2023高考數(shù)學(xué)真題分類(lèi)匯編：第03章 函數(shù) (含解析)

第三章函數(shù)第二節(jié)函數(shù)的基本性質(zhì)1.（2023全國(guó)甲卷理科13，文科14）若SKIPIF1<0為偶函數(shù)，則SKIPIF1<0.【分析】利用偶函數(shù)的性質(zhì)得到SKIPIF1<0，從而求得SKIPIF1<0，再檢驗(yàn)即可得解.

【解析】因?yàn)镾KIPIF1<0為偶函數(shù)，定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，

則SKIPIF1<0，故a=2，

此時(shí)SKIPIF1<0，

所以SKIPIF1<0，

又定義域?yàn)镾KIPIF1<0，故SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0.

故答案為2.2.（2023全國(guó)乙卷理科4，文科5）已知SKIPIF1<0是偶函數(shù)，則SKIPIF1<0（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【分析】根據(jù)偶函數(shù)的定義運(yùn)算求解.

【解析】因?yàn)镾KIPIF1<0為偶函數(shù)，則SKIPIF1<0，

又因?yàn)镾KIPIF1<0不恒為0，可得SKIPIF1<0，即SKIPIF1<0，

則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.

故選D.3.（2023新高考I卷11）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，則（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0是偶函數(shù) D.SKIPIF1<0為SKIPIF1<0的極小值點(diǎn)【解析】選項(xiàng)A，令SKIPIF1<0，則SKIPIF1<0，故A正確；選項(xiàng)B，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，故B正確；選項(xiàng)C，令SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，故C正確；選項(xiàng)D，對(duì)式子兩邊同時(shí)除以SKIPIF1<0，得到SKIPIF1<0，故可以設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增.又SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.SKIPIF1<0的圖像如圖所示，所以SKIPIF1<0為SKIPIF1<0的極大值點(diǎn)，故D錯(cuò)誤.故選ABC.4.（2023新高考II卷4）若SKIPIF1<0為偶函數(shù)，則SKIPIF1<0（）A. SKIPIF1<0B. 0C. SKIPIF1<0D. SKIPIF1<0【解析】SKIPIF1<0，則SKIPIF1<0.因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，所以有SKIPIF1<0，得SKIPIF1<0.故選B.5.（2023北京卷4）下列函數(shù)中，在區(qū)間SKIPIF1<0上單調(diào)遞增的是（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【分析】利用基本初等函數(shù)的單調(diào)性，結(jié)合復(fù)合函數(shù)的單調(diào)性判斷ABC，舉反例排除D即可.【解析】對(duì)于A，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A錯(cuò)誤；對(duì)于B，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故B錯(cuò)誤；對(duì)于C，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；對(duì)于D，因?yàn)镾KIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上不單調(diào)，D錯(cuò)誤.故選C.6.（2023北京卷15）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在最大值；③設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；④設(shè)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0存在最小值，則SKIPIF1<0的取值范圍是SKIPIF1<0.其中所有正確結(jié)論的序號(hào)是.【分析】先分析SKIPIF1<0圖像，再逐一分析各結(jié)論；對(duì)于①，取SKIPIF1<0，結(jié)合圖像即可判斷；對(duì)于②，分段討論SKIPIF1<0的取值范圍，從而得以判斷；對(duì)于③，結(jié)合圖像可知SKIPIF1<0的范圍；對(duì)于④，取SKIPIF1<0，結(jié)合圖像可知此時(shí)SKIPIF1<0存在最小值，從而得以判斷.【解析】依題意，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像為一條端點(diǎn)取不到值的單調(diào)遞增的射線(xiàn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像是，圓心為SKIPIF1<0，半徑為SKIPIF1<0的圓在SKIPIF1<0軸上方的圖像（即半圓）；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像是一條端點(diǎn)取不到值的單調(diào)遞減的曲線(xiàn)；對(duì)于①，取SKIPIF1<0，則SKIPIF1<0的圖像如下，顯然，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故①錯(cuò)誤；對(duì)于②，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0顯然取得最大值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，綜上：SKIPIF1<0取得最大值SKIPIF1<0，故②正確；對(duì)于③，結(jié)合圖像，易知在SKIPIF1<0，SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0的距離最小，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0，此時(shí)，SKIPIF1<0，故③正確；對(duì)于④，取SKIPIF1<0，則SKIPIF1<0的圖像如下，因?yàn)镾KIPIF1<0，結(jié)合圖像可知，要使SKIPIF1<0取得最小值，則點(diǎn)SKIPIF1<0在SKIPIF1<0上，點(diǎn)SKIPIF1<0在SKIPIF1<0，同時(shí)SKIPIF1<0的最小值為點(diǎn)SKIPIF1<0到SKIPIF1<0的距離減去半圓的半徑SKIPIF1<0，此時(shí)，因?yàn)镾KIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，故直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上，滿(mǎn)足SKIPIF1<0取得最小值，即SKIPIF1<0也滿(mǎn)足SKIPIF1<0存在最小值，故SKIPIF1<0的取值范圍不僅僅是SKIPIF1<0，故④錯(cuò)誤.故答案為：②③.【評(píng)注】本題解決的關(guān)鍵是分析得SKIPIF1<0的圖像，特別是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖像為半圓，解決命題④時(shí)，可取特殊值進(jìn)行排除即可.第三節(jié)冪函數(shù)1.（2023天津卷3）若SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【分析】根據(jù)對(duì)應(yīng)冪、指數(shù)函數(shù)的單調(diào)性判斷大小關(guān)系即可.【解析】由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0.所以SKIPIF1<0.故選D.第四節(jié)指數(shù)與指數(shù)函數(shù)1.（2023天津卷3）若SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【分析】根據(jù)對(duì)應(yīng)冪、指數(shù)函數(shù)的單調(diào)性判斷大小關(guān)系即可.【解析】由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0.所以SKIPIF1<0.故選D.2.（2023全國(guó)甲卷文科11）已知函數(shù)SKIPIF1<0.記SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（）A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【分析】利用作差法比較自變量的大小，再根據(jù)指數(shù)函數(shù)的單調(diào)性及二次函數(shù)的性質(zhì)判斷即可.【解析】令SKIPIF1<0，則SKIPIF1<0開(kāi)口向下，對(duì)稱(chēng)軸為SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0由二次函數(shù)性質(zhì)知SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，綜上，SKIPIF1<0，又SKIPIF1<0為增函數(shù)，故SKIPIF1<0，即SKIPIF1<0.故選A.3.（2023新高考I卷4）設(shè)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，則SKIPIF1<0的取值范圍是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【解析】令SKIPIF1<0，要使得SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，需要滿(mǎn)足SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選D.4.（2023北京卷11）已知函數(shù)SKIPIF1<0，則SKIPIF1<0.【分析】根據(jù)給定條件，把SKIPIF1<0代入，利用指數(shù)、對(duì)數(shù)運(yùn)算計(jì)算作答.【解析】函數(shù)SKIPIF1<0，所以SKIPIF1<0.故答案為1.第五節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)1.（2023北京卷11）已知函數(shù)SKIPIF1<0，則SKIPIF1<0.【分析】根據(jù)給定條件，把SKIPIF1<0代入，利用指數(shù)、對(duì)數(shù)運(yùn)算計(jì)算作答.【解析】函數(shù)SKIPIF1<0，所以SKIPIF1<0.故答案為1.2.（2023新高考I卷10）噪聲污染問(wèn)題越來(lái)越受到重視，用聲壓級(jí)來(lái)度量聲音的強(qiáng)弱，定義聲壓級(jí)SKIPIF1<0，其中常數(shù)SKIPIF1<0是聽(tīng)覺(jué)下限閾值，SKIPIF1<0是實(shí)際聲壓.下表為不同聲源的聲壓級(jí)：聲源與聲源的距離/m聲壓級(jí)/dB燃油汽車(chē)1060~90混合動(dòng)力汽車(chē)1050~60電動(dòng)汽車(chē)1040已知在距離燃油汽車(chē)、混合動(dòng)力汽車(chē)、電動(dòng)汽車(chē)10m處測(cè)得實(shí)際聲壓分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（）SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【解析】選項(xiàng)A，SKIPIF1<0，所以SKIPIF1<0，所以A正確；選項(xiàng)B，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故B錯(cuò)誤；選項(xiàng)C，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故C正確；選項(xiàng)D，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故D正確.故選ACD.第六節(jié)函數(shù)的圖像及應(yīng)用1.（2023全國(guó)甲卷理科10，文科12）已知SKIPIF1<0為函數(shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位所得函數(shù)，則SKIPIF1<0與SKIPIF1<0交點(diǎn)個(gè)數(shù)為（）A.SKIPIF1<0B.SKIPIF1<0C.3D.4【解析】因?yàn)楹瘮?shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位可得SKIPIF1<0而SKIPIF1<0過(guò)SKIPIF1<0與SKIPIF1<0兩點(diǎn)，分別作出SKIPIF1<0與SKIPIF1<0的圖像如圖所示，考慮SKIPIF1<0，即SKIPIF1<0處SKIPIF1<0與SKIPIF1<0的大小關(guān)系，結(jié)合圖像可知有3個(gè)交點(diǎn).故選C.【評(píng)注】本題考查了三角函數(shù)的圖像與性質(zhì)，畫(huà)出圖像，不難得到答案.2.（2023北京卷15）設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0存在最大值；③設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；④設(shè)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0存在最小值，則SKIPIF1<0的取值范圍是SKIPIF1<0.其中所有正確結(jié)論的序號(hào)是.【分析】先分析SKIPIF1<0圖像，再逐一分析各結(jié)論；對(duì)于①，取SKIPIF1<0，結(jié)合圖像即可判斷；對(duì)于②，分段討論SKIPIF1<0的取值范圍，從而得以判斷；對(duì)于③，結(jié)合圖像可知SKIPIF1<0的范圍；對(duì)于④，取SKIPIF1<0，結(jié)合圖像可知此時(shí)SKIPIF1<0存在最小值，從而得以判斷.【解析】依題意，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像為一條端點(diǎn)取不到值的單調(diào)遞增的射線(xiàn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像是，圓心為SKIPIF1<0，半徑為SKIPIF1<0的圓在SKIPIF1<0軸上方的圖像（即半圓）；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，易知其圖像是一條端點(diǎn)取不到值的單調(diào)遞減的曲線(xiàn)；對(duì)于①，取SKIPIF1<0，則SKIPIF1<0的圖像如下，顯然，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故①錯(cuò)誤；對(duì)于②，當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0顯然取得最大值SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，綜上：SKIPIF1<0取得最大值SKIPIF1<0，故②正確；對(duì)于③，結(jié)合圖像，易知在SKIPIF1<0，SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0的距離最小，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0且接近于SKIPIF1<0處，SKIPIF1<0，此時(shí)，SKIPIF1<0，故③正確；對(duì)于④，取SKIPIF1<0，則SKIPIF1<0的圖像如下，因?yàn)镾KIPIF1<0，結(jié)合圖像可知，要使SKIPIF1<0取得最小值，則點(diǎn)SKIPIF1<0在SKIPIF1<0上，點(diǎn)SKIPIF1<0在SKIPIF1<0，同時(shí)SKIPIF1<0的最小值為點(diǎn)SKIPIF1<0到SKIPIF1<0的距離減去半圓的半徑SKIPIF1<0，此時(shí)，因?yàn)镾KIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，故直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上，滿(mǎn)足SKIPIF1<0取得最小值，即SKIPIF1<0也滿(mǎn)足SKIPIF1<0存在最小值，故SKIPIF1<0的取值范圍不僅僅是SKIPIF1<0，故④錯(cuò)誤.故答案為：②③.【評(píng)注】本題解決的關(guān)鍵是分析得SKIPIF1<0的圖像，特別是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖像為半圓，解決命題④時(shí)，可取特殊值進(jìn)行排除即可.3.（2023天津卷4）函數(shù)SKIPIF1<0的圖象如下圖所示，則SKIPIF1<0的解析式可能為（

）

A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【分析】由圖知函數(shù)為偶函數(shù)，先判斷函數(shù)的奇偶性排除選項(xiàng)；再判斷函數(shù)在SKIPIF1<0上的函數(shù)符號(hào)排除選項(xiàng)，即得答案.【解析】由圖知：函數(shù)圖象關(guān)于y軸對(duì)稱(chēng)，其為偶函數(shù)，且SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，且定義域?yàn)镾KIPIF1<0，即A，B中函數(shù)為奇函數(shù)，排除選項(xiàng)A，B；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即C中SKIPIF1<0上函數(shù)值為正，排除選項(xiàng)C；故選D.第七節(jié)函數(shù)與方程1.（2023全國(guó)甲卷理科10，文科12）已知SKIPIF1<0為函數(shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位所得函數(shù)，則SKIPIF1<0與SKIPIF1<0交點(diǎn)個(gè)數(shù)為（）A.SKIPIF1<0B.SKIPIF1<0C.3D.4【解析】因?yàn)楹瘮?shù)SKIPIF1<0向左平移SKIPIF1<0個(gè)單位可得SKIPIF1<0而SKIPIF1<0過(guò)SKIPIF1<0與SKIPIF1<0兩點(diǎn)，分別作出SKIPIF1<0與SKIPIF1<0的圖像如圖所示，考慮SKIPIF1<0，即SKIPIF1<0處SKIPIF1<0與SKIPIF1<0的大小關(guān)系，結(jié)合圖像可知有3個(gè)交點(diǎn).故選C.【評(píng)注】本題考查了三角函數(shù)的圖像與性質(zhì)，畫(huà)出圖像，不難得到答案.2.（2023天津卷15）若函數(shù)SKIPIF1<0有且僅有兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為_(kāi)______．【分析】根據(jù)絕對(duì)值的意義，去掉絕對(duì)值，求出零點(diǎn)，再根據(jù)根存在的條件即可判斷SKIPIF1<0的取值范圍．【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，若SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIP