高聯(lián)基礎(chǔ)-2講義-秋奢華版

目標(biāo)2017高中數(shù)賽基礎(chǔ)班函數(shù)-2:函數(shù)的性質(zhì)主講教師:(1PUZZLE-xyxy1.f(x,y x+=[x][y]+[x]+[y]+([xx的最大整數(shù)證明任何定義域關(guān)于原點(diǎn)對(duì)稱的函數(shù)都可以表示為一個(gè)奇函數(shù)和一個(gè)偶函數(shù)的y=f(xxf(3+x)=f(3?f(x06個(gè)不同的實(shí)根.6個(gè)實(shí)根的和f(xlogax[2π1.a的值f(x(?∞+∞上的偶函數(shù)f(x(?∞0)上是增函數(shù),f(2a2af(3a22a1).a的取值范圍xyRxy的值

(x?1)3+2017(x?1)=(y?1)3+2017(y?1)=y=f(xax2bxcy軸為對(duì)稱軸.ab1, (x,y)在y=f(x)的圖像上,則點(diǎn)x,y2+1在函數(shù)g(x)= g(x)之解析式F(xg(xλf(xλF(x

p?∞,?

內(nèi)是減函數(shù)( 2?22

證明:函數(shù)f(x)=3x2可以表示為兩個(gè)單調(diào)遞增的多項(xiàng)式函數(shù)f(x)x滿足f(2?x)=f(2+x),f(7?x)=f(7+f(00.求證:f(x[?303013個(gè)零點(diǎn)f(x10為周期的函f(x)的定義域關(guān)于原點(diǎn)對(duì)稱,0;x,在定義域x1x2,x=x1?x2,f(x1?=f(x2),且滿足以下三個(gè)條件:x1x2f(x定義域中的數(shù),f(x1?=f(x20|x1x2|2af(x

x)=f(x1)f(x2)+112f(a1a是一個(gè)正常數(shù)0x2a時(shí),f(xf(x是奇函數(shù)f(x)是周期函數(shù),并求出其周期f(x)(04a)內(nèi)為減函數(shù)

f(x2)?ff(x),x(?∞+∞f(xTkf(xkT均為正常數(shù).求證:f(x=ax?(x),a為正常數(shù),?(x)T為周期的函數(shù).a為何值時(shí)loga

x2+ax+5+

·

x2+ax+

loga30x滿足條件

f(x+a)=1

f(x)?[f證明:f(x是周期函數(shù)f(xx5ax3bxc3x8abcRf(?2)10.f(2的值f(x)具有如下性質(zhì):x,f(x)+f(x?1)=f(19)94,那么f(94)1000的余數(shù)是多少f(x在定義域(?11)內(nèi)單調(diào)遞減當(dāng)m取何值時(shí),f(1?m)f(1m2)x[?1,1]時(shí),

f(x)

x4+4x3+17x2+26x+x2+2x+xyzabcr0.證明x+y+a+ y+z+b+ x+z+a+ x+y+a+b+c+ y+z+a+b+ x+z+a+b+c+f(x)(0+∞),且單調(diào)遞增,f(2f(xy)=f(x)+f(y證明:f(1)ff(xf(x3)2,x的范圍a為