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高一数学上册课堂练习题(含答案) 5 c 1312 一、选择题 1.函数 f(x)=2x+6 x∈[1,2]x+7 x∈[-1,1],则 f(x)的 最大值、最小值分别为( ) A.10,6B.10,8 c.8,6D.以上都不对 [答案] A [解析] 分段函数的最大值为各段上最大值中的最大者,最小值 为各段上最小值中的最小者. 当 1≤x≤2 时,8≤2x+6≤10, 当-1≤x≤1 时,6≤x+7≤8 ∴f(x)in=f(-1)=6, f(x)ax=f(2)=10 故选 A 2.函数=x|x|的图象大致是( ) [答案] A [解析] =x2 x≥0-x2 x 0,故选 A 3.某司在甲乙两地同时销售一种品牌车,利润(单位万元)分别 为 L1=-x2+21x 和 L2=2x(其中销售量 x 单位辆).若该司在两地 共销售 15 辆,则能获得的最大利润为( ) A.90 万元 B.60 万元 c.120 万元 D.12025 万元 [答案] c [解析] 设司在甲地销售 x 辆(0≤x≤15,x 为正整数),则在乙 地销售(15-x)辆, ∴司获得利润 L=-x2+21x+2(15-x) =-x2+19x+30 ∴当 x=9 或 10 时,L 最大为 120 万元. 故选 c [点评] 列函数关系式时,不要出现=-x2+21x+2x 的错误. 4.已知 f(x)在 R 上是增函数,对实数 a、b 若 a+b>0,则有( ) A.f(a)+f(b)>f(-a)+f(-b) B.f(a)+f(b)<f(-a)+f(-b) c.f(a)-f(b)>f(-a)-f(-b) D.f(a)-f(b)<f(-a)+f(-b) [答案] A [解析] ∵a+b>0∴a>-b 且 b>-a,又=f(x)是增函数 ∴f(a)>f(-b)且 f(b)>f(-a)故选 A 5.(河南郑州市智林学校 2018~2018 高一期末)若 f(x)=-x2 +2ax 与 g(x)=ax+1 在区间[1,2]上都是减函数,则 a 的取值范围 是( ) A.(-1,0)∪(0,1)B.(-1,0)∪(0,1] c.(0,1)D.(0,1] [答案] D [解析] ∵f(x)=-x2+2ax=-(x-a)2+a2 在[1,2]上是减函 数,∴a≤1, 又∵g(x)=ax+1 在[1,2]上是减函数, ∴a 0,∴0 a≤1 6.函数=3x+2x-2(x≠2)的值域是( ) A.[2,+∞)B.(-∞,2] c.{|∈R 且≠2}D.{|∈R 且≠3} [答案] D [解析] =3x+2x-2=3(x-2)+8x-2=3+8x-2,由于 8x- 2≠0,∴≠3,故选 D 7.函数=f(x)的图象关于原点对称且函数=f(x)在区间[3,7] 上是增函数,最小值为 5,那么函数=f(x)在区间[-7,-3]上( ) A.为增函数,且最小值为-5 B.为增函数,且最大值为-5 c.为减函数,且最小值为-5 D.为减函数,且最大值为-5 [答案] B [解析] 由题意画出示意图,如下图,可以发现函数=f(x)在区 间[-7,-3]上仍是增函数,且最大值为-5 8.函数=|x-3|-|x+1|有( ) A.最大值 4,最小值 0 B.最大值 0,最小值-4 c.最大值 4,最小值-4 D.最大值、最小值都不存在 [答案] c [解析] =|x-3|-|x+1| =-4 (x≥3)2-2x (-1<x<3)4 (x≤-1),因此∈[-4,4], 故选 c 9.已知函数 f(x)=x2+bx+c 的图象的对称轴为直线 x=1,则 () A.f(-1) f(1) f(2) B.f(1) f(2) f(-1) c.f(2) f(-1) f(1) D.f(1) f(-1) f(2) [答案] B [解析] 因为二次函数图象的对称轴为直线 x=1,所以 f(-1) =f(3). 又函数 f(x)的图象为开口向上的抛物线,知 f(x)在区间[1,+ ∞)上为增函数,故 f(1) f(2) f(3)=f(-1).故选 B 10.(08 重庆理)已知函数=1-x+x+3 的最大值为,最小值为, 则的值为 () A14 B12 c22 D32 [答案] c [解析] ∵≥0,∴=1-x+x+3 =4+2(x+3)(1-x) (-3≤x≤1), ∴当 x=-3 或 1 时,in=2,当 x=-1 时,ax=22,即=2,= 22,∴=22 二、填空题 11 . 函 数 = - x2 - 10x + 11 在 区 间 [ - 1,2] 上 的 最 小 值 是 ________. [答案] -13 [解析] 函数=-x2-10x+11=-(x+5)2+36 在[-1,2]上为 减函数,当 x=2 时,in=-13 12.已知函数 f(x)在 R 上单调递增,经过 A(0,-1)和 B(3,1) 两点,那么使不等式|f(x+1)| 1 成立的 x 的集合为________. [答案] {x|-1 x 2} [解析] 由|f(x+1)| 1 得-1 f(x+1) 1,即 f(0) f(x+1) f(3), ∵f(x)在 R 上是增函数, ∴0 x+1 3∴-1 x 2 ∴使不等式成立的 x 的集合为{x|-1 x 2}. 13.如果函数 f(x)=-x2+2x 的定义域为[,n],值域为[-3,1], 则|-n|的最小值为________. [答案] 2 [解析] ∵f(x)=-x2+2x=-(x-1)2+1,当≤x≤n 时,- 3≤≤1,∴1∈[,n], 又令-x2+2x=-3 得,x=-1 或 x=3, ∴-1∈[,n]或 3∈[,n], 要使|-n|最小,应取[,n]为[-1,1]或[1,3],此时|-n|=2 三、解答题 14.求函数 f(x)=-x2+|x|的单调区间.并求函数=f(x)在[- 1,2]上的最大、小值. [解析] 由于函数解析式含有绝对值符号

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