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第二章 函数概念与基本初等函数(Ⅰ)第 5 课 函数的单调性与最值 课时分层训练 A 组 基础达标 (建议用时:30 分钟) 一、填空题 1.函数 y=(2k+1)x+b 在 R 上是减函数,则 k 的取值范围是________. 【导学号:62172026】 ???-∞,-12??? [由题意知 2k+1<0,得 k<-12.] 2.给定函数:①y=x;②y=log12(x+1);③y=|x-1|;④y=2x+1,其中在区间(0,1) 上单调递减的函数序号是________. ②③ [①y=x 在区间(0,1)上单调递增;②y=log12(x+1)在区间(0,1)上单调递减; ③y=|x-1|=???x-1,x≥1, 在区间(0,1)上单调递减;④y=2x+1 在区间(0,1)上单调递 ??1-x,x<1, 增.] 3.已知函数 f(x)=|x+a|在(-∞,-1)上是单调函数,则 a 的取值范围是________. 【导学号:62172027】 (-∞,1] [函数 f(x)=???x+a,x≥-a, 即函数 f(x)在(-∞,-a)上是减函数, ??-x-a,x<-a, 在[-a,+∞)上是增函数,要使函数 f(x)在(-∞,-1)上单调递减,则-a≥-1,即 a≤1.] 4.函数 f(x)=x2+x1在[1,2]上的最大值和最小值分别是________. 43,1 [f(x)=x2+x1= + x+1 -2=2-x+2 1在[1,2]上是增函数,∴f(x)max=f(2)=43, f(x)min=f(1)=1.] 5.设函数 f(x)=ln(1+|x|)-1+1x2,则使得 f(x)>f(2x-1)成立的 x 的取值范围为 ________. ???13,1??? [由已知得函数 f(x)为偶函数,所以 f(x)=f(|x|), 由 f(x)>f(2x-1),可得 f(|x|)>f(|2x-1|). 当 x>0 时,f(x)=ln(1+x)-1+1x2,因为 y=ln(1+x)与 y=-1+1x2在(0,+∞)上都 单调递增,所以函数 f(x)在(0,+∞)上单调递增. 由 f(|x|)>f(|2x-1|),可得|x|>|2x-1|, 两边平方可得 x2>(2x-1)2,整理得 3x2-4x+1<0,解得13<x<1. 所以符合题意的 x 的取值范围为???13,1???.] 6.函数 f(x)=-(x-3)|x|的递增区间是________. ???0,32??? [f(x)=-(x-3)|x|=???-x2+3x,x>0, ??x2-3x,x≤0. 作出该函数的图象,观察图象知递增区间为???0,32???.] 7.函数 f(x)=???log12x,x≥1, ??2x,x<1 的值域为________. (-∞,2) [当 x≥1 时,f(x)=log12x≤log121=0. 当 x<1 时,f(x)=2x∈(0,2), ∴f(x)的值域为(-∞,2).] ?? - ,x≥2, 8 . 已 知 函 数 f(x) = ??????12???x-1,x<2, 满 足 对 任 意 的 实 数 x1≠x2 , 都 有 - x1-x2 <0 成立,则实数 a 的取值范围为________. ???-∞,183??? [ 由 - x1-x2 <0 可 知 f(x) 在 R 上 是 减 函 数 , 故 ??a-2<0, ??????12???2- -, 解得 a≤183.] 9.已知函数 y=f(x)的图象关于 x=1 对称,且在(1,+∞)上单调递增,设 a=f???-12???, b=f(2),c=f(3),则 a,b,c 的大小关系为________. 【导学号:62172028】 b<a<c [∵y=f(x)的图象关于 x=1 对称, ∴f???-12???=f???52???. 又 5 2<2<3,且 f(x)在(1,+∞)上单调递增, ∴f(2)<f???52???<f(3), ∴f(2)<f???-12???<f(3), 即 b<a<c.] 10.f(x)是定义在(0,+∞)上的单调增函数,满足 f(xy)=f(x)+f(y),f(3)=1,则 不等式 f(x)+f(x-8)≤2 的解集为________. (8,9] [因为 2=1+1=f(3)+f(3)=f(9),由 f(x)+f(x-8)≤2 可得 f[x(x- ?? x>0, 8)]≤f(9),f(x)是定义在(0,+∞)上的增函数,所以有?x-8>0, ?? - , 解得 8<x≤9.] 二、解答题 11.(2017·苏州模拟)已知函数 f(x)=1a-1x(a>0,x>0), (1)求证:f(x)在(0,+∞)上是增函数; (2)若 f(x)在???12,2???上的值域是???12,2???,求 a 的值. [解] (1)证明:任取 x1>x2>0, 则 f(x1)-f(x2)=1a-x11-1a+x12=xx11-x2x2,∵x1>x2>0,∴x1-x2>0,x1x2>0,∴f(x1)- f(x2)>0,即 f(x1)>f(x2),∴f(x)在(0,+∞)上是增函数. (2)由(1)可知 f(x)在???12,2???上为增函数,∴f???12???=1a-2=12,f(2)=1a-12=2,解得 a= 25. 12.已知 f(x)=x-x a(x≠a). (1)若 a=-2,试证 f(x)在(-∞,-2)上单调递增; (2)若 a>0 且 f(x)在(1,+∞)上单调递减,求 a 的取值范围. 【导学号:62172029】 [解] (1)证明:设 x1<x2<-2, 则 f(x1)-f(x2)=x1x+1 2-x2x+2 2 = - + + . ∵(x1+2)(x2+2)>0,x1-x2<0, ∴f(x

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