2022届高考数学统考一轮复习-第2章-函数-第4节-函数性质的综合问题教案-理-新人教版.doc

2022届高考数学统考一轮复习 第2章 函数 第4节 函数性质的综合问题教案 理 新人教版2022届高考数学统考一轮复习 第2章 函数 第4节 函数性质的综合问题教案 理 新人教版年级：姓名：函数性质的综合问题 考点一函数的奇偶性与单调性 函数的单调性与奇偶性的综合问题解题思路(1)解决比较大小、最值问题应充分利用奇函数在关于原点对称的两个区间上具有相同的单调性，偶函数在关于原点对称的两个区间上具有相反的单调性(2)解决不等式问题时一定要充分利用已知的条件，把已知不等式转化成f (x1)f (x2)或f (x1)f (x2)的形式，再根据函数的奇偶性与单调性，列出不等式(组)，要注意函数定义域对参数的影响典例1(1)(2019全国卷)设f (x)是定义域为R的偶函数，且在(0，)单调递减，则()(2)函数f (x)在(，)上单调递减，且为奇函数若f (1)1，则满足1f (x2)1的x的取值范围是()A2,2B1,1C0,4 D1,3(1)C(2)D(1)f (x)是定义域为R的偶函数，f (x)f (x)f f (log34)f (log34)又log34log331，且1220，log34220.f (x)在(0，)上单调递减，f (2)f (2)f (log34)f .故选C(2)f (x)为奇函数，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(，)上单调递减，1x21，1x3.故选D点评：解答此类题目时，奇偶性的作用是把不在同一单调区间的自变量转化到同一单调区间上1函数yf (x)在0,2上单调递增，且函数f (x2)是偶函数，则下列结论成立的是()Af (1)f f Bf f (1)f Cf f f (1)Df f (1)f B函数yf (x)在0,2上单调递增，且函数f (x2)是偶函数，函数yf (x)在2,4上单调递减，且在0,4上函数yf (x)满足f (2x)f (2x)，f (1)f (3)，f f (3)f ， 即f f (1)f .2已知f (x)是定义在2b,1b上的偶函数，且在2b,0上为增函数，则f (x1)f (2x)的解集为()A BC1,1 DBf (x)是定义在2b,1b上的偶函数，2b1b0，b1.f (x)在2b,0上为增函数，即函数f (x)在2,0上为增函数，故函数f (x)在(0,2上为减函数，则由f (x1)f (2x)，可得|x1|2x|，即(x1)24x2，解得1x.由于定义域为2,2， 解得1x，故选B3已知奇函数f (x)在x0时单调递增，且f (1)0，若f (x1)0，则x的取值范围为()Ax|0x1或x2Bx|x0或x2Cx|x0或x3Dx|x1或x1A奇函数f (x)在(0，)上单调递增，且f (1)0，函数f (x)在(，0)上单调递增，且f (1)0，则1x0或x1时，f (x)0；x1或0x1时，f (x)0.不等式f (x1)0即1x10或x11，解得0x1或x2，故选A 考点二函数的奇偶性与周期性 利用函数的奇偶性和周期性求值的策略已知f (x)是周期函数且为偶(奇)函数，求函数值，常利用奇偶性及周期性进行转换，将所求函数值的自变量转化到已知解析式的函数定义域内，把未知区间上的函数性质转化为已知区间上的函数性质求解典例2(1)已知f (x)是定义在R上的偶函数，并且f (x2)，若当2x3时，f (x)x，则f (105.5)_.(2)设f (x)是定义在R上的奇函数，并且f (x)f (x2)，若在区间2,0)(0,2上，f (x)则f (2 022)_.(1)2.5(2)0(1)由f (x2)得f (x4)f (x2)2f (x)，函数f (x)是周期为4的周期函数f (105.5)f (4272.5)f (2.5)f (2.5)2.5.(2)由f (x)f (x2)，得f (x4)f (x2)2f (x2)f (x)，函数f (x)是周期为4的周期函数又f (x)是奇函数，则有即解得f (x)f (2 022)f (2)210.设定义在R上的函数f (x)同时满足以下条件：f (x)f (x)0；f (x1)f (x1)；当0x1时，f (x)2x1，则f f (1)f f (2)f _.1由f (x)f (x)0得f (x)f (x)，即函数f (x)是奇函数，由f (x1)f (x1)得f (x2)f (x)，即函数f (x)是周期为2的周期函数所以f (0)0，f f ，f (2)f (0)0，f f .又f (1)f (12)f (1)f (1)，所以f (1)0.所以f f (1)f f (2)f f f f f 211. 考点三函数性质的综合应用 函数的奇偶性、周期性与对称性的解题技法(1)函数的奇偶性、周期性、对称性，一般是知二得一，特别是已知奇偶性和对称性，一般要先确定周期性(2)奇函数在x0处有意义，则一定有f (0)0，偶函数一定有f (|x|)f (x)，要注意这两个结论在解题中的应用(3)如果f (x)的图象关于点(a,0)对称，且关于直线xb对称，则函数f (x)的周期T4|ab|.(类比ysin x的图象)(4)如果f (x)的图象关于点(a,0)对称，且关于点(b,0)对称，则函数f (x)的周期T2|ab|.(类比ysin x的图象)(5)若函数f (x)关于直线xa与直线xb对称，那么函数的周期是2|ba|.(类比ysin x的图象)典例3(1)已知函数f (x)对任意的xR都满足f (x)f (x)0，f 为偶函数，当0x时，f (x)x，则f (2 021)f (2 022)_.(2)已知f (x)是定义域为(，)的奇函数，满足f (1x)f (1x)若f (1)2，则f (1)f (2)f (3)f (50)_.(1)1(2)2(1)由f (x)f (x)0得f (x)f (x)，即函数f (x)是奇函数，由f 为偶函数知f f ，结合f (x)是奇函数，可得f f ，f (x3)f (x)f (x6)f (x)，即函数f (x)是周期为6的周期函数f (2 021)f (1)f (1)1，f (2 022)f (0)0，f (2 021)f (2 022)1.(2)法一：(直接法)f (x)在(，)上是奇函数，f (x)f (x)，f (1x)f (x1)由f (1x)f (1x)，得f (x1)f (x1)，f (x2)f (x)，f (x4)f (x2)f (x)，函数f (x)是周期为4的周期函数由f (x)为奇函数得f (0)0.又f (1x)f (1x)，f (x)的图象关于直线x1对称，f (2)f (0)0，f (2)0.又f (1)2，f (1)2，f (1)f (2)f (3)f (4)f (1)f (2)f (1)f (0)20200，f (1)f (2)f (3)f (4)f (49)f (50)012f (49)f (50)f (1)f (2)202.法二：(特例法)由题意可设f (x)2sin，作出f (x)的部分图象如图所示由图可知，f (x)的一个周期为4，所以f (1)f (2)f (3)f (50)12f (1)f (2)f (3)f (4)f (49)f (50)120f (1)f (2)2.点评：求和问题一般是先求一个周期的和，再求总和1已知奇函数f (x)的定义域为R，若f (x1)为偶函数，且f (1)2，则f (2 021)f (2 022)()A2 B1 C0 D2D由f (x1)为偶函数得f (x1)f (x1)，又函数f (x)是奇函数，则f (x1)f (x1)，即f (x2)f (x)，f (x4)f (x)，即函数f (x)的周期为4，f (2 021)f (1)2，f (2 022)f (2)f (0)0，f (2 021)f (2 022)2，故选D2定义在R上的函数f (x)满足f (x)f (2x)及f (x)f (x)，且在0,1上有f (x)x2，则f ()A B C DD函数f (x)的定义域是R，f (x)f (x)，所以函数f (x)是奇函数又f (x)f (2x)，所以f (x)f (2x)f (x)，所以f (4x)f (2x)f (x)，故函数f (x)是以4为周期的奇函数，所以f f f f .因为在0,1上有f (x)x2，所以f ，故f ，故选D核心素养2用数学思维思考世界用活函数性质中的三个结论数学运算是解决数学问题的基本手段，通过运算能够促进学生数学思维的发展通过常见的“二级结论”解决数学问题，可优化数学运算的过程，使学生逐步形成规范化、程序化的思维品质，养成一丝不苟、严谨求实的科学精神.奇函数的最值性质已知函数f (x)是定义在区间D上的奇函数，则对任意的xD，都有f (x)f (x)0.特别地，若奇函数f (x)在D上有最值，则f (x)maxf (x)min0，且若0D，则f (0)0.设函数f (x)的最大值为M，最小值为m，则Mm_.2显然函数f (x)的定义域为R，f (x)1，设g(x)，则g(x)g(x)，g(x)为奇函数，由奇函数图象的对称性知g(x)maxg(x)min0，Mmg(x)1maxg(x)1min2g(x)maxg(x)min2.已知函数f (x)的最大值为M，最小值为m，则Mm等于()A0 B2 C4 D8Cf (x)2，设g(x)，因为g(x)定义域为R，关于原点对称，且g(x)g(x)，所以g(x)为奇函数，所以g(x)maxg(x)min0.因为Mf (x)max2g(x)max，mf (x)min2g(x)min，所以Mm2g(x)max2g(x)min4.抽象函数的周期性(1)如果f (xa)f (x)(a0)，那么f (x)是周期函数，其中的一个周期T2a.(2)如果f (xa)(a0)，那么f (x)是周期函数，其中的一个周期T2a.(3)如果f (xa)f (x)c(a0)，那么f (x)是周期函数，其中的一个周期T2a.已知函数f (x)为定义在R上的奇函数，当x0时，有f (x3)f (x)，且当x(0,3)时，f (x)x1，则f (2 017)f (2 018)()A3B2C1D0C因为函数f (x)为定义在R上的奇函数，所以f (2 017)f (2 017)，因为当x0时，有f (x3)f (x)，所以f (x6)f (x3)f (x)，即当x0时，自变量的值每增加6，对应函数值重复出现一次又当x(0,3)时，f (x)x1，f (2 017)f (33661)f (1)2，f (2 018)f (33662)f (2)3.故f (2 017)f (2 018)f (2 017)31.已知f (x)是定义在R上的函数，且满足f (x2)，当2x3时，f (x)x，则f _.f (x2)，f (x4)f (x)，f f ，又2x3时，f (x)x，f ，f .抽象函数的对称性已知函数f (x)是定义在R上的函数(1)若f (ax)f (bx)恒成立，则yf (x)的图象关于直线x对称，特别地，若f (ax)f (ax)恒成立，则yf (x)的图象关于直线xa对称(2)若函数yf (x)满足f (ax)f (ax)0，即f (x)f (2ax)，则f (x)的图象关于点(a,0)对称函数yf (x)对任意xR都有f (x2)f (x)成立，且函数yf (x1)的图象关于点(1,0)对称，f (1)4，则f (2 016)f (2 017)f (2 018)的值为_4因为函数yf (x1)的图象关于点(1,0)对称，所以函数yf (x)的图象关于原点对称，所以f (x)是R上的奇函数，则f (x2)f (x)f (x)，所以f (x4)f (x2)f (x)，故f (x)的周期为4.所以f (2 017)f (50441)f (1)4，f (2 016)f (5044)f (0)0，f (2 018)f (50442)f (2)f (0)0，所以f (2 016)f (2 017)f (2 018)4.已知函数f (x)(xR)满足f (x)f (2x)，若函数y|x22x3|与yf (x)图象的交点为(x1，y1)，(x2，y2)，(xm，ym)，则xi()A0BmC2mD4mB函数f (x)(xR)满足f (x)f (2x)，故函数f (x)的图象关于直线x1对称，函数y|x22x3|的图象也关于直线x1对称，故函数y|x22x3|与yf (x)图象的交点也关于直线x1对称，且相互对称的两点横坐标和为2.当f (x)不过点(1,4)时，xi2m，当f (x)的图象过点(1,4)时，xi21m.综上，xim.