng)0b1,即2b02M

2(1)

(b)b1)24 2b用max{ab}表示ab中的較大者.當(dāng)1b1,即2b22Mmax{

(1),f(1)}f(b f2(1)f2(1)|f2(1)f2(1)| b2b1c|b| 4

(b)1|b|)242證法一xf(x在1,1內(nèi)的唯一零點(diǎn)(n f(x)xn

1, )xn1

10,

1

fn(xn)0fn1(xn1)又由(1)

(x在1,1x

(n2)

,,

證法二xf(x在1,1

(xn)

[來 xn1x1xnx1 fn1(xxn1在(xn,1xnxn1(n2,,

1klnf(x)

lnxkf(x)ex ex

f(10,即1k0,解得k1e11lnf(x) f(x)0x1,ex當(dāng)0x1f(x)11lnx0x1f(x)11lnx0 f(x在區(qū)間(0,1內(nèi)為增函數(shù);在(1,11lng(x)(x2x)x 1 ,exx1時(shí),1x20lnx0x2x0ex0g(x01e211ln當(dāng)0x1時(shí),要證g(x)(x2x) 1e2.ex

x1ex

1xp(x)

,q(x)1x(1lnx), 則p(x)x0,q(x)2lnx,x ex則當(dāng)0x1p(x)x1p(0)1q(x2lnx0xe20,1x0e2時(shí)q(x)0xe21q(x0則當(dāng)0x1q(x)1x(1lnx)q(e21e2,且q(x)01則

1e21

1,于是可知當(dāng)0x1x1ex

11x(1ln綜合(1)(2x>0,g(x)1e21:p(x)x1x0,1p(x)x0x則當(dāng)0x1p(x)ex

exp(0)111ln于是當(dāng)0x1時(shí),要證g(x)(x2x) x(11lnx)1e2,ex x11lnx)1e2xq(x1x(1lnxx0,1q(x)1x(1lnxq(x2lnx0xe20,1x0e2時(shí)q(x)0xe21q(x0則當(dāng)0x1q(x)1x(1lnxq(e21e211ln于是可知當(dāng)0x1時(shí)(x2x) 1e2成立ex綜合(1)(2x>0,g(x)1e22:根據(jù)重要不等式當(dāng)0x1時(shí)ln(x1xx1ex11ln于是不等式g(x)(x2x) x(11lnx)1e2,ex q(x1x(1lnxx0,1q(x)1x(1lnxq(x2lnx0xe20,1x0e2時(shí)q(x)0xe21q(x0則當(dāng)0x1q(x)1x(1lnxq(e21e211ln于是可知當(dāng)0x1時(shí)(x2x) 1e2成立.exf(x)yf(x)y3x在(0,0ab2數(shù)的單調(diào)性或者均值不等式證明f(x)

x

【答案】解:(1f(xx3ax2bxf'(x3x22axb∵11f(xx3ax2bx∴f'(132ab=0f'(132ab=0a=0，b3∵由(1f(xx33x g(xf(x2=x33x2=x12x2x=x=1，x x<2g(x0;當(dāng)2x<1g(x0x2g(x∵當(dāng)2x<1x>1g(x0g(x)的極值點(diǎn)是-f(x)=th(xf(tc

x=1g(xxf(x)=dd2,d=2時(shí),由(2)可知,f(x2I2,注意到f(xf(x)=2的兩2.d2f(1d=f(2)d=2d0f(1d=f(2)d2d021,1,2f(x)=d的根.由(1)知f'(x)=3x1x1.

f(xf(x>f(2)=2f(1d0f(2)d0y=f(xdf(x)=d在(1,2f(x)=d2If(1d0,f(1d0y=f(xdf(x)=d1,1d=2f(x)=dx1，x2x1=1，x2=2d<2f(x)=dx3，x1，x5xi2，i=3,4,5yh(xic=2f(t)=c有兩個(gè)根t1，t2t1=1，t2=2f(x)=t1f(x)=t2yh(x511c2f(t)=c有三個(gè)不同的根t3，t4，t5

<2，i=3,4,5f(x)=tii=3,4,5yh(x9c=2yh(x5c<2yh(x9【解析】(1y

f(x11y

f(x由(1)f(x)x33xg(x)g(x)=0d=2d<2xf(x)=dyh(x【解析】(Ⅰ)若a0,則對(duì)一切x0,f(x)eaxx1,這與題設(shè),又a0a0 而f(x)aeax1,令f(x)0,得x x1ln1f(x0,f(xx1ln1f(x0,f(xx1ln 時(shí),f(x)取最小值f(1ln1)1 xR,f(x1 g(tttlntg(tln當(dāng)0t1g(t)0g(t單調(diào)遞增;當(dāng)t1g(t0g(t故當(dāng)t1g(tg(1111即a1af(x)f(x

eax2(Ⅱ)由題意知,k 1 x2 x2 eax2令(x

k

x2(x1)

a(x2x1)

a(xx) (x2

1ex2e

x2)F(t)ett1F(t)et1當(dāng)t0F(t0F(t單調(diào)遞減;當(dāng)t0F(t0F(t單調(diào)遞增.故當(dāng)t0F(t)F(0)0即ett10.從而

a(x2x1

a(x2x1)10,

a(x1x2

a(x1x210

x2

x2

所以(x10(x2y(x)在區(qū)間x1x2x0(x1x2使(x0 2

eax2(x)a

0,(x)單調(diào)遞增,故這樣的c是唯一的,且c

a(x2x1

x

eax2

時(shí),f(xk(

x),x2 x0(x1x2f(x0kx0 eax2(

x),x2) 論思想、函數(shù)與方程思想,轉(zhuǎn)化與劃歸思想等數(shù)學(xué)思想方法.第一問利用導(dǎo)函數(shù)法求出f(x)取最小值 f(

x∈R,f(x)1f

考點(diǎn)分析:本題主要利用導(dǎo)數(shù)求函數(shù)的最值,并結(jié)合推理,數(shù)學(xué)歸納法,對(duì)考生的歸納推理能力有f(xrrxr1r(1xr1f(x)0x當(dāng)0 時(shí),f(x)0,所以f(x)在(0,1)內(nèi)是減函數(shù)當(dāng)x 時(shí),f(x)0,所以f(x)在(1,)內(nèi)是增函數(shù)故函數(shù)f(x)在x1處取得最小值f(1)0 (Ⅱ)由(Ⅰ)知,當(dāng)x(0,)時(shí),有f(x)f(1)0,即xrrx(1r) 若a,a中有一個(gè)為0,則abababab成立; 1 2若a1,a2均不為0,又b1b21,可得b21b1,于是[來 xa1rb,可得a1bba11baa ) aa 即aba1baba(1b,亦即abababab 1 1 2綜上,對(duì)a0a0bb為正有理數(shù)且bb1abababab 1 2(Ⅲ)(Ⅱ)中命題的推廣形式為:[來源:ababab 1 2an設(shè)a1ababab 1 2an若b1

1,則ab 11當(dāng)n1b11,有a1a1ababab 1 2ak假設(shè)當(dāng)nk時(shí),③成立,即若a1,ababab 1 2ak且b1

1,則ab1當(dāng)n 時(shí),已知a1,a2 ,ak,ak1為非負(fù)實(shí)數(shù),b1,b2 ,bk,bk1為正有理數(shù)1且b1b2 bkbk11,此時(shí)0bk11,即1bk10,于

ab

abab(ab

ab)ab=(a1ba1b

a1b)1bab1

k

k因

b

b

ababa1ba1b

a1ba a

11 22 12k 12k

aba

a

ababkk1

1 2 kk1

bk又因(1bk1bk11

k aba

a

aba

a 1 2 kk ab1 2 kk

k

1ababk akbkakababk從而ab

a

aba 1故當(dāng)n

2

k k1k由(1)(2)可知,對(duì)一切正整數(shù)n,所推廣題成立說明:(Ⅲ)中如果推廣形式中③式對(duì)n2成立,則后續(xù)證明中不需討論n1的情況解析:(Ⅰ)考慮不等式2x231ax6a01a10BRDA03②當(dāng)a1時(shí),0,此時(shí)Bxx1,D 1,3a

,此時(shí)2x231ax6a

有兩根,設(shè)為x

x

xxx1

331a3a33a

,x2

331a3a33a4Bxxx1或xx2.當(dāng)0a1時(shí),xx31a0,xx3a0,所以xx0,此時(shí)D0,x x,;當(dāng)a 1 x1x23a0x10x20Dx2綜上所述,當(dāng)1a13

DA0,

a3

D

0a131a3a33a x2;當(dāng)

a0時(shí),

.其中

x1 31a31a3a33afx6x261ax6afx0可得xax10a1fx0m1am21,且m1m21a1DA0fx0D內(nèi)有兩根mam1 xa1f+0-0+fxfxD1,極小值點(diǎn)a②當(dāng)a1時(shí),D Z。X。X。x 3 1313 f+0-+fxfxD內(nèi)只有極小值點(diǎn)a③當(dāng)0a1時(shí),D0,x x,,此時(shí)0ax1

fx0D 只有一根m1axaf+0-+fxfxD內(nèi)只有極小值點(diǎn)a④當(dāng)a0時(shí),Dx2,,此時(shí)x21,于是fx在 1a1fxD1,極小值點(diǎn)a;當(dāng)0a1fxD 值點(diǎn)a,沒有極大值點(diǎn).當(dāng)a0fxD解

f(x)ex2axe,k

f(12a0a0f(xexx1f(x0x1f(x0f(x的增區(qū)間為(1,減區(qū)間為(2)P(x0y0yf(x0)(xx0f(x0g(xf(xf(x0)(xx0f(x0),因?yàn)橹挥幸粋€(gè)切點(diǎn),所以函數(shù)g(x)就只有一個(gè)零點(diǎn),因?yàn)間(x0) g(x)f(x)f(x)exex02a(xx),若a g(x) g(xg(x00P的任意性知a0不合題意a0,令h(x)exex02a(xxh(x)0 h(xex2aP(ln(2a),f(ln2aa的取值范圍為a0f(xasinxx[0,,所以0sinx1a1f(x0f(xx[0,上為單調(diào)遞增函數(shù);a0f(x0f(xx[0,上為單調(diào)遞減函數(shù);當(dāng)0

時(shí),由f(x)0得sin f(x0得0xf(x0得arcsina

或 .所以當(dāng)0

時(shí)f

[0arcsin

[arcsina,

[arcsina,arcsina]f(x1sinxaxcosx1sinxax1sinxcosx001sin0cos00當(dāng)0xax1sinxcosxa1sinxcosxa

1sinxcosxg(x)

1sinxcosx

(0x

g(x)(cosxsinx)x1sinxcosx(1x)cosx(x1)sinx 又令c(x1xcosxx1sinx1,則c(x)cosx(1x)sinxsinx(x1)cosxx(sinxcosx(03sinxcosx0,故c(x0c(x4x(3,sinxcosx0,故c(x0c(x42所以c(x)在x(0,