（江蘇專用）高考數(shù)學(xué) 專題2 函數(shù)概念與基本初等函數(shù) 8 函數(shù)的奇偶性與周期性 理.doc

【步步高】（江蘇專用）2017版高考數(shù)學(xué) 專題2 函數(shù)概念與基本初等函數(shù) 8 函數(shù)的奇偶性與周期性 理訓(xùn)練目標(1)函數(shù)奇偶性的概念；(2)函數(shù)周期性.訓(xùn)練題型(1)判定函數(shù)的奇偶性；(2)函數(shù)奇偶性的應(yīng)用(求函數(shù)值，求參數(shù))；(3)函數(shù)周期性的應(yīng)用.解題策略(1)判斷函數(shù)的奇偶性首先要考慮函數(shù)定義域是否關(guān)于原點對稱；(2)根據(jù)奇偶性求參數(shù)，可先用特殊值法求出參數(shù)，然后驗證；(3)理解并應(yīng)用關(guān)于周期函數(shù)的重要結(jié)論：如f(x)滿足f(xa)f(x)，則f(x)的周期t2|a|.1(2015煙臺模擬)下列函數(shù)中既不是奇函數(shù)也不是偶函數(shù)的是_y2|x|；ylg(x)；y2x2-x；ylg.2(2015嘉興一模)已知函數(shù)yf (x)x是偶函數(shù)，且f (2)1，則f (2)_.3下面四個結(jié)論：偶函數(shù)的圖象一定與y軸相交；奇函數(shù)的圖象一定過原點；偶函數(shù)的圖象關(guān)于y軸對稱；沒有一個函數(shù)既是奇函數(shù)，又是偶函數(shù)其中正確的個數(shù)是_4設(shè)函數(shù)f (x)是定義在r上的偶函數(shù)，當x0時，f (x)2x1.若f (a)3，則實數(shù)a的值為_5已知函數(shù)f (x)是r上的偶函數(shù)，g(x)是r上的奇函數(shù)，且g(x)f (x1)，若f (2)2，則f (2 018)的值為_6函數(shù)f (x)是周期為4的偶函數(shù)，當x0,2時，f (x)x1，則不等式xf (x)0在1,3上的解集為_7函數(shù)f (x)是_函數(shù)(填“奇”或“偶”)8如果函數(shù)g(x)是奇函數(shù)，則f (x)_.9(2014湖南改編)已知f (x)，g(x)分別是定義在r上的偶函數(shù)和奇函數(shù)，且f (x)g(x)x3x21，則f(1)g(1)_.10已知f (x)是定義在r上且以3為周期的奇函數(shù)，當x(0,1.5)時，f (x)ln(x2x1)，則函數(shù)f (x)在區(qū)間0,6上與x軸的交點的個數(shù)是_11設(shè)a0，f (x)是r上的偶函數(shù)，則a_.12(2015保定三模)若偶函數(shù)f (x)對定義域內(nèi)任意x都有f (x)f (2x)，且當x(0,1時，f (x)log2x，則f ()_.13已知函數(shù)f (x)(a，b，cr，a0)是奇函數(shù)，若f (x)的最小值為，且f (1)，則b的取值范圍是_14定義x表示不超過x的最大整數(shù)，例如，1.51，1.52.若f (x)sin(xx)，則下列結(jié)論中：f (x)為奇函數(shù)；f (x)是周期函數(shù)，周期為2；f (x)的最小值為0，無最大值；f (x)無最小值，最大值為sin 1.其中說法正確的序號是_答案解析1解析對，函數(shù)定義域為(1，)，不關(guān)于原點對稱，故ylg不是奇函數(shù)也不是偶函數(shù)，選項為偶函數(shù)，選項為奇函數(shù)，選項為偶函數(shù)25解析yf (x)x是偶函數(shù)，f (x)xf (x)x，f (2)f (2)221225.31解析函數(shù)y是偶函數(shù)，但不與y軸相交，故錯；函數(shù)y是奇函數(shù)，但不過原點，故錯；函數(shù)f (x)0既是奇函數(shù)又是偶函數(shù)，故錯4152解析由已知得，g(x)f (x1)，f (x)為偶函數(shù)，g(x)為奇函數(shù)，g(x)f (x1)，又g(x)f (x1)，f (x1)f (x1)，f (x2)f (x)，f (x4)f (x2)f (x)，f (2 018)f (45042)f (2)2.6(1,0)(1,3)解析f (x)的圖象如圖由圖象可知，不等式xf (x)0在1,3上的解集為(1,0)(1,3)7偶解析當x0，則f (x)(x)2(x)x2xf (x)；當x0時，x0，則f (x)(x)2(x)x2xf (x)，所以f (x)f (x)，即f (x)是偶函數(shù)82x3解析令x0，g(x)2x3.又g(x)為奇函數(shù)，g(x)2x3.f(x)2x3.91解析f (x)g(x)x3x21，f (x)g(x)x3x21.f (x)是偶函數(shù)，g (x)是奇函數(shù)，f (x)f (x)，g (x)g (x)f (x)g (x)x3x21，f (1)g (1)1111.109解析當x(0,1.5)時，f (x)ln (x2x1)，令f (x)0，則x2x11，解得x1.函數(shù)f (x)是定義域為r的奇函數(shù)，在區(qū)間1.5,1.5上，f (1)f (1)0，f (0)0，f (1.5)f (1.53)f (1.5)f (1.5)，f (1)f (1)f (0)f (1.5)f (1.5)0.函數(shù)f (x)是周期為3的周期函數(shù)，函數(shù)f (x)在區(qū)間0,6上的零點為0,1,1.5,2,3,4,4.5,5,6，共9個111解析方法一f (x)是r上的偶函數(shù)，f (x)f (x)在r上恒成立，即，(a21)e2x1a20對任意的x恒成立，解得a1.方法二f (x)是r上的偶函數(shù)，f (1)f (1)，ae，(a)e(a)0，(a)(e21)0，a0.又a0，a1.經(jīng)驗證，當a1時，有f (x)f (x)，a1.121解析f (x)是偶函數(shù)，f (x)f (x)f (x2)f 2(x2)f (x)f(x)，函數(shù)f (x)的周期為2，f ()f (8)f ()f ()log2 1.13.解析f (x)f (x)0，c0，f (x).x0時，f (x) (a0)，f (x)min，當且僅當x時，f (x)取得最小值bab2.又f (1)，2b25b20，b2.14解析f (1.5)sin (1.51.5 )sin 0.5，f (1.5)sin (1.51.5 )sin 0.5，則f (1.5)f (1.5)，故錯；f (x1)sin (x1x