江蘇專用2025版高考數(shù)學(xué)總復(fù)習(xí)第二章函數(shù)第二節(jié)函數(shù)的單調(diào)性與最值練習(xí)含解析

PAGEPAGE13其次節(jié)函數(shù)的單調(diào)性與最值學(xué)習(xí)要求:1.借助函數(shù)圖象,會(huì)用符號(hào)語(yǔ)言表達(dá)函數(shù)的單調(diào)性、最大值、最小值.2.理解函數(shù)的單調(diào)性、最大值、最小值的作用和實(shí)際意義.1.函數(shù)的單調(diào)性(1)增函數(shù)與減函數(shù)的定義:增函數(shù)減函數(shù)定義一般地,設(shè)函數(shù)y=f(x)的定義域?yàn)镮,且D?I,假如對(duì)隨意x1,x2∈D當(dāng)x1<x2時(shí),都有①f(x1)<f(x2),那么就稱y=f(x)在區(qū)間D上是增函數(shù)當(dāng)x1<x2時(shí),都有②f(x1)>f(x2),那么就稱y=f(x)在區(qū)間D上是減函數(shù)圖象描述自左向右看圖象是③上升的自左向右看圖象是④下降的(2)單調(diào)區(qū)間的定義:若函數(shù)y=f(x)在區(qū)間D上⑤單調(diào)遞增或單調(diào)遞減,則稱函數(shù)y=f(x)在這一區(qū)間具有(嚴(yán)格的)單調(diào)性,區(qū)間D叫做y=f(x)的單調(diào)區(qū)間.

?提示(1)求函數(shù)的單調(diào)區(qū)間或探討函數(shù)的單調(diào)性必需先求函數(shù)的定義域.(2)一個(gè)函數(shù)的同一種單調(diào)區(qū)間用“和”或“,”連接,不能用“∪”連接.(3)“函數(shù)的單調(diào)區(qū)間為M”與“函數(shù)在區(qū)間N上單調(diào)”是兩個(gè)不同的概念,明顯N?M.2.函數(shù)的最值前提設(shè)函數(shù)y=f(x)的定義域?yàn)镮,假如存在實(shí)數(shù)M滿意條件(1)對(duì)于隨意的x∈I,都有⑥f(x)≤M;(2)存在x0∈I,使得⑦f(x0)=M(1)對(duì)于隨意的x∈I,都有⑧f(x)≥M;(2)存在x0∈I,使得⑨f(x0)=M結(jié)論M為函數(shù)y=f(x)的最大值M為函數(shù)y=f(x)的最小值學(xué)問(wèn)拓展1.單調(diào)性定義的等價(jià)形式設(shè)隨意的x1,x2∈[a,b],x1≠x2.(1)若有(x1-x2)[f(x1)-f(x2)]>0或f(x1)-f(x2)x1-x2(2)若有(x1-x2)[f(x1)-f(x2)]<0或f(x1)-f(x2)x1-x22.復(fù)合函數(shù)的單調(diào)性函數(shù)y=f(u),u=φ(x),在函數(shù)y=f(φ(x))的定義域上,假如y=f(u)與u=φ(x)的單調(diào)性相同,那么y=f(φ(x))單調(diào)遞增;假如y=f(u)與u=φ(x)的單調(diào)性相反,那么y=f(φ(x))單調(diào)遞減.3.函數(shù)單調(diào)性的常用結(jié)論(1)若f(x),g(x)均為區(qū)間A上的增(減)函數(shù),則f(x)+g(x)也是區(qū)間A上的增(減)函數(shù).(2)若k>0,則kf(x)與f(x)的單調(diào)性相同;若k<0,則kf(x)與f(x)的單調(diào)性相反.(3)函數(shù)y=f(x)(f(x)>0)與y=-f(x),y=1f((4)函數(shù)y=x+ax(a>0)的增區(qū)間為(-∞,-a]和[a,+∞),減區(qū)間為(-a,0)和(0,a)1.推斷正誤(正確的打“√”,錯(cuò)誤的打“?”).(1)若定義在R上的函數(shù)f(x)滿意f(-1)<f(3),則函數(shù)f(x)在R上為增函數(shù).()(2)若函數(shù)y=f(x)在[1,+∞)上是增函數(shù),則函數(shù)f(x)的單調(diào)遞增區(qū)間是[1,+∞). ()(3)函數(shù)y=1x的單調(diào)遞減區(qū)間是(-∞,0)∪(0,+∞). ((4)全部的單調(diào)函數(shù)都有最值. ()(5)閉區(qū)間上的單調(diào)函數(shù),其最值肯定在區(qū)間端點(diǎn)處取到. ()答案(1)?(2)?(3)?(4)?(5)√2.(新教材人教A版必修第一冊(cè)P79T3改編)下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞減的是()A.y=1x-B.y=x2-xC.y=lnx-xD.y=ex答案A3.(新教材人教A版必修第一冊(cè)P85T1改編)已知函數(shù)f(x)的圖象如圖所示,則函數(shù)g(x)=log12f(x)的單調(diào)遞增區(qū)間為 (A.(-∞,-3],[0,3]B.[-3,0],[3,+∞)C.(-∞,-5),[0,1)D.(-1,0],(5,+∞)答案C4.(新教材人教A版必修第一冊(cè)P81例5改編)函數(shù)y=2x-1在區(qū)間[2,3]答案25.函數(shù)f(x)=1x2-答案(-∞,-1)確定函數(shù)的單調(diào)性(區(qū)間)角度一確定不含參函數(shù)的單調(diào)性(區(qū)間)典例1(1)(2024湖北荊州高三期末)設(shè)max{a,b}=a,a≥b,b,a<b,則函數(shù)f(x)=max{x2A.[-1,0],12B.(-∞,-1],0C.-∞,-12D.-1(2)(2024黑龍江大慶高三模擬)函數(shù)f(x)=x2+x-A.(-∞,-3)B.[2,+∞)C.[0,2)D.[-3,2]答案(1)D(2)B解析(1)由x2-x=1-x2得2x2-x-1=0,解得x=1或x=-12當(dāng)x≥1或x≤-12時(shí),f(x)=max{x2-x,1-x2}=x2-x,此時(shí)函數(shù)f(x)的單調(diào)遞增區(qū)間為當(dāng)-12<x<1時(shí),f(x)=max{x2-x,1-x2}=1-x2,此時(shí)函數(shù)f(x)的單調(diào)遞增區(qū)間為-12,0.綜上所述,函數(shù)f(x(2)要使函數(shù)有意義,則x2+x-6≥0,解得x≥2或x≤-3,易知y=x2+x-6在區(qū)間(-∞,-3]上單調(diào)遞減,在區(qū)間[2,+∞)上單調(diào)遞增,y=x在定義域內(nèi)單調(diào)遞增,結(jié)合復(fù)合函數(shù)的單調(diào)性可得函數(shù)f(x)=x2+x角度二確定含參函數(shù)的單調(diào)性(區(qū)間)典例2(1)試探討函數(shù)f(x)=axx-1(a≠0)在(2)已知f(x)=xx-a(a∈R,x≠①若a=-2,試證明f(x)在(-∞,-2)上單調(diào)遞增;②若a>0,且f(x)在(1,+∞)上單調(diào)遞減,求a的取值范圍.解析(1)f(x)=a·x-任取x1,x2∈(-1,1),且x1<x2,則f(x1)-f(x2)=a1+1因?yàn)?1<x1<x2<1,所以x2-x1>0,x1-1<0,x2-1<0,故當(dāng)a>0時(shí),f(x1)-f(x2)>0,即f(x1)>f(x2),函數(shù)f(x)在(-1,1)上單調(diào)遞減;當(dāng)a<0時(shí),f(x1)-f(x2)<0,即f(x1)<f(x2),函數(shù)f(x)在(-1,1)上單調(diào)遞增.(2)①證明:任取x1,x2∈(-∞,-2),且x1<x2,則f(x1)-f(x2)=x=2(易知(x1+2)(x2+2)>0,x1-x2<0,所以f(x1)-f(x2)<0,即f(x1)<f(x2),所以f(x)在(-∞,-2)上單調(diào)遞增.②任取x1,x2∈(1,+∞),且x1<x2,則f(x1)-f(x2)=x=a(因?yàn)閍>0,x2-x1>0,所以要使f(x1)-f(x2)>0恒成立,只需(x1-a)(x2-a)>0恒成立,所以a≤1.綜上所述,0<a≤1.名師點(diǎn)評(píng)1.求函數(shù)的單調(diào)區(qū)間時(shí),應(yīng)先求函數(shù)的定義域,在定義域內(nèi)求單調(diào)區(qū)間,單調(diào)區(qū)間不能用集合或不等式表示,且圖象不連續(xù)的單調(diào)區(qū)間要用“和”或“,”連接.2.(1)函數(shù)單調(diào)性的推斷方法:①定義法;②圖象法;③利用已知函數(shù)的單調(diào)性;④導(dǎo)數(shù)法.(2)函數(shù)y=f[g(x)]的單調(diào)性應(yīng)依據(jù)外層函數(shù)y=f(t)和內(nèi)層函數(shù)t=g(x)的單調(diào)性推斷,遵循“同增異減”的原則.推斷并證明函數(shù)f(x)=ax2+1x(其中1<a<3)在x∈[1,2]上的單調(diào)性解析函數(shù)f(x)在[1,2]上單調(diào)遞增.證明如下:任取x1,x2∈[1,2],且x1<x2,則f(x2)-f(x1)=ax22+1x2-ax12-1x1=(x2-x1)a(x1+x2)-1x1x2,由1≤因?yàn)?<a<3,所以2<a(x1+x2)<12,得a(x1+x2)-1x從而f(x2)-f(x1)>0,即f(x2)>f(x1),故當(dāng)a∈(1,3)時(shí),f(x)在[1,2]上單調(diào)遞增.函數(shù)單調(diào)性的應(yīng)用角度一利用單調(diào)性比較大小典例3(2024河南鄭州模擬)已知函數(shù)f(x)=ex-e-x,x>0,-x2,x≤0,若a=50.01A.f(b)>f(a)>f(c)B.f(c)>f(a)>f(b)C.f(a)>f(c)>f(b)D.f(a)>f(b)>f(c)答案D當(dāng)x>0時(shí),f(x)=ex-e-x單調(diào)遞增,且f(0)=0;當(dāng)x≤0時(shí),f(x)=-x2單調(diào)遞增,且f(0)=0,所以函數(shù)f(x)在R上單調(diào)遞增,因?yàn)閍=50.01>1,0<b=log322<1,c=log30.9<0,所以a>b>c,所以f(a)>f(b)>f(c).角度二利用單調(diào)性解不等式典例4(2024山東聊城三模)已知函數(shù)f(x)=3e-x,x≤0,-4x+3,x>0,若A.(-∞,1]B.(-∞,-3]∪[1,+∞)C.(-∞,1]∪[3,+∞)D.[-3,1]答案D當(dāng)x≤0時(shí),f(x)=3e-x單調(diào)遞減;當(dāng)x>0時(shí),f(x)=-4x+3單調(diào)遞減.又3e0=-4×0+3=3,所以函數(shù)y=f(x)在R上連續(xù),則函數(shù)y=f(x)在R上單調(diào)遞減.作出函數(shù)y=f(x)的圖象如圖所示.由f(a2-3)≥f(-2a)可得a2-3≤-2a,即a2+2a-3≤0,解得-3≤a≤1.故實(shí)數(shù)a的取值范圍是[-3,1].角度三利用函數(shù)的單調(diào)性求參數(shù)的取值范圍典例5(2024廣西柳州試驗(yàn)中學(xué)高三開(kāi)學(xué)考試)已知函數(shù)f(x)=logax,0<x<1,(4a-1)x+2a,x≥1滿意對(duì)隨意x1≠A.0C.0,1答案B因?yàn)楹瘮?shù)f(x)對(duì)隨意x1≠x2且x1,x2∈(0,+∞),都有f(x1)-f(x2)x1-x名師點(diǎn)評(píng)1.(1)比較函數(shù)值的大小時(shí),應(yīng)將自變量轉(zhuǎn)化到同一個(gè)單調(diào)區(qū)間內(nèi),然后利用函數(shù)的單調(diào)性解決;(2)求解與抽象函數(shù)有關(guān)的不等式時(shí),往往利用函數(shù)的單調(diào)性脫去“f”,使其轉(zhuǎn)化為求解詳細(xì)的不等式.此時(shí)應(yīng)特殊留意函數(shù)的定義域.2.利用單調(diào)性求參數(shù)的值(取值范圍)的思路:依據(jù)函數(shù)的單調(diào)性干脆構(gòu)建參數(shù)滿意的方程(組)(不等式(組)),或先得到其圖象的升降,再結(jié)合圖象求解.對(duì)于分段函數(shù),要留意連接點(diǎn)的取值.1.已知函數(shù)f(x)的圖象向左平移1個(gè)單位長(zhǎng)度后關(guān)于y軸對(duì)稱,當(dāng)x2>x1>1時(shí),[f(x2)-f(x1)](x2-x1)<0恒成立,設(shè)a=f-12,b=f(2),c=f(3),則a,b,c的大小關(guān)系為 (A.c>a>bB.c>b>aC.a>c>bD.b>a>c答案D由于函數(shù)f(x)的圖象向左平移1個(gè)單位長(zhǎng)度后得到的圖象關(guān)于y軸對(duì)稱,故函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱,所以a=f-12=f52.當(dāng)x2>x1>1時(shí),[f(x2)-f(x1)](x2-x1)<0恒成立等價(jià)于函數(shù)f(x)在(1,+∞)上單調(diào)遞減,2.假如函數(shù)f(x)=ax+1x+2a在區(qū)間(-2,+∞)上是增函數(shù),那么答案[1,+∞)解析f(x)=ax+2∵函數(shù)f(x)在區(qū)間(-2,+∞)上是增函數(shù),∴2即a2>12,求函數(shù)的最值(值域)典例6(1)(2024安徽六安一中高三月考)若函數(shù)f(x)=2x2+31+x2,則f(A.(-∞,3]B.(2,3)C.(2,3]D.[3,+∞)(2)已知函數(shù)f(x)=x+2x-3,x≥1,lg(x2+1答案(1)C(2)0;22-3解析(1)∵f(x)=2x2+31+x2=2(x2+1∴f(x)的值域?yàn)?2,3],故選C.(2)∵f(-3)=lg[(-3)2+1]=lg10=1,∴f[f(-3)]=f(1)=0.當(dāng)x≥1時(shí),f(x)=x+2x-3≥22-3,當(dāng)且僅當(dāng)x=2時(shí),取等號(hào),此時(shí)f(x當(dāng)x<1時(shí),f(x)=lg(x2+1)≥lg1=0,當(dāng)且僅當(dāng)x=0時(shí),取等號(hào),此時(shí)f(x)min=0.∴函數(shù)f(x)的最小值為22-3.名師點(diǎn)評(píng)求函數(shù)最值的五種常用方法(1)單調(diào)性法:先確定函數(shù)的單調(diào)性,再由單調(diào)性求最值.(2)圖象法:先作出函數(shù)的圖象,再視察其最高點(diǎn)、最低點(diǎn),求出最值.(3)基本不等式法:先對(duì)解析式變形,使之具備“一正二定三相等”的條件后用基本不等式求出最值.(4)導(dǎo)數(shù)法:先求導(dǎo),然后求出在給定區(qū)間上的極值,最終結(jié)合端點(diǎn)值,求出最值.(5)換元法:對(duì)比較困難的函數(shù)可通過(guò)換元將其轉(zhuǎn)化為熟識(shí)的函數(shù),再用相應(yīng)的方法求最值.定義max{a,b,c}為a,b,c中的最大值,設(shè)函數(shù)M=max{2x,2x-3,6-x},則M的最小值是()A.2B.3C.4D.6答案C畫(huà)出函數(shù)M=max{2x,2x-3,6-x}的圖象如圖,由圖可知,函數(shù)在A(2,4)處取得最小值4.A組基礎(chǔ)達(dá)標(biāo)1.(2024海安校級(jí)月考)已知函數(shù)f(x)=x2-4,則函數(shù)的值域?yàn)锳.(0,+∞)B.[0,+∞)C.(2,+∞)D.[2,+∞)答案B2.(多選題)下列函數(shù)中,在區(qū)間(0,1)上是增函數(shù)的是 ()A.y=|x|B.y=x+3C.y=1xD.y=-x2答案AB3.函數(shù)f(x)=-x+1x在-2A.32B.-8答案A4.(多選題)(2024山東棗莊模擬)已知函數(shù)f(x)=bx+3ax+2在區(qū)間(-2,+∞)上單調(diào)遞增,則a、b的取值可以是A.a=1,b>32B.0<a≤1,bC.a=-1,b=2D.a=12,b答案ABD由題意知,不等式ax+2≠0對(duì)隨意的x∈(-2,+∞)恒成立.①當(dāng)a=0時(shí),f(x)=b2x+32在區(qū)間(-2,+∞)上單調(diào)遞增,則b②當(dāng)a>0時(shí),由ax+2≠0,可得x≠-2a,則-2a≤-2,解得0<a則f(x)=bx+3∵函數(shù)f(x)在區(qū)間(-2,+∞)上單調(diào)遞增,∴3-2ba<0,∴b>當(dāng)a=1時(shí),b>3a2=當(dāng)0<a≤1時(shí),b=2>3a2恒成立,B當(dāng)a=12時(shí),b=1>3a2恒成立③當(dāng)a<0時(shí),-2a>0,函數(shù)f(x)在x=-2a時(shí)沒(méi)有定義,C5.(2024山東臨沂高三月考)已知函數(shù)f(x)=2x(1)用定義法證明f(x)在區(qū)間[1,+∞)上是增函數(shù);(2)求該函數(shù)在區(qū)間[2,4]上的最大值與最小值.解析(1)證明:任取x1,x2∈[1,+∞),且x1<x2,則f(x1)-f(x2)=2x∵1≤x1<x2,∴x1-x2<0,(x1+1)(x2+1)>0,∴f(x1)-f(x2)<0,即f(x1)<f(x2),故函數(shù)f(x)在區(qū)間[1,+∞)上是增函數(shù).(2)由(1)知函數(shù)f(x)在區(qū)間[2,4]上是增函數(shù),∴f(x)max=f(4)=2×4+14+1B組實(shí)力拔高6.(多選題)(2024山東嘉祥一中高三月考)假如對(duì)于函數(shù)f(x)的定義域內(nèi)隨意的兩個(gè)自變量的值x1,x2,當(dāng)x1<x2時(shí),都有f(x1)≤f(x2),且存在兩個(gè)不相等的自變量的值y1,y2,使得f(y1)=f(y2),那么就稱f(x)為定義域上的“不嚴(yán)格的增函數(shù)”.下列所給出的四個(gè)函數(shù)中為“不嚴(yán)格的增函數(shù)”的是 ()A.f(x)=x,x≥10,-1C.f(x)=1,x≥10,-1答案ACf(x)=x,x≥1,0,-1<x<f(x)=1,x=-π2,sinx,-π2<x≤π2,當(dāng)x1=-π2,x2∈f(x)=1,x≥1,0,-1<xf(x)=x,x≥1,x+1,x<1,當(dāng)x1=12,x2∈1,32時(shí),f7.(多選題)(2024海南華僑中學(xué)高三模擬)已知函數(shù)f(x)=2x,g(x)=x2+ax(其中a∈R).對(duì)于不相等的實(shí)數(shù)x1,x2,設(shè)m=f(x1)-f(x2)A.對(duì)于隨意不相等的實(shí)數(shù)x1,x2,都有m>0B.對(duì)于隨意的a及隨意不相等的實(shí)數(shù)x1,x2,都有n>0C.對(duì)于隨意的a,存在不相等的實(shí)數(shù)x1,x2,使得m=nD.對(duì)于隨意的a,存在不相等的實(shí)數(shù)x1,x2,使得m=-n答案AD對(duì)于A,由指數(shù)函數(shù)的單調(diào)性可得函數(shù)f(x)在R上遞增,即有m>0,則A中說(shuō)法正確;對(duì)于B,由二次函數(shù)的單調(diào)性可得函數(shù)g(x)在-∞,-a2上遞減,在-a2,+∞上遞增,則n對(duì)于C,若m=n,則f(x1)-f(x2)=g(x1)-g(x2),即g(x1)-f(x1)=g(x2)-f(x2),設(shè)h(x)=x2+ax-2x,則應(yīng)有h(x1)=h(x2),而h′(x)=2x+a-2xln2,當(dāng)a→-∞時(shí),h′(x)小于0,函數(shù)h(x)單調(diào)遞減,則C中說(shuō)法錯(cuò)誤;對(duì)于D,若m=-n,則f(x1)-f(x2)=-[g(x1)-g(x2)],即f(x1)+g(x1)=f(x2)+g(x2)設(shè)p(x)=x2+ax+2x,則應(yīng)有p(x1)=p(x2),而p′(x)=2x+a+2xln2,對(duì)于隨意的a,p′(x)不恒大于0或小于0,即p(x)在定義域上有增有減,則D中說(shuō)法正確.8.已知函數(shù)f(x)=x2-2ax+a在區(qū)間(-∞,1)上有最小值,則函數(shù)g(x)=f(x)x在區(qū)間(1,+∞)上A.有最小值B.有最大值C.是減函數(shù)D.是增函數(shù)答案D因?yàn)楹瘮?shù)f(x)=x2-2ax+a=(x-a)2+a-a2在區(qū)間(-∞,1)上有最小值,所以函數(shù)f(x)圖象