033i.net:2018-2019学年高中新创新一轮复习理数通用版:课时达标检测(五) 函数的单调性与最值 Word版含解析

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课时达标检测(五) 对点练(一) 函数的单调性 函数的单调性与最值 [小题对点练——点点落实] 1.(2018· 阜阳模拟)给定函数①y=x ,②y=log 1 (x+1),③y=|x-1|,④y=2x 1.其 + 1 2 2 中在区间(0,1)上单调递减的函数序号是( A.①② C.③④ 1 2 ) B.②③ D.①④ 1 解析: 选 B ①y=x 在(0,1)上递增; ②∵t=x+1 在(0,1)上递增, 且 0< <1, 故 y=log 1 2 2 (x+1)在(0,1)上递减; ③结合图象可知 y=|x-1|在(0,1)上递减; ④∵u=x+1 在(0,1)上递增, 且 2>1,故 y=2x +1 在(0,1)上递增.故在区间(0,1)上单调递减的函数序号是②③. 2.(2018· 天津模拟)若函数 f(x)满足“对任意 x1,x2∈(0,+∞),当 x1<x2 时,都有 f(x1)>f(x2)”,则 f(x)的解析式可以是( A.f(x)=(x-1)2 1 C.f(x)= x ) B.f(x)=ex D.f(x)=ln(x+1) 解析:选 C 根据条件知,f(x)在(0,+∞)上单调递减.对于 A,f(x)=(x-1)2 在(1, +∞)上单调递增,排除 A;对于 B,f(x)=ex 在(0,+∞)上单调递增,排除 B;对于 C,f(x) 1 =x在(0,+∞)上单调递减,C 正确;对于 D,f(x)=ln(x+1)在(0,+∞)上单调递增,排除 D. 3.(2018· 宜春模拟)函数 f(x)=log3(3-4x+x2)的单调递减区间为( A.(-∞,2) C.(-∞,1) B.(-∞,1),(3,+∞) D.(-∞,1),(2,+∞) ) 解析: 选 C 由 3-4x+x2>0 得 x<1 或 x>3.易知函数 y=3-4x+x2 的单调递减区间为(- ∞,2),函数 y=log3x 在其定义域上单调递增,由复合函数的单调性知,函数 f(x)的单调递 减区间为(-∞,1),故选 C. 4.(2018· 贵阳模拟)下列四个函数中,在定义域上不是单调函数的是( A.y=-2x+1 C.y=lg x 1 B.y=x D.y=x3 ) 解析:选 B y=-2x+1 在定义域上为单调递减函数;y=lg x 在定义域上为单调递增 1 函数;y=x3 在定义域上为单调递增函数; y=x在(-∞,0)和(0, +∞)上均为单调递减函数, 但在定义域上不是单调函数.故选 B. 5.若函数 f(x)=8x2-2kx-7 在[1,5]上为单调函数,则实数 k 的取值范围是( A.(-∞,8] C.(-∞,8]∪[40,+∞) B.[40,+∞) D.[8,40] ) k 解析:选 C 由题意知函数 f(x)=8x2-2kx-7 的图象的对称轴为 x= ,因为函数 f(x) 8 k k =8x2-2kx-7 在[1,5]上为单调函数,所以 ≤1 或 ≥5,解得 k≤8 或 k≥40,所以实数 k 8 8 的取值范围是(-∞,8]∪[40,+∞).故选 C. ?a b?=ad-bc,若函数 f(x)=?x-1 2 ?在(-∞,m)上单调递减,则 6.定义运算? ? ? ? ?c d ? ?-x x+3? 实数 m 的取值范围是( A.(-2,+∞) C.(-∞,-2) 解析:选 D ) B.[-2,+∞) D.(-∞,-2] ?a b?=ad-bc,∴f(x)=?x-1 2 ?=(x-1)(x+3)-2×(-x)=x2 ∵? ? ? ? ?c d ? ?-x x+3? +4x-3=(x+2)2-7, ∴f(x)的单调递减区间为(-∞,-2), ∵函数 f(x)在(-∞,m)上单调递减, ∴(-∞,m)?(-∞,-2),即 m≤-2.故选 D. 对点练(二) 函数的最值 2 018x 1+2 016 1.已知 a>0,设函数 f(x)= (x∈[-a,a])的最大值为 M,最小值为 N, 2 018x+1 + 那么 M+N=( A.2 016 C.4 032 ) B.2 018 D.4 034 2 018 +2 016 2 =2 018- .∵y=2 018x+1 在[-a, 2 018x+1 2 018x+1 x+1 解析: 选 D 由题意得 f(x)= 2 a]上是单调递增的,∴f(x)=2 018- 在[-a,a]上是单调递增的,∴M=f(a),N= 2 018x+1 f(-a),∴M+N=f(a)+f(-a)=4 036- 2 2 - =4 034. - 2 018a+1 2 018 a+1 f?x? 2. 已知函数 f(x)=x2-2ax+a 在区间(-∞, 1)上有最小值, 则函数 g(x)= x 在区间(1, +∞)上一定( ) B.有最大值 D.是增函数 A.有最小值 C.是减函数 a 解析:选 D 由题意知 a<1,又函数 g(x)=x+ -2a 在[ |a|,+∞)上为增函数,故选 x D. ?-x+6,x≤2, ? 3.(2018· 湖南雅礼中学月考)若函数 f(x)=? (a>0 且 a≠1)的值域是[4, ? ?3+logax,x>2 +∞),则实数 a 的取值范围是( A.(1,2] C.[2,+∞) ) B.(0,2] D.(1,2 2 ] ?3+logax≥4, ? 解析:选 A 当 x≤2 时,-x+6≥4.当 x>2 时,? ∴a∈(1,2],故选 A. ? ?a>1, 4.(2018· 安徽合肥模拟)已知函数 f(x)=(x2-2x)· sin(x-1)+x+1 在[-1,3]上的最大值 为 M,最小值为 m,则 M+m=( A.4 ) B.2 C.1 D.0 解析:选 A 设 t=x-1,则 y=(x2-2x)sin(x-1)+x+1=(t2-1)sin t+t+2,t∈[- 2,2].记 g(t)=(t2-1)sin

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