2024年高考數(shù)學(xué)真題分類匯編02：不等式與不等關(guān)系（含答案解析）

試卷第=page11頁，共=sectionpages33頁不等式與不等關(guān)系一、單選題1．（2024·全國1卷）已知函數(shù)為SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，則下列結(jié)論中一定正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2024·全國1卷）已知函數(shù)為SKIPIF1<0，在R上單調(diào)遞增，則a取值的范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2024·全國2卷）已知命題p：SKIPIF1<0，SKIPIF1<0；命題q：SKIPIF1<0，SKIPIF1<0，則（

）A．p和q都是真命題 B．SKIPIF1<0和q都是真命題C．p和SKIPIF1<0都是真命題 D．SKIPIF1<0和SKIPIF1<0都是真命題4．（2024·全國2卷）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．15．（2024·全國甲卷文）若實(shí)數(shù)SKIPIF1<0滿足約束條件SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2024·北京）已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07．（2024·北京）記水的質(zhì)量為SKIPIF1<0，并且d越大，水質(zhì)量越好．若S不變，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的關(guān)系為（

）A．SKIPIF1<0B．SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0；D．若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0；8．（2024·北京）已知SKIPIF1<0，SKIPIF1<0是函數(shù)SKIPIF1<0圖象上不同的兩點(diǎn)，則下列正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．（2024·天津）若SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題10．（2024·上海）已知SKIPIF1<0則不等式SKIPIF1<0的解集為．三、解答題11．（2024·全國甲卷文）已知函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0時(shí)，證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立．12．（2024·全國甲卷理）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．答案第=page11頁，共=sectionpages22頁參考答案：1．B【分析】代入得到SKIPIF1<0，再利用函數(shù)性質(zhì)和不等式的性質(zhì)，逐漸遞推即可判斷.【解析】因?yàn)楫?dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則依次下去可知SKIPIF1<0，則B正確；且無證據(jù)表明ACD一定正確.故選：B.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題的關(guān)鍵是利用SKIPIF1<0，再利用題目所給的函數(shù)性質(zhì)SKIPIF1<0，代入函數(shù)值再結(jié)合不等式同向可加性，不斷遞推即可.2．B【分析】根據(jù)二次函數(shù)的性質(zhì)和分界點(diǎn)的大小關(guān)系即可得到不等式組，解出即可.【解析】因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，則需滿足SKIPIF1<0，解得SKIPIF1<0，即a的范圍是SKIPIF1<0.故選：B.3．B【分析】對于兩個命題而言，可分別取SKIPIF1<0、SKIPIF1<0，再結(jié)合命題及其否定的真假性相反即可得解.【解析】對于SKIPIF1<0而言，取SKIPIF1<0，則有SKIPIF1<0，故SKIPIF1<0是假命題，SKIPIF1<0是真命題，對于SKIPIF1<0而言，取SKIPIF1<0，則有SKIPIF1<0，故SKIPIF1<0是真命題，SKIPIF1<0是假命題，綜上，SKIPIF1<0和SKIPIF1<0都是真命題.故選：B.4．C【分析】解法一：由題意可知：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，分類討論SKIPIF1<0與SKIPIF1<0的大小關(guān)系，結(jié)合符號分析判斷，即可得SKIPIF1<0，代入可得最值；解法二：根據(jù)對數(shù)函數(shù)的性質(zhì)分析SKIPIF1<0的符號，進(jìn)而可得SKIPIF1<0的符號，即可得SKIPIF1<0，代入可得最值.【解析】解法一：由題意可知：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，令SKIPIF1<0解得SKIPIF1<0；令SKIPIF1<0解得SKIPIF1<0；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，此時(shí)SKIPIF1<0，不合題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，此時(shí)SKIPIF1<0，不合題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，此時(shí)SKIPIF1<0；可知若SKIPIF1<0，符合題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，此時(shí)SKIPIF1<0，不合題意；綜上所述：SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號成立，所以SKIPIF1<0的最小值為SKIPIF1<0；解法二：由題意可知：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，令SKIPIF1<0解得SKIPIF1<0；令SKIPIF1<0解得SKIPIF1<0；則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0；故SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號成立，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：C.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：分別求SKIPIF1<0、SKIPIF1<0的根，以根和函數(shù)定義域?yàn)榕R界，比較大小分類討論，結(jié)合符號性分析判斷.5．D【分析】畫出可行域后，利用SKIPIF1<0的幾何意義計(jì)算即可得.【解析】實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，作出可行域如圖：由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0的幾何意義為SKIPIF1<0的截距的SKIPIF1<0，則該直線截距取最大值時(shí)，SKIPIF1<0有最小值，此時(shí)直線SKIPIF1<0過點(diǎn)SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0.故選：D.6．A【分析】直接根據(jù)并集含義即可得到答案.【解析】由題意得SKIPIF1<0，故選：A.7．C【分析】根據(jù)題意分析可得SKIPIF1<0，討論SKIPIF1<0與1的大小關(guān)系，結(jié)合指數(shù)函數(shù)單調(diào)性分析判斷.【解析】由題意可得SKIPIF1<0，解得SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0；結(jié)合選項(xiàng)可知C正確，ABD錯誤；故選：C.8．A【分析】根據(jù)指數(shù)函數(shù)和對數(shù)函數(shù)的單調(diào)性結(jié)合基本不等式分析判斷AB；舉例判斷CD即可.【解析】由題意不妨設(shè)SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0是增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，對于選項(xiàng)AB：可得SKIPIF1<0，即SKIPIF1<0，根據(jù)函數(shù)SKIPIF1<0是增函數(shù)，所以SKIPIF1<0，故A正確，B錯誤；對于選項(xiàng)C：例如SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，故C錯誤；對于選項(xiàng)D：例如SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，故D錯誤，故選：A.9．B【分析】利用指數(shù)函數(shù)和對數(shù)函數(shù)的單調(diào)性分析判斷即可.【解析】因?yàn)镾KIPIF1<0在SKIPIF1<0上遞增，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上遞增，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故選：B10．SKIPIF1<0【分析】求出方程SKIPIF1<0的解后可求不等式的解集.【解析】方程SKIPIF1<0的解為SKIPIF1<0或SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0，故答案為：SKIPIF1<0.11．(1)見解析(2)見解析【分析】（1）求導(dǎo)，含參分類討論得出導(dǎo)函數(shù)的符號，從而得出原函數(shù)的單調(diào)性；（2）先根據(jù)題設(shè)條件將問題可轉(zhuǎn)化成證明當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0即可.【解析】（1）SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減.綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.（2）SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，下證SKIPIF1<0即可.SKIPIF1<0，再令SKIPIF1<0，則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上遞增，故SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，問題得證12．(1)極小值為SKIPIF1<0，無極大值.(2)SKIPIF1<0【分析】（1）求出函數(shù)的導(dǎo)數(shù)，根據(jù)導(dǎo)數(shù)的單調(diào)性和零點(diǎn)可求函數(shù)的極值.（2）求出函數(shù)的二階導(dǎo)數(shù)，就SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分類討論后可得參數(shù)的取值范圍.【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，而SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF