專(zhuān)題212含絕對(duì)值的函數(shù)問(wèn)題研究與拓展

fxf1x對(duì)所有實(shí)數(shù)成立的充要條件（p1p2表示ab為兩實(shí)數(shù)，abp1,p2ab,fab

2

（閉區(qū)間mn的長(zhǎng)度定義為n3x3x（Ⅰ）fx

x恒成立

f1x

x

3xp1x

xp1x

log32（*）xp1x

xp1xp2

p1

p1

fxf1x

p1

log31°p1p2log32xp1faxp1b

,22°

p1

p

log2時(shí).

x

，

x

3p1x,xa,p

3p2xlog32,xa,pf1

pplog xp1,b

f2

32

3

因?yàn)閒1x0,f2x0

f1xf2x，故fx

f1

pplog xap2

f2

31

3

因?yàn)閒1x0f2x0，所以f1xf2x，故2fxfx=3p2xlog32fafb，所以3bp13p2alog32，所以bp

alog2abp1p2log3

xppfx

x，則3px3xplog2x

p1p2log32

2xp,p1p2log32fx

xfx

x=3xp2log3 xp1p2log32,pfx

xfx

1 fx在區(qū)間a,b上的單調(diào)增區(qū)間的長(zhǎng)度和b

p1p2log32 =bp1p2log32babb

plog2時(shí).

x

，

x

3p1x,xa,p

3p2xlog32,xa,pf1

pplog 當(dāng)xp2,b

f2

32

3

因?yàn)閒1x0f2x0，所以f1xf2x，故2fxfx=3xp2log32f1

pplog 當(dāng)xap1

f2

31

3

因?yàn)閒1x0,f2x

，所以f1xf2xfxf1x=3pfa

b，所以3p1a3bp2log32，所以abpplog xppfx

x，則3x

3p2xlog32x

p1p2log32 xp,p1p2log32

fx

xfx

xp1p2log32,pfx

xfx

x=3p2xlog3

1 fx在區(qū)間a,b上的單調(diào)增區(qū)間的長(zhǎng)度和b

p1p2log32=bp1p2log32babb

b21cdcdcdcd的區(qū)間長(zhǎng)度均為dcab x

x

1的區(qū)間長(zhǎng)度之和 x2(ab2)x(ab

(xa)(x

0ab 2f(x)|3x1|, f(x)f1(x),f1(x)f2

|a3x9|a0xRf2(x),f1(x)f2fa=1f(x在（1）f(xm04個(gè)不等的實(shí)根，求實(shí)數(shù)m2a9f(xf2nml的最大值2解:(1)當(dāng)a1時(shí),f(x)|3x9|2

所對(duì)應(yīng)的自變量取值區(qū)間的長(zhǎng)度為l（閉區(qū)間[m,n]3xf1(x)

x

3xf2(x)

x13x,xxlog35f1(x)

f2

93x,x3x9,x f(x)9

,log35x

531,0xlog313x,xf(x)myf(xym由數(shù)形結(jié)合可知，當(dāng)m(0,1)時(shí)方程f(x)m0有4個(gè)不等的實(shí) 9x

9時(shí),因?yàn)閍3x903x103

(xf(xa3x93x1a1)3x80,x

3a1從而當(dāng)

9x3

3a

時(shí),f(x

當(dāng)0x

9時(shí),因?yàn)閍3x903x103

(xf(x9a3x3x110a1)3x0,x

3a1從而當(dāng)

3a

x

9時(shí),f(x3

x0時(shí),f(xf(x9a3x13x8a1)3x0 f(xf2

[log3a1,log3a

]時(shí),f(x)f2(x) 141 故 log3a log3a log3[5 a從而當(dāng)a2時(shí)l取得最大值為

165 3f(x)e|x2a1|,f(x)e|xa|1xR 若a2,f(x)

f1(x+f2(x)x[2，3]x[a時(shí)

f2(xf1(x,ag(x)

f1(xf2(x|f1(xf2(x|x[1，6]上的最小值 【解答】(1)因?yàn)閍2,x[2，3],e3e 3 e3ef(x)

2e x=2時(shí)取等號(hào),f(xx[2，3]上的最小值為3e由題意知,x[a時(shí)e|x2a1|e|xa|1,即|x2a1||xa|1恒成立．所以|x2a1|xa1,即2ax3a22ax[a)恒成立, 2a則由2a23a22a,a的取值范圍是0a2h1(x|x(2a1|h2(x)|xa|1,h1(xh2(x的圖象分別是以(2a-1,0)和(a,1)為頂點(diǎn)V型線,且射線的斜率均為1.設(shè)h1(xH(ah2(x的最小值為G(a

H(aG(a

2a,aH(a)0,1a

G(a1,1a6minH(a),G(a)的圖像如下（粗體線部分a,a2a7,a

綜上所述,g(xx[1，6]

g(x)minee

a0a 1a e e .如果存在實(shí)數(shù)a和函數(shù)h(x)其中x1,h(x)>0f'(x)h(x)(x2ax1f(xP(af(x)lnxb2(x1，其中bx（i）f(xP(b；（ii）f(x（2）gxP(2.x1x2(1x1x2mmx11m)x21m)x1mx2，且11若|g(g(|<|g(x

|，求m的取值范圍（1）(i)

1x

b2

(x2bxx(xx1h(x)

x(x

0f(xP(b2

b (ii)（方法一）設(shè)(xxbx1x)1

xf'(x 當(dāng)1 0,2b2時(shí)，(x)0，f'(x)0，故此時(shí)f(x)在區(qū)間(1,)上遞增4當(dāng)b2x1f'(x)0f(x在區(qū)間(1,當(dāng)b2時(shí)，xxb1，而(0)2x1，總有x)0f'(x)0f(x在區(qū)間(1,（方法二）當(dāng)b2x1(xx2bx1x2f'(x)0f(x在區(qū)間(1,bbbb2bb2當(dāng)b2時(shí)，(x)圖像開(kāi)口向上對(duì)稱(chēng)軸x 方程(x)0的兩根為bbb2bb2

而

b

b2bb2bb2當(dāng)x )時(shí)，(x)0，f'(x)bb2bb2

bb2理得：f(x)在區(qū)bb22綜上所述，當(dāng)b2f(x在區(qū)間(1,bb2bb2當(dāng)b2時(shí)，bb2bb2 2

,2(2)（方法一）g'(x)h(x)(x22x1)h(x)(x又h(x)x1,都有h(x)x1,g(x)0gx在(1)上遞增。又x1x2,2m1)(x1x2。當(dāng)m1m1時(shí)，，且xm1)x1m)xx1m)xm1)x （0，1（方法二由題設(shè)知，gxg'(x)h(x)(x22x1h(x)0xx1g'(xh(x)(x1)20gx在區(qū)間(1,mx11m)x2mx21m)x2x2，得(x1x2(x1x2gx的單調(diào)g()g()(g(x1),g(x2，從而有|g()g(|<|g(x

②當(dāng)m0時(shí)，mx11m)x2mx21m)x2x2

，