函數(shù)的單調(diào)性 同步練習(xí)（二）-高一上學(xué)期數(shù)學(xué)蘇教版（2019）必修第一冊(cè)

《第三節(jié)函數(shù)的單調(diào)性》同步練習(xí)一、選擇題1.函數(shù)f(x)=?x2?2A.[-3,-1] B.[-1,1]C.(-∞,-3] D.[-1,+∞)2.max{a,b}表示取a,b中的較大者,例如max{0,1,-2}=1,max{2,2}=2,則函數(shù)f(x)=max{x+1,4-2x}的最小值為()A.1 B.2 C.3 D.43.已知函數(shù)f(x)=4?x2,若0<x1<x2<x3≤2,則f(x1)xA.fB.fC.fD.f4.[2022山西晉中新一雙語學(xué)校高三上月考]已知函數(shù)f(x)的定義域?yàn)镽,對(duì)任意的x1<x2,都有f(x1)-f(x2)<x1-x2,且f(3)=4,則f(2x-1)>2x的解集為()A.(2,+∞) B.(1,+∞)C.(0,+∞) D.(-1,+∞)5.(多選)[2022江蘇省通州高級(jí)中學(xué)高一期中]已知函數(shù)f(x)=x|x|,若對(duì)任意的x∈[t,t+1],不等式f(x+t)≥3f(x)恒成立,則整數(shù)t的取值可以是()A.-1 B.1 C.3 D.5二、填空題6.[2022廣東汕頭河溪中學(xué)高一上期中]已知函數(shù)f(x)=ax?1,x<1,x2?2ax,x≥1,對(duì)?x1,x2∈R且x1≠7.[2022江蘇南京六校聯(lián)合體高一上期中]已知函數(shù)f(x)=x2-4x,g(x)=ax(a<0),對(duì)?x1∈[-2,-1],?x2∈[-3,-1],使f(x1)=g(x2)成立,則實(shí)數(shù)a的取值范圍是8.對(duì)任意實(shí)數(shù)a,b,函數(shù)F(a,b)=12(a+b-|a-b|),如果函數(shù)f(x)=-x2+2x+3,g(x)=x+1,那么函數(shù)G(x)=F(f(x),g(x))的最大值為9.已知函數(shù)y=x+ax有如下性質(zhì):若常數(shù)a>0,那么函數(shù)在(0,a]上是減函數(shù),在[a,+∞)上是增函數(shù).若函數(shù)f(x)=|x+4x-m|+m在區(qū)間[1,4]上的最小值為7,則實(shí)數(shù)m的值是三、解答題10.已知函數(shù)f(x)=x|x-m|+m2.(1)若函數(shù)f(x)在[1,2]上是增函數(shù),求實(shí)數(shù)m的取值范圍;(2)若函數(shù)f(x)在[1,2]上的最小值為7,求實(shí)數(shù)m的值.11.[2021重慶育才中學(xué)高一上期中]已知定義在R上的函數(shù)f(x)對(duì)任意x,y∈R都有f(x+y)=f(x)+f(y)-1,且當(dāng)x>0時(shí),f(x)>1.(1)求證:f(x)在R上是增函數(shù);(2)若f(3)=4,關(guān)于x的不等式f(x+2+t)+f(2?x)>3有解,求實(shí)數(shù)t12.[2022江蘇揚(yáng)州大學(xué)附屬中學(xué)高一上期中]已知函數(shù)f(x)=xx(1)求證:函數(shù)f(x)在區(qū)間[0,+∞)上是增函數(shù);(2)若?a∈[-2,2],?x∈[12,2]使不等式f(x)≤m2-2am+103成立,求實(shí)數(shù)m參考答案一、選擇題1.B由-x2-2x+3≥0,得-3≤x≤1,所以函數(shù)f(x)的定義域?yàn)閇-3,1].f(x)可以看成由y=t及t=-x2-2x+3復(fù)合而成.因?yàn)楹瘮?shù)t=-x2-2x+3在[-3,-1]上是增函數(shù),在[-1,1]上是減函數(shù),函數(shù)y=t在[0,+∞)上是增函數(shù),所以根據(jù)復(fù)合函數(shù)單調(diào)性“同增異減”的判斷方法,可知函數(shù)f(x)的減區(qū)間是[-1,1],故選B.2.B當(dāng)x+1≥4-2x,即x≥1時(shí),f(x)=x+1;當(dāng)x+1<4-2x,即x<1時(shí),f(x)=4-2x.所以f(x)=4?2x,x<1,x+1,x≥1.顯然f(x)在(-∞,1)上單調(diào)遞減,在[1,+∞)上單調(diào)遞增,且4-2≤1+1,所以f3.C由題意可得f(x)x=4?x2x=4x4.A5.CD二、填空題6.(0,237.[-15,-12]8.39.6三、解答題10.(1)由題意,得f(x)=x①若m=0,則當(dāng)x≥0時(shí),f(x)=x2,此時(shí)函數(shù)f(x)在[1,2]上是增函數(shù);②若m<0,則[1,2]?[m,+∞).二次函數(shù)y=x2-mx+m2的圖象的開口向上,對(duì)稱軸為直線x=m2如圖1所示,則函數(shù)y=f(x)在[1,2]上是增函數(shù);③若m>0,如圖2所示,若函數(shù)f(x)在[1,2]上是增函數(shù),則m2≥2或m≤1,所以0<m≤1或m≥4綜上所述,實(shí)數(shù)m的取值范圍是(-∞,1]∪[4,+∞).(2)①由(1),知當(dāng)m≤1時(shí),函數(shù)f(x)在[1,2]上是增函數(shù),所以f(x)min=f(1)=1-m+m2=7,即m2-m-6=0,解得m=3(舍去)或m=-2;②由(1),知當(dāng)m≥4時(shí),函數(shù)f(x)在[1,2]上是增函數(shù),所以f(x)min=f(1)=-1+m+m2=7,即m2+m-8=0,所以m=?1±332③當(dāng)2<m<4時(shí),函數(shù)f(x)在[1,m2]上是增函數(shù),在[m2,2]因?yàn)閒(1)=-1+m+m2,f(2)=m2+2m-4,所以f(2)-f(1)=m-3.當(dāng)3≤m<4時(shí),f(2)≥f(1),所以f(x)min=f(1)=7,即m2+m-8=0,解得m=?1±332當(dāng)2<m<3時(shí),f(2)<f(1),則f(x)min=f(2)=7,即m2+2m-11=0,解得m=23-1或m=-23-1(舍去);④當(dāng)1<m≤2時(shí),m2若1<m<2,則函數(shù)f(x)在[1,m]上是減函數(shù),在[m,2]上是增函數(shù),若m=2,則函數(shù)f(x)在[1,m]上為減函數(shù),所以f(x)min=f(m)=m2=7,得m=±7(均舍去).綜上,m=-2或m=23-1.11.(1)任取x1,x2∈R,且x1<x2,則x2-x1>0,所以f(x2-x1)>1.又f(x2)=f(x1)+f(x2-x1)-1,所以f(x2)-f(x1)=f(x2-x1)-1>0,即f(x2)>f(x1).故f(x)在R上是增函數(shù).(2)因?yàn)閒(3)=f(1)+f(2)-1=f(1)-1+f(1)+f(1)-1=3f(1)-2=4,所以f(1)=2.因?yàn)閒(x+2+t)+f(2?x)=f(x+2所以f(x+2+t)+f(2?x)>3即f(x+2+2?x+t)>2=故x+2+2?x+t>1有解,即x令y=x+2+2?x則y2=4+24?因?yàn)間(x)=4+24?x2在(-2,0)上單調(diào)遞增,在所以g(x)max=4+24=8,又y>0,所以ymax=22,因此1-t<22,所以t>1-22,故實(shí)數(shù)t的取值范圍為(1-22,+∞).12.(1)f(x)=xx+1=1-令x2>x1≥0,則f(x2)-f(x1)=1-1x2+1-1+1x1+1=x2?x1故f(x)在區(qū)間[0,+∞)上是增函數(shù).(2)令g(a)=-2am+m2+103,則問題可轉(zhuǎn)化為a∈[-2,2],x∈[12,2]時(shí)g(a)min≥f(x)因?yàn)閒(x)=1-1x+1在[12所以f(x)min=f(12)=1當(dāng)m<0時(shí),g(a)在[-2,2]上單調(diào)遞增,則g(a)min=g(-2)=m2+4m+103≥f(x)min=1整理得(m+3)(m+1)≥0,所以m≥-1或m≤-3,所以m∈(-∞,-3]∪[