9、,4,5,6,7,8,9,10,11,12,答案,解析,1,解析由題意知,f(2)541,f(1)e01, 所以f(f(2))1.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,3,3.已知函數(shù)f(x)在(,)上單調(diào)遞減,且為奇函數(shù).若f(1)1,則滿足1f(x2)1的x的取值范圍是________.,解析f(x)為奇函數(shù),f(x)f(x). f(1)1,f(1)f(1)1. 故由1f(x2)1,得f(1)f(x2)f(1). 又f(x)在(,)單調(diào)遞減, 1x21, 1x3.,4.如果函數(shù)f(x)ax22x3在區(qū)間(,4)上是單調(diào)遞增的,則實數(shù)a的 取值范圍是____
10、______.,解析當a0時,f(x)2x3,在定義域R上是單調(diào)遞增的,故在(,4)上單調(diào)遞增;,因為f(x)在(,4)上單調(diào)遞增,,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,cab,5.已知定義在R上的函數(shù)f(x)2|xm|1(m為實數(shù))為偶函數(shù).記af(log0.53),bf(log25),cf(2m),則a,b,c的大小關系為__________. 解析由f(x)2|xm|1是偶函數(shù),得m0,則f(x)2|x|1. 當x0,)時,f(x)2x1單調(diào)遞增, 又af(log0.53)f(|log0.53
11、|)f(log23),cf(0),且0log23log25, 則f(0)f(log23)f(log25), 即cab.,為___________________.,解析因為f(4)2a3,所以a1.,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,所以f(x)在區(qū)間3,4上單調(diào)遞減,,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,0,解析由題意知,f(x2)f(x2), f(x)f(x4), 又f(x)f(x2),f(x4)f(x2), f(x2)f(x),f(x4)f(x), f(x)的周
12、期為4, 故f(2 018)f(2 0162)f(2)f(0)0.,,1 009,解析由所給函數(shù)知,,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,(,2)(2,),解析當a0時,a2a3(a)0a22a0a2; 當a0時,3a(a)2(a)0a<2. 綜上,實數(shù)a的取值范圍為(,2)(2,).,12.能夠把圓O:x2y216的周長和面積同時分為相等的兩部分的函數(shù)稱為圓O的“和諧函數(shù)”,下
13、列函數(shù)是圓O的“和諧函數(shù)”的是________.(填序號) f(x)exex;,1,2,3,4,5,6,7,8,9,10,11,12,答案,解析,,解析由“和諧函數(shù)”的定義知,若函數(shù)為“和諧函數(shù)”, 則該函數(shù)為過原點的奇函數(shù), 中,f(0)e0e02,所以f(x)exex的圖象不過原點, 故f(x)exex不是“和諧函數(shù)”;,1,2,3,4,5,6,7,8,9,10,11,12,1,2,3,4,5,6,7,8,9,10,11,12,中,f(0)tan 00,f(x)的定義域為x|x2k,kZ,,中,f(0)0,且f(x)的定義域為R,f(x)為奇函數(shù), 故f(x)4x3x為“和諧函數(shù)”, 所以中的函數(shù)都是“和諧函數(shù)”.,本課結(jié)束,