第7讲 函数的单调性与最值期末复习作业-2024-2025学年高一上学期数学人教A版（2019）必修第一册

第7讲　函数的单调性与最值 1.下列函数在其定义域内是增函数的是 (　 ) A.y=2x B.y=-log2x C.y=- D.y=tan x 2.[2024·广东四校联考] 函数y=的单调递增区间是 (　 ) A.(-∞,1] B.[1,2] C. D. 3.已知函数f(x)=在R上单调递减,则实数a的取值范围是 (　 ) A. B. C. D. 4.“a>2”是“函数y=|x-a|在(-∞,2]上单调递减”的 (　 ) A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件 5.函数f(x)=x+2cos x在区间上的最小值是　　　　.  6.已知函数f(x)是定义在[0,+∞)上的增函数,则满足f(2x-1)<f的x的取值范围是　　　　.  7.已知函数f(x)=ex+4x+1,a=f(ln 4),b=f(ln 3),c=f(1),则a,b,c的大小关系为 (　 ) A.b>c>a B.c>b>a C.b>a>c D.a>b>c 8.已知函数f(x)=|ln x-a|+a(a>0)在[1,e2]上的最小值为1,则a的值为 (　 ) A.1 B.2 C. D. 9.已知函数f(x)=且f(2)是f(x)的最小值,则实数a的取值范围是 (　 ) A.[-2,1] B.[0,1] C.[0,6] D.[1,6] 10.(多选题)已知函数f(x)=且满足对于任意的x1≠x2,都有<0成立,则a的值可能是 (　 ) A.1 B.2 C.3 D.4 11.(多选题)[2023·长春模拟] 已知函数f(x)=|x|+sin2x,设x1,x2∈R,则f(x1)>f(x2)成立的一个充分条件可以是 (　 ) A.|x1|>x2 B.x1+x2>0 C.> D.|x1|>|x2| 12.函数f(x)=log2的单调递增区间为　　　　　　　　.  13.已知函数f(x)是定义在R上的单调函数,且f[f(x)-2x-2x]=10,则f(x)在[-2,2]上的最大值为　　　　.  14.已知函数f(x)=且f(1)=5,f(2)=6. (1)求f(x)的解析式; (2)写出f(x)的单调递增区间和单调递减区间. 15.定义在R上的函数f(x)满足:对于任意x,y∈R,都有f(x+y)=f(x)+f(y),当x<0时,f(x)>0恒成立. (1)求f(0)的值; (2)判断并证明f(x)的单调性; (3)当a>0时,解关于x的不等式f(ax2)-f(x)>-f(-a2x)+f(-a). 16.已知cos5θ-sin5θ<7(sin3θ-cos3θ),θ∈[0,2π),则θ的取值范围是 (　 ) A. B. C. D. 17.[2023·湖南永州三模] 若函数y=f(x)和y=f(-x)在区间[m,n]上的单调性相同,则把区间[m,n]叫作y=f(x)的“稳定区间”.已知区间[1,2024]为函数y=的“稳定区间”,则实数a的值可能为 (　 ) A.- B.- C. D. 参考答案 1.A　[解析] 对于A,y=2x是指数函数,在其定义域内是增函数,符合题意;对于B,y=-log2x=log_(1/2)x是对数函数,在其定义域内是减函数,不符合题意;对于C,y=-1/x是反比例函数,在其定义域内不具有单调性,不符合题意;对于D,y=tan x是正切函数,在其定义域内不具有单调性,不符合题意.故选A. 2.D　[解析] y=(1/2)^u在R上单调递减,由复合函数的单调性可知,只需求出u=x2-3x+2的单调递减区间.易知u=x2-3x+2=(x"-" 3/2)^2-1/4的单调递减区间为("-∞," 3/2],故y=(1/2)^(x^2 "-" 3x+2)的单调递增区间是("-∞," 3/2].故选D. 3.B　[解析] 因为函数f(x)={■("(" 3a"-" 2")" x+3a"," x<1"," @log_a x"," x≥1)┤在R上单调递减,所以{■(3a"-" 2<0"," @0<a<1"," @"(" 3a"-" 2")" +3a≥0"," )┤解得1/3≤a<2/3,故实数a的取值范围是[1/3 "," 2/3). 4.A　[解析] y=|x-a|={■("-" x+a"," x<a"," @x"-" a"," x≥a"," )┤显然函数y=|x-a|的单调递减区间为(-∞,a).当a>2时,函数y=|x-a|在(-∞,2]上单调递减;若函数y=|x-a|在(-∞,2]上单调递减,则a≥2.所以“a>2”是“函数y=|x-a|在(-∞,2]上单调递减”的充分不必要条件.故选A. 5.-π/2　[解析] 由y=x和y=2cos x在区间["-" π/2 "," 0]上均单调递增,可知f(x)=x+2cos x在区间["-" π/2 "," 0]上单调递增,故f(x)=x+2cos x在区间["-" π/2 "," 0]上的最小值是f("-" π/2)=-π/2+2cos("-" π/2)=-π/2. 6.[1/2 "," 2/3)　[解析] 因为函数f(x)是定义在[0,+∞)上的增函数,所以不等式f(2x-1)<f(1/3)可化为0≤2x-1<1/3,解得1/2≤x<2/3. 7.D　[解析] 函数f(x)=ex+4x+1的定义域为R,且f(x)是增函数,因为ln 4>ln 3>1,所以f(ln 4)>f(ln 3)>f(1),即a>b>c.故选D. 8.A　[解析] 由x∈[1,e2],得ln x∈[0,2].当a≥2时,f(x)=2a-ln x在[1,e2]上单调递减,所以f(x)的最小值为f(e2)=2a-2=1,解得a=3/2,舍去;当0<a<2时,f(x)={■(2a"-" lnx"," x"∈[" 1"," e^a ")," @lnx"," x"∈[" e^a "," e^2 "]," )┤f(x)在[1,ea)上单调递减,在[ea,e2]上单调递增,所以f(x)的最小值为f(ea)=a=1,符合题意.故a=1. 9.D　[解析] 当x≤2时,若2a<2,即a<1,有f(x)={■(2a"-" x"," x≤2a"," @x"-" 2a"," 2a<x≤2"," )┤f(x)在(-∞,2a]上单调递减,在(2a,2]上单调递增,则f(2a)<f(2),与f(2)是f(x)的最小值矛盾;若2a≥2,即a≥1,有f(x)=2a-x在(-∞,2]上单调递减,则f(x)≥f(2),则a≥1.当x>2时,函数f(x)=x-2+1/(x"-" 2)+a+2≥2√("(" x"-" 2")·" 1/(x"-" 2))+a+2=a+4,当且仅当x-2=1/(x"-" 2),即x=3时取等号,因为f(2)是f(x)的最小值,所以{■(a≥1"," @2a"-" 2≤a+4"," )┤解得1≤a≤6.所以实数a的取值范围是[1,6].故选D. 10.BC　[解析] 因为(f"(" x_1 ")-" f"(" x_2 ")" )/(x_1 "-" x_2 )<0,所以f(x)在R上单调递减,则{■(a>0"," @"-" ("-" a)/2≥1"," @1^2 "-" a+5≥a/1 "," )┤解得2≤a≤3.故选BC. 11.CD　[解析] 函数f(x)=|x|+sin2x的定义域为R,且f(-x)=|-x|+sin2(-x)=|x|+sin2x=f(x),所以函数f(x)是R上的偶函数.当x≥0时,f(x)=x+sin2x,则f'(x)=1+2sin xcos x=1+sin 2x≥0,故函数f(x)在[0,+∞)上单调递增.对于A,取x1=2,x2=-3,满足|x1|>x2,而f(2)<f(3)=f(-3),故A错误;对于B,取x1=1,x2=2,满足x1+x2>0,而f(1)<f(2),故B错误;对于C,D,x_1^2>x_2^2⇔|x1|>|x2|,则f(|x1|)>f(|x2|),因为函数f(x)是偶函数,所以f(x1)>f(x2),故C,D正确.故选CD. 12.(kπ"-" π/6 "," kπ+π/12),k∈Z　[解析] 令t=cos(2x"-" π/6),由t>0,可得cos(2x"-" π/6)>0,所以2kπ-π/2<2x-π/6<2kπ+π/2,k∈Z,解得kπ-π/6<x<kπ+π/3,k∈Z,所以函数f(x)的定义域为(kπ"-" π/6 "," kπ+π/3),k∈Z.由余弦函数的性质可知,t=cos(2x"-" π/6)在(kπ"-" π/6 "," kπ+π/12),k∈Z上单调递增,在(kπ+π/12 "," kπ+π/3),k∈Z上单调递减,又因为y=log2t在定义域上为增函数,所以由复合函数的单调性可知,函数f(x)=log2[cos(2x"-" π/6)]的单调递增区间为(kπ"-" π/6 "," kπ+π/12),k∈Z. 13.10　[解析] 因为f(x)是定义在R上的单调函数,所以存在唯一的t∈R,使得f(t)=10,则f(x)-2x-2x=t,令x=t,则f(t)-2t-2t=t,即f(t)=2t+3t=10,因为函数y=2t+3t为增函数,且22+3×2=10,所以t=2,则f(x)=2x+2x+2.易知f(x)在[-2,2]上单调递增,所以f(x)在[-2,2]上的最大值为f(2)=10. 14.解:(1)因为f(x)={■(2"-" x"," x<0"," @"-" x^2+bx+c"," x≥0"," )┤且f(1)=5,f(2)=6,所以{■(f"(" 1")" ="-" 1+b+c=5"," @f"(" 2")" ="-" 4+2b+c=6"," )┤解得{■(b=4"," @c=2"," )┤所以f(x)={■(2"-" x"," x<0"," @"-" x^2+4x+2"," x≥0"." )┤ (2)画出函数f(x)的图象,如图所示. 由图可知,函数f(x)的单调递减区间是(-∞,0),[2,+∞),单调递增区间是[0,2). 15.解:(1)令x=y=0,则f(0+0)=f(0)+f(0),可得f(0)=0. (2)f(x)在R上单调递减,证明如下: 对于任意x,y∈R,f(x+y)=f(x)+f(y),f(0)=0, 令y=-x,则f(x-x)=f(x)+f(-x)=0,所以对于任意x∈R,有f(-x)=-f(x),故f(x)是奇函数,任取x1,x2∈R且x1<x2,则x1-x2<0,由已知得f(x1-x2)>0,所以f(x1-x2)=f(x1)+f(-x2)=f(x1)-f(x2)>0,所以f(x1)>f(x2),所以f(x)在R上单调递减. (3)因为1/2f(ax2)-f(x)>-1/2f(-a2x)+f(-a),所以1/2f(ax2)-f(x)>1/2f(a2x)-f(a), 所以f(ax2)-f(a2x)>2[f(x)-f(a)], 即f(ax2-a2x)>2f(x-a)=f(2x-2a), 因为f(x)在R上单调递减, 所以ax2-a2x<2(x-a), 即(x-a)(ax-2)<0, 又a>0,所以(x-a)(x"-" 2/a)<0,当0<a<2/a,即0<a<√2时,原不等式的解集为{x├|a<x<2/a┤}; 当0<a=2/a,即a=√2时,原不等式的解集为⌀;当0<2/a<a,即a>√2时,原不等式的解集为{x├|2/a<x<a┤}. 综上所述,当0<a<√2时,原不等式的解集为{x├|a<x<2/a┤};当a=√2时,原不等式的解集为⌀; 当a>√2时,原不等式的解集为{x├|2/a<x<a┤}. 16.B　[解析] 不等式cos5θ-sin5θ<7(sin3θ-cos3θ)等价于sin3θ+1/7sin5θ>cos3θ+1/7cos5θ.令f(x)=x3+1/7x5,则f(x)是R上的增函数,所以sin θ>cos θ,故2kπ+π/4<θ<2kπ+5π/4(k∈Z).因为θ∈[0,2π),所以θ的取值范围是(π/4 "," 5π/4).故选B. 17.B　[解析] 令f(x)=|(1/2)^x+a|,则f(-x)=|(1/2)^("-" x)+a|=|2x+a|,由题意得f(x)=|(1/2)^x+a|与f(-x)=|2x+a|在区间[1,2024]上的单调性相同.若两函数都单调递增,则{■((1/2)^x+a≤0"," @2^x+a≥0)┤在区间[1,2024]上恒成立,即{■(a≤"-" 1/2 "," @a≥"-" 2"," )┤所以-2≤a≤-1/2.若两函数都单调递减,则{■((1/2)^x+a≥0"," @2^x+a≤0)┤在区间[1,2024]上恒成立,即{■(a≥"-" (1/2)^2024 "," @a≤"-" 2^2024 "," )┤无解.综上,实数a的取值范围是["-" 2",-" 1/2].故选 学科网（北京）股份有限公司 $$