新高考數(shù)學(xué)一輪復(fù)習(xí)8.4 單調(diào)性（精練）（基礎(chǔ)版）（解析版）

8.4單調(diào)性（精練）（基礎(chǔ)版）題組一題組一性質(zhì)法1．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意，可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，二次函數(shù)SKIPIF1<0的對稱軸為SKIPIF1<0，且在SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0，根據(jù)復(fù)合函數(shù)的單調(diào)性，可知函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0.故選：B.2．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0．因為函數(shù)SKIPIF1<0是關(guān)于SKIPIF1<0的遞減函數(shù)，且SKIPIF1<0時，SKIPIF1<0為增函數(shù)，所以SKIPIF1<0為減函數(shù)，所以函數(shù)SKIPIF1<0的單調(diào)減區(qū)間是SKIPIF1<0．故選：C.3．（2022·全國·高三專題練習(xí)）（多選）下列函數(shù)中在區(qū)間SKIPIF1<0上單調(diào)遞減的函數(shù)有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】A選項:根據(jù)冪函數(shù)SKIPIF1<0中SKIPIF1<0時在SKIPIF1<0上單調(diào)遞增，故此選項不符合題意；B選項:將SKIPIF1<0圖像向左平移一個單位，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以符合題意；C選項:保留SKIPIF1<0圖像在SKIPIF1<0軸上方的部分，SKIPIF1<0軸下方圖像翻折到SKIPIF1<0軸的上方，根據(jù)圖像可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0上單調(diào)遞增，符合題意；D選項：SKIPIF1<0的圖像由指數(shù)函數(shù)SKIPIF1<0圖像向左平移一個單位得到，且底數(shù)大于1，所以SKIPIF1<0在R上單調(diào)遞增，所以不符合題意。故選：BC4．（2021·浙江·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為___________.【答案】SKIPIF1<0【解析】由SKIPIF1<0得，函數(shù)的定義域是R，設(shè)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上是減函數(shù)，在SKIPIF1<0上是增函數(shù)，∵SKIPIF1<0在定義域上減函數(shù)，∴函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<0故答案為：SKIPIF1<05．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是_____．【答案】SKIPIF1<0【解析】SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，對稱軸為SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0為單調(diào)遞增；在SKIPIF1<0上單調(diào)遞減.所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0.故答案為：SKIPIF1<06．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為__________.【答案】SKIPIF1<0或SKIPIF1<0【解析】由題意得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，（SKIPIF1<0），令SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，因為SKIPIF1<0在SKIPIF1<0上遞減，所以SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，故答案為：SKIPIF1<0或SKIPIF1<07．（2022·全國·高三專題練習(xí)）函數(shù)f(x)=lg(-SKIPIF1<0)的單調(diào)增區(qū)間____________.【答案】SKIPIF1<0【解析】令t=-SKIPIF1<0＞0，求得0＜x＜2，故函數(shù)的定義域為{x|0＜x＜2}，根據(jù)y=g（t）=lgt，本題即求函數(shù)t在定義域內(nèi)的增區(qū)間，再利用二次函數(shù)的性質(zhì)求得函數(shù)t在定義域內(nèi)的增區(qū)間為SKIPIF1<0，故答案為：SKIPIF1<0．8．（2021·江蘇省阜寧中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是_________，值域是______．【答案】

SKIPIF1<0

SKIPIF1<0【解析】令SKIPIF1<0，則由SKIPIF1<0，可得SKIPIF1<0．又因為SKIPIF1<0為減函數(shù)，而函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．故SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．易知SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為SKIPIF1<0，故SKIPIF1<0的值域為SKIPIF1<0．故答案為：SKIPIF1<0；SKIPIF1<0題組二題組二圖像法1．（2022·江蘇南通·高三期末）（多選）下列函數(shù)在區(qū)間SKIPIF1<0上單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】對于A：SKIPIF1<0為開口向上的拋物線，對稱軸為SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，故選項A不正確；對于B：SKIPIF1<0的定義域為SKIPIF1<0，將SKIPIF1<0的圖象向右平移一個單位可得SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，向右平移一個單位可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，故選項B正確；對于C：SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，故選項C正確；對于D：SKIPIF1<0是由SKIPIF1<0和SKIPIF1<0復(fù)合而成，因為SKIPIF1<0單調(diào)遞減，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，故選項D不正確；故選：BC.2．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的遞減區(qū)間是____.【答案】SKIPIF1<0【解析】由題意SKIPIF1<0，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0單調(diào)遞增；綜上所述，函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故答案為：SKIPIF1<0.3．（2022·全國·高三專題練習(xí)（文））函數(shù)SKIPIF1<0的單調(diào)減區(qū)間是_______．【答案】SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0∵SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減作出SKIPIF1<0的圖象由圖象可以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減故答案為：SKIPIF1<0.4．（2022·全國·高三專題練習(xí)）函數(shù)y=|-x2+2x+1|的單調(diào)遞增區(qū)間是_________；單調(diào)遞減區(qū)間是_________．【答案】

SKIPIF1<0，SKIPIF1<0

SKIPIF1<0，SKIPIF1<0【解析】作出函數(shù)y=|-x2+2x+1|的圖像，如圖所示，觀察圖像得，函數(shù)y=|-x2+2x+1|在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，所以原函數(shù)的單調(diào)增區(qū)間是SKIPIF1<0，SKIPIF1<0，單調(diào)遞減區(qū)間是SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0題組三題組三導(dǎo)數(shù)法1．（2022福建）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0和SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0.因此，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0.故選：D.2．（2022北京）（多選）下列函數(shù)中，既是奇函數(shù)又在區(qū)間SKIPIF1<0上單調(diào)遞增的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AB【解析】易知A，B，D均為奇函數(shù)，C為偶函數(shù)，所以排除C；對于A，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；對于B，SKIPIF1<0（不恒為零），所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；對于D，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．故選：AB．3．（2022河北）函數(shù)f(x)＝lnx－x的單調(diào)增區(qū)間是________.【答案】(0，1)【解析】f′(x)＝－1，令f′(x)>0，又x>0，∴0<x<1，則f(x)的單調(diào)增區(qū)間是(0，1).4．（2022湖南）函數(shù)f(x)＝2x3－9x2＋12x＋1的單調(diào)減區(qū)間是________.【答案】(1，2)【解析】f′(x)＝6x2－18x＋12，令f′(x)＜0，即6x2－18x＋12＜0，解得1＜x＜2.故答案為：(1，2)5（2022北京）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是________.【答案】SKIPIF1<0，SKIPIF1<0【解析】SKIPIF1<0的定義域是SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0，SKIPIF1<0故答案為：SKIPIF1<0，SKIPIF1<0題組四題組四已知單調(diào)性求參數(shù)1．（2022·江西·二模（文））已知函數(shù)SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】依題意，SKIPIF1<0解得a＝－1，故SKIPIF1<0，可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增故選：D2．（2022·陜西·武功縣普集高級中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0．因為SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以方程SKIPIF1<0的兩個根分別位于區(qū)間SKIPIF1<0和SKIPIF1<0上，所以SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0.故選：A．3．（2022·浙江·舟山中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0都有SKIPIF1<0成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】當(dāng)SKIPIF1<0時，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，因為SKIPIF1<0時SKIPIF1<0為增函數(shù)，又SKIPIF1<0為奇函數(shù)，SKIPIF1<0為SKIPIF1<0上單調(diào)遞增函數(shù)，SKIPIF1<0的圖象如下，由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0都成立，即SKIPIF1<0，解得SKIPIF1<0.故選：D.4．（2021·福建龍巖·高三期中）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍是___________【答案】[3，SKIPIF1<0）【解析】由題意，SKIPIF1<0，而函數(shù)SKIPIF1<0的對稱軸為：SKIPIF1<0，根據(jù)復(fù)合函數(shù)單調(diào)性“同增異減”的原則，函數(shù)SKIPIF1<0的增區(qū)間為：SKIPIF1<0