《(完整word版)千题百炼——高考数学100个热点问题(一)：第12炼复合函数零点问题.pdf》由会员分享，可在线阅读，更多相关《(完整word版)千题百炼——高考数学100个热点问题(一)：第12炼复合函数零点问题.pdf（7页珍藏版）》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
1、第二章第 12 炼 复合函数零点问题函数及其性质第 12 炼 复合函数零点问题一、基础知识：1、复合函数定义：设yf t，tg x，且函数g x的值域为f t定义域的子集，那么y通过t的联系而得到自变量x的函数，称y是x的复合函数，记为yfg x2、复合函数函数值计算的步骤：求ygfx函数值遵循“由内到外”的顺序，一层层求出函数值。例如：已知22,xfxg xxx，计算2gf解：2224f241 2gfg3、已知函数值求自变量的步骤：若已知函数值求x的解，则遵循“由外到内”的顺序，一层层拆解直到求出x的值。例如：已知2xfx，22g xxx，若0gfx，求x解：令tfx，则2020g ttt解
2、得0,2tt当0020 xtfx，则x当2222xtfx，则1x综上所述：1x由上例可得，要想求出0gfx的根，则需要先将fx视为整体，先求出fx的值，再求对应x的解，这种思路也用来解决复合函数零点问题，先回顾零点的定义：4、函数的零点：设fx的定义域为D，若存在0 xD，使得00fx，则称0 xx为fx的一个零点5、复合函数零点问题的特点：考虑关于x的方程0g fx根的个数，在解此类问题时，要分为两层来分析，第一层是解关于fx的方程，观察有几个fx的值使得等式成立；第二层是结合着第一层fx的值求出每一个fx被几个x对应，将x的个数汇总后即为0gfx的根的个数6、求解复合函数ygfx零点问题的
3、技巧：第二章第 12 炼 复合函数零点问题函数及其性质（1）此类问题与函数图象结合较为紧密，在处理问题的开始要作出,fxg x的图像（2）若已知零点个数求参数的范围，则先估计关于fx的方程0gfx中fx解的个数，再根据个数与fx的图像特点，分配每个函数值ifx被几个x所对应，从而确定ifx的取值范围，进而决定参数的范围复合函数：二、典型例题例1：设 定 义 域 为R的 函 数1,111,1xxfxx，若 关 于x的 方 程20fxbfxc由 3 个不同的解123,x xx，则222123xxx_ 思路：先作出fx的图像如图：观察可发现对于任意的0y，满足0yfx的x的个数分别为 2 个（000
4、,1yy）和 3 个（01y），已知有3 个解，从而可得1fx必为20fxbfxc的根，而另一根为1或者是负数。所以1ifx，可解得：1230,1,2xxx，所以2221235xxx答案：5 例 2：关于x的方程22213120 xx的不相同实根的个数是（）A.3 B.4 C.5 D.8 思路：可将21x视为一个整体，即21t xx，则方程变为2320tt可解得：1t或2t，则只需作出21t xx的图像，然后统计与1t与2t的交点总数即可，共有 5 个答案：C 例3：已 知 函 数11()|f xxxxx，关 于x的 方 程2()()0fxa fxb（,a bR）恰有 6 个不同实数解，则a的
5、取值范围是文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文
6、档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE
7、4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T
8、6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD
9、10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3
10、A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10
11、ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3第二章第 12 炼 复合函数零点问题函数及其性质思路：所解方程2()()0fxa f xb可视为20fxa fxb，故考虑作出fx的图像：2,12,012,1
12、02,1xxxxfxxxxx，则fx的图像如 图，由 图 像 可 知，若 有6个 不 同 实 数 解，则 必 有122,02fxfx，所以122,4afxfx，解得42a答案：42a例 4：已知定义在R上的奇函数，当0 x时，121,0212,22xxfxfxx，则关于x的方程2610fxfx的实数根个数为（）A.6B.7C.8D.9思路：已知方程2610fxfx可解，得1211,23fxfx，只需统计11,23yy与yfx的交点个数即可。由奇函 数 可 先 做 出0 x的 图 像，2x时，122fxfx，则2,4x的 图 像 只需 将0,2x的图像纵坐标缩为一半即可。正半轴图像完成后可再利用
13、奇函数的性质作出负半轴图像。通过数形结合可得共有7 个交点答案：B 小炼有话说：在作图的过程中，注意确定分段函数的边界点属于哪一段区间。例 5：若函数32fxxaxbxc有极值点12,x x，且11fxx，则关于x的方程2320fxafxb的不同实根的个数是（）A3 B4 C5 D6 文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10
14、ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C
15、7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文
16、档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE
17、4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T
18、6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD
19、10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3
20、A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3第二章第 12 炼 复合函数零点问题函数及其性质思路：232fxxaxb由极值点可得：12,x x为2320 xaxb的两根，观察到方程与2320fxafxb结构完全相同，所以可得2320fxafxb的两根为1122,fxxfxx，其中111fxx，若12xx，可 判 断 出1x是 极 大 值 点，2x是 极 小 值 点。且2211fxxxfx，所以1yfx与fx有两个交点，而2fx与fx有一个交点，共计3 个；若12xx，可判断出1x是极小值点，2x是极大值点。且2211fx
21、xxfx，所以1yfx与fx有两个交点，而2fx与fx有一个交点，共计3 个。综上所述，共有3 个交点答案：A 例 6：已知函数243f xxx，若方程20fxbfxc恰有七个不相同的实根，则实数b的取值范围是（）A.2,0B.2,1C.0,1D.0,2思路：考虑通过图像变换作出fx的图像（如图），因为20fxbfxc最多只能解出2 个fx，若要出七个根，则121,0,1fxfx，所以121,2bfxfx，解得：2,1b答案：B 例 7：已知函数xxfxe，若关于x的方程210fxmf xm恰有 4 个不相等的实数根，则实数m的取值范围是（）A.1,22,eeB.1,1eC.11,1eD.1,
22、ee文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码
23、:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5
24、Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1
25、 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P
26、5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z
27、10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1
28、K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3第二章第 12 炼 复合函数零点问题函数及其性质思路：,0,0 xxxxefxxxe，分析fx的图像以便于作图，0 x时，1xfxx e，从而fx在0,1单调递增，
29、在1,单调递减，11fe，且当,0 xy，所以x正半轴为水平渐近线；当0 x时，1xfxxe，所以fx在,0单调递减。由此作图，从图像可得，若恰有 4 个不等实根，则关于fx的方程210fxmfxm中，12110,fxfxee，从而将问题转化为根分布问题，设tfx，则210tmtm的两根12110,ttee，设21g ttmtm，则有20010111100gmmmgeee，解得11,1me答案：C 小炼有话说：本题是作图与根分布综合的题目，其中作图是通过分析函数的单调性和关键点来进行作图，在作图的过程中还要注意渐近线的细节，从而保证图像的准确。例 8：已知函数21,0log,0axxfxx x
30、，则下列关于函数1yffx的零点个数判断正确的是（）A.当0a时，有 4 个零点；当0a时，有 1 个零点B.当0a时，有 3 个零点；当0a时，有 2 个零点C.无论a为何值，均有2 个零点D.无论a为何值，均有4 个零点思路：所求函数的零点，即方程1ffx的解的个数，先作出fx的图像，直线1yax为过定点0,1的一条直线，但需要对a的符号进行分类讨论。当0a时，图像如图所示，先拆外层可得12210,2fxfxa，而1fx有两个对应的x，2fx也文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 Z
31、U1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7
32、Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档
33、编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4
34、G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6
35、P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD1
36、0P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A
37、5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3第二章第 12 炼 复合函数零点问题函数及其性质有两个对应的x，共计 4 个；当0a时，fx的图像如图所示，先拆外层可得12fx，且12fx只有一个满足的x，所以共一个零点。结合选项，可判断出A正确答案：A 例9：已知 函数232211,0231,31,0 xxfxxxg xxx，则方程0gfxa（a为正实数）的实
38、数根最多有_个思 路：先 通 过 分 析,fxg x的 性 质 以 便 于 作 图，23632fxxxx x，从而fx在,0,2,单增，在0,2单减，且01,23ff，g x为分段函数，作出每段图像即可，如图所示，若要实数根最多，则要优先选取fx能对应x较多的情况，由fx图像可得，当3,1fx时，每个fx可对应 3 个x。只需判断gfxa中，fx能在3,1取得的值的个数即可，观察g x图像可得，当51,4a时，可以有2 个3,1fx，从而能够找到6 个根，即最多的根的个数答案：6 个文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6
39、P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD1
40、0P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A
41、5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 Z
42、U1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7
43、Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档
44、编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4
45、G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3第二章第 12 炼 复合函数零点问题函数及其性质例 10：已知函数yfx和yg x在2,2的图像如下，给出下列四个命题：（1）方程0fg x有且只有6 个根（2）方程0gfx有且只有3 个根（3）方程0ffx有且只有5 个根（4）方程0g g x有且只有 4 个根则正确命题的个数
46、是（）A.1 B.2 C.3 D.4 思路：每个方程都可通过图像先拆掉第一层，找到内层函数能取得的值，从而统计出x的总数。（1）中可得1232,1,0,1,2gxgxgx，进而1gx有 2 个对应的x，2gx有 3 个，3gx有 2 个，总计7 个，（1）错误；（2）中可得122,1,0,1fxfx，进而1fx有 1 个对应的x，2fx有 3 个，总计 4 个，（2）错误；（3）中可得1232,1,0,1,2fxfxfx，进而1fx有 1 个对应的x，2fx有 3 个，3fx有 1 个，总计5 个，（3）正确；（4）中可得：122,1,0,1gxgx，进而1gx有 2 个对应的x，2gx有 2
47、个，共计4 个，（4）正确则综上所述，正确的命题共有2个答案：B 文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P
48、1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10
49、P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5
50、Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU1K5C7Z3Z3文档编码:CE4G5Q2T6P1 HD10P5V3A5Z10 ZU