11、1,12,由于y|2x4|在(,2上單調(diào)遞減,在2,)上單調(diào)遞增, 所以f(x)在(,2上單調(diào)遞增,在2,)上單調(diào)遞減.,1,2,3,4,5,6,7,8,9,10,11,12,3.函數(shù)f(x)ln xex(e為自然對數(shù)的底數(shù))的零點所在的區(qū)間是,,解析,答案,解析函數(shù)f(x)ln xex在(0,)上單調(diào)遞增, 因此函數(shù)f(x)最多只有一個零點.,,,,函數(shù)f(x)ln xex(e為自然對數(shù)的底數(shù))的零點所在的區(qū)間是,解析y (0 x3), 當0 x3時,3(x1)211, e3 e1, 即e3ye, 函數(shù)的值域是(e3,e.,1,2,3,4,5,6,7,8,9,10,11
12、,12,4.函數(shù)y (0 x3)的值域是 A.(0,1 B.(e3,e C.e3,1 D.1,e,,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,5.函數(shù)f(x)axloga(x1)在0,1上的最大值和最小值之和為a,則a的值為,,解析,答案,解析當a1時,由aloga21a, 得loga21,,當0a1時,由1aloga2a,,1,2,3,4,5,6,7,8,9,10,11,12,,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,解析當x0時,f(x)2x1.令f(x)0,解得x ; 當x0時,f(x)exa,此時函數(shù)f(x)exa在(,0上有且僅
13、有一個零點,等價轉(zhuǎn)化為方程exa在(,0上有且僅有一個實根, 而函數(shù)yex在(,0上的值域為(0,1,所以0a1, 解得1a0.故選D.,7. 已知函數(shù)f(x) (aR),若函數(shù)f(x)在R上有兩個零點,則a的取值范圍是 A.(,1) B.(,0) C.(1,0) D.1,0),1,2,3,4,5,6,7,8,9,10,11,12,,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,A.a0,b0,c0,c0 C.a0,c<0 D.a<0,b<0,c<0,,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,當f(x)0時,axb0,,1,2,3,4,5
14、,6,7,8,9,10,11,12,9.已知冪函數(shù)f(x)(n22n2) (nZ)的圖象關(guān)于y軸對稱,且在 (0,)上是減函數(shù),那么n的值為___.,解析,答案,1,解析由于f(x)為冪函數(shù),所以n22n21, 解得n1或n3,經(jīng)檢驗,只有n1符合題意.,10.函數(shù)f(x)的定義域為實數(shù)集R,f(x) 對于任意xR都有f(x2) ,若在區(qū)間5,3內(nèi)函數(shù)g(x)f(x)mxm恰 有三個不同的零點,則實數(shù)m的取值范圍是__________.,1,2,3,4,5,6,7,8,9,10,11,12,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,解析f(x2)
15、f(x2),f(x)f(x4),f(x)是以4為周期的函數(shù), 若在區(qū)間5,3上函數(shù)g(x)f(x)mxm恰有三個不同的零點, 則f(x)和ym(x1)的圖象在5,3上有三個不同的交點,ym(x1)恒過點C(1,0),畫出函數(shù)f(x)在5,3上的圖象,如圖所示,,1,2,3,4,5,6,7,8,9,10,11,12,11.設(shè)函數(shù)f(x) 則函數(shù)yf(f(x))1的零點個數(shù)為____.,解析,答案,2,解析當x0時,yf(f(x))1f(2x)1log22x1x1,令x10, 則x1,顯然與x0矛盾, 所以當x0時,yf(f(x))1無零點. 當x0時,分兩種情況:當x1時,log2x
16、0,yf(f(x))1f(log2x)1log2(log2x)1, 令log2(log2x)10, 得log2x2,解得x4; 當0 x1時,log2x0,yf(f(x))1f(log2x)1 1x1, 令x10,解得x1. 綜上,函數(shù)yf(f(x))1的零點個數(shù)為2.,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,解析如圖,畫出函數(shù)f(x)和g(x)在0,4上的圖象, 可知有4個交點,并且關(guān)于點(2,0)對稱, 所以y1y2y3y40,x1x2x3x48, 所以f(y1y2y3y4)g(x1x2x3x4),