培優(yōu)2教師版函數(shù)性質(zhì)專題

“函數(shù)的性質(zhì)”專題培優(yōu)1（2009湖南卷文）yf(x在區(qū)間[abyf在區(qū)間[abAy y b b b bA yf(x．yf(x在區(qū)間[ab上是增函數(shù)，即在區(qū)間[ab上k是遞增的，由圖易知選A.注意Cyk為常數(shù)噢.22、

2

（A）關(guān)于原點(diǎn)對(duì) （B）關(guān)于主線yx對(duì)（C）關(guān)于y軸對(duì) （D）關(guān)于直線yx對(duì)

exeexe

yy1 xyy1yy11OxOxBCy1O1xA【解析】函數(shù)有意義需使exex0其定義域?yàn)閤|x0排除C,D又因?yàn)?yexe2

e2x1

22exe答案

4、(2009Rf(xf(x4f(x,且在區(qū)間 上是增函數(shù)若方程f(x)=m(m>0)在區(qū)間8,8上有四個(gè)不同的根xxx x1x2x3x4 【解析】:因?yàn)槎x在R滿足f(x4)f(x) f(x4f(x,所以,f(x為奇函數(shù)

xx2f(0)0,f(x4)f(xf(x8)f(x),所以函數(shù)是以8為周期的周期函數(shù),f(x在區(qū)間[0,2]上是增函數(shù),f(x在區(qū)間[- 不妨設(shè)x1x2x3 由對(duì)稱性知x1x2 x3x4 所 x2 x3 x142 與f(x1)都是奇函數(shù)則 (A)f(x)是偶函 (B)f(x)是奇函(C)f(x)f(x (D)f(x3)解 f(x 與f(x1)都是奇函數(shù)，f(x1)f(x1),f(x1)f(x1)f(x關(guān)于點(diǎn)(1,0)，及點(diǎn)(1,0f(x是周期T2[114的周期函數(shù).f(x14)f(x14f(x3)f(x3)f(x3)的值為

log2(1x),xf(x1)f(x2x0f（ B. D.【解析】:f(1)log221,f(0)0,f(1)f(0)f(1)f(2)f(1)f(0)1,f(3)f(2)f(1)1(1)0f(4)f(3)f(2)0(1)1,f(5)f(4)f(3)1,f(6)f(5)f(4)0答案7、(2009Rf(xf(x4f(x,且在區(qū)間 A.f(25)f(11)f B.f(80)f(11)fC.f(11)f(80)f D.f(25)f(80)f【解析】:f(xf(x4f(x,f(x8)f(x,8為周期的周期函數(shù),f(25)f(1,f(80f(0,f(11f(3),f(x在Rf(00,f(80f(00,f(25f(1f(1,f(x4f(xf(11f(3f(3f(14f(1)又因?yàn)閒(x)在區(qū)間[0,2]上是增函數(shù)所以f(1f(00,所以f(10,f(25f(80f(11,故選答案8（2009江西卷文）f(x)()上的偶函數(shù)，若對(duì)于x0f(x2）f(xx[0,2)fxlog2(x，f(2008)f(2009)為

C. x24x6,x9 卷文設(shè)函數(shù)f(x)x6,x B(3,1)

f(xf(1)的解集是（ D(,3)x0f(x2f(13f(xx1x3x0x63xf(xf(13，解得3x1或x5x都有xf(x1)1xf(xf(2 【答案】

C. D. 1【解析】若x≠0，則有f(x1) f(x)，取xx

21 f( f() 1) f()f()f()（∵

f(x) 2

f(1)f(1) 1f()2 5f()

3f 1)

1 22 22 312222

f(

f

1 ]f()5f()113x

， (B)［

(C)（

(D)［， 1

3

得3

解得 12f(xx3bx2cxx=2對(duì)稱解:（Ⅰ）f(x3x22bxc.f(x對(duì)稱，所以6

2，于是b（Ⅱ）由（Ⅰ）f(xx36x2cxf(x3x212xc3(x2)2c12c12f(x0f(xc<12f(x0有兩個(gè)互異實(shí)x1x2.x1x2x1＜2＜x2xx1f(x0，f(x在區(qū)間(x1內(nèi)為增函數(shù)；x1＜x＜x2f(x0f(x在區(qū)間(x1x2內(nèi)為減函數(shù);xx2時(shí)，f(x)0，f(x)在區(qū)間(x2內(nèi)為增函數(shù).f(xxx1xx2處取極小值.因此，當(dāng)且僅當(dāng)c12f(xxx2處存在唯一極小值，所以tx22g(t的定義域?yàn)?2.由f(t)3t212tc0得c3t212tg(tf(tt36t2ct2t36t2t當(dāng)t2g(t6t212t6t(2t0,g在區(qū)間(2,g(t的值域?yàn)?,8).13（1）x、yt的方程t22ata60的兩個(gè)實(shí)根，則(x1)2y1)2A、4

3 D4 ，即a或a∴(x1)2(y1)2x2y22(xy)(xy)22xy2(xy)4a22(a6)4a 4a26a104(a32 由此可知，當(dāng)a3(x1)2y1)2A、 B、解析：當(dāng)a0時(shí)，方程化為2x1

C、 D、1x ，符合題意，故排除12a1x22x1a1x22x1

2x 214（1）設(shè)m

f(x)

x2x

[1,

f(x1(x1)22

∴f(m)m，

m2m3mm24m3 m3或m1（舍去a取值范圍 f(xx2axa23yf(xx軸在(0,2) 2a 0a 0 3f(0)a233f(2)a22a13 a3

a

3或a答案 3,f(x)ax2bxc(b0f(x2)f(x)(xR)A、f(3x)f(2x B、f(3x)f(2xCf(3xf(2xf(x2)f

Df(3xf(2xb∴yf(x)的圖象關(guān)于直線x1對(duì)稱，從而 bb0,a0.yf(x在(,1]上是減函數(shù)，在[1,上是增函數(shù)∵當(dāng)x0時(shí)，3x2x ∴f(3x)f(2x)x03x2xx03x2x

∴f(3x)f(2x)∴f(3x)f(2x15f(xRx1xx

[0,]f(x1x2f(x1f(x2f(1)a(a0) f(f( f(x

x,x 都有f(xx)f(x)f(x) f(f(x)f(x f(