奇偶性（第二課時(shí)） 同步練習(xí)-高一上學(xué)期人教A版2019數(shù)學(xué)必修第一冊(cè)

1、 .2.2 奇偶性（第二課時(shí)）（同步練習(xí)）一、選擇題1.若函數(shù)f (x)是奇函數(shù)，且當(dāng)x0時(shí)，f (x)x3x1，則當(dāng)x0時(shí)，f (x)的解析式為()Af (x)x3x1 Bf (x)x3x1Cf (x)x3x1 Df (x)x3x12.設(shè)函數(shù)f (x)eq blcrc (avs4alco1(x2x，x0，,gx，x0，)且f (x)為偶函數(shù)，則g(2)等于()A6 B6C2 D23.已知f (x)是偶函數(shù)，且在區(qū)間0，)上單調(diào)遞增，則f (0.5)，f (1)，f (0)的大小關(guān)系是()Af (0.5)f (0)f (1) Bf (1)f (0.5)f (0)Cf (0)f (0.5)f (

2、1) Df (1)f (0)f (0.5)4.若函數(shù)f (x)ax2(2a)x1是偶函數(shù)，則函數(shù)f (x)的單調(diào)遞增區(qū)間為()A(，0B0，)C(，)D1，)5.一個(gè)偶函數(shù)定義在區(qū)間7，7上，它在0，7上的圖象如圖，下列說(shuō)法正確的是()A.這個(gè)函數(shù)僅有一個(gè)單調(diào)增區(qū)間 B.這個(gè)函數(shù)有兩個(gè)單調(diào)減區(qū)間C.這個(gè)函數(shù)在其定義域內(nèi)有最大值是7 D.這個(gè)函數(shù)在其定義域內(nèi)有最小值是76.設(shè)奇函數(shù)f (x)在(0，)上單調(diào)遞減，且f (1)0，則不等式eq f(fxfx,x)0的解集為()A(1，0)(1，) B(，1)(0，1)C(，1)(1，) D(1，0)(0，1)7.(多選)設(shè)f (x)為偶函數(shù)，且在區(qū)

3、間(，0)內(nèi)單調(diào)遞增，f (2)0，則下列區(qū)間中使得xf (x)0，,fx，x0)是奇函數(shù)，則F (1)_，f (x)_12.已知f (x)是偶函數(shù)，g(x)是奇函數(shù)，且f (x)g(x)x2x2，則f (x)_，g(x)_三、解答題13.已知f (x)是定義在(1，1)上的奇函數(shù)，且f (x)在(1，1)上是減函數(shù)，解不等式f (1x)f (12x)014.已知yf (x)是奇函數(shù)，它在(0，)上單調(diào)遞增，且f (x)0，試問(wèn)F (x)eq f(1,fx)在(，0)上是單調(diào)遞增還是單調(diào)遞減？證明你的結(jié)論15.我們知道，函數(shù)yf (x)的圖象關(guān)于坐標(biāo)原點(diǎn)成中心對(duì)稱的充要條件是函數(shù)yf (x)為

4、奇函數(shù)，有同學(xué)發(fā)現(xiàn)可以將其推廣為：函數(shù)yf (x)的圖象關(guān)于點(diǎn)P(a，b)成中心對(duì)稱的充要條件是函數(shù)yf (xa)b為奇函數(shù)(1)求函數(shù)f (x)x33x2圖象的對(duì)稱中心；(2)請(qǐng)利用函數(shù)f (x)x33x2的對(duì)稱性求f (2 018)f (2 017)f (0)f (1)f (2)f (2 019)f (2 020)的值；(3)類比上述推廣結(jié)論，寫出“函數(shù)yf (x)的圖象關(guān)于x軸成軸對(duì)稱的充要條件是函數(shù)yf (x)為偶函數(shù)”的一個(gè)推廣結(jié)論參考答案：一、選擇題1.A 2.A 3.C 4.A 5.C 6.C 7.CD二、填空題8.答案：eq r(x)1 9.答案：2 022 10.答案：f (

5、2)f (1)f (0)11.答案：1，2x3 12.答案：x22，x三、解答題13.解：f (x)是定義在(1，1)上的奇函數(shù)，由f (1x)f (12x)0，得f (1x)f (12x)，f (1x)f (2x1)又f (x)在(1，1)上是減函數(shù)，eq blcrc (avs4alco1(11x1，,112x2x1，)解得0 xeq f(2,3)，原不等式的解集為eq blc(rc)(avs4alco1(0，f(2,3)14.解：F (x)在(，0)上單調(diào)遞減證明如下：任取x1，x2(，0)，且x1x20因?yàn)閥f (x)在(0，)上單調(diào)遞增，且f (x)0，所以f (x2)f (x1)f

6、(x1)0于是F (x1)F (x2)eq f(fx2fx1,fx1fx2)0，即F (x1)F (x2)，所以F (x)eq f(1,fx)在(，0)上單調(diào)遞減15.解：(1)設(shè)f (x)x33x2的對(duì)稱中心為點(diǎn)P(a，b)，設(shè)g(x)f (xa)b(xa)33(xa)2b，則g(x)為奇函數(shù)，依題可知，g(x)f (xa)b且g(x)g(x)，故f (xa)bbf (xa)，即f (xa)f (xa)2b，即(xa)33(xa)2(xa)33(xa)22b，(6a6)x22a36a22b0，eq blcrc (avs4alco1(6a60，,2a36a22b0，)解得eq blcrc (avs4alco1(a1，,b2.)函數(shù)f (x)x33x2的圖象的對(duì)稱中心為(1，2)(2)由(1)知函數(shù)f (x)x33x2的圖象的對(duì)稱中心為(1，2)，f (x1)f (x1)4，f (2 018)f (2 020)f (2 017)f (2 019)f (0)f (2)4，且f (1)2，f (2 018)f (2 017)f (0)f (1)f (2)f (2