兩年(22-23)高考數(shù)學(xué)真題專題分類匯編專題三 函數(shù)（教師版）

專題三函數(shù)真題卷題號(hào)考點(diǎn)考向2023新課標(biāo)1卷4函數(shù)的基本性質(zhì)復(fù)合函數(shù)的單調(diào)性、已知函數(shù)單調(diào)性求參10對(duì)數(shù)運(yùn)算、對(duì)數(shù)函數(shù)對(duì)數(shù)運(yùn)算、對(duì)數(shù)函數(shù)解決實(shí)際問(wèn)題11函數(shù)的基本性質(zhì)、函數(shù)的極值抽象函數(shù)的奇偶性、求抽象函數(shù)的函數(shù)值、極值點(diǎn)定義2023新課標(biāo)2卷4函數(shù)的基本性質(zhì)利用奇偶性求參2022新高考1卷12函數(shù)的基本性質(zhì)對(duì)稱性、周期性的綜合應(yīng)用2022新高考2卷8函數(shù)的基本性質(zhì)奇偶性、周期性的綜合應(yīng)用2021新高考1卷13函數(shù)的基本性質(zhì)利用奇偶性求參2021新高考2卷7比較大小利用對(duì)數(shù)函數(shù)的單調(diào)性比較大小8函數(shù)的基本性質(zhì)奇偶性、周期性的綜合應(yīng)用14函數(shù)的基本性質(zhì)基本初等函數(shù)的性質(zhì)2020新高考1卷6指數(shù)運(yùn)算、對(duì)數(shù)運(yùn)算指數(shù)、對(duì)數(shù)運(yùn)算解決實(shí)際問(wèn)題8函數(shù)的基本性質(zhì)單調(diào)性、奇偶性的綜合應(yīng)用2020新高考2卷7函數(shù)的單調(diào)性與最值利用單調(diào)性求參數(shù)的取值范圍8函數(shù)的基本性質(zhì)單調(diào)性、奇偶性的綜合應(yīng)用12對(duì)數(shù)函數(shù)新定義問(wèn)題、對(duì)數(shù)運(yùn)算、對(duì)數(shù)函數(shù)的性質(zhì)、不等式的性質(zhì)【2023年真題】1.（2023·新課標(biāo)I卷第4題）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【答案】D

【解析】【分析】本題考查復(fù)合函數(shù)的單調(diào)性，為較易題.【解答】解：結(jié)合復(fù)合函數(shù)單調(diào)性的性質(zhì)，易得SKIPIF1<0，所以a的取值范圍是SKIPIF1<0故選SKIPIF1<02.（2023·新課標(biāo)=2\*ROMANII卷第4題）若SKIPIF1<0為偶函數(shù)，則SKIPIF1<0(

)A.SKIPIF1<0 B.0 C.SKIPIF1<0 D.1【答案】B

【解析】【分析】本題考查利用函數(shù)的奇偶性求解函數(shù)的解析式，為基礎(chǔ)題.【解答】解：SKIPIF1<0為偶函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故選SKIPIF1<03.（2023·新課標(biāo)=1\*ROMANI卷第10題）（多選）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，p是實(shí)際聲壓SKIPIF1<0下表為不同聲源的聲壓級(jí)：聲源與聲源的距離SKIPIF1<0聲壓級(jí)SKIPIF1<0燃油汽車10

SKIPIF1<0

混合動(dòng)力汽車10

SKIPIF1<0

電動(dòng)汽車10

40

已知在距離燃油汽車、混合動(dòng)力汽車、電動(dòng)汽車10m處測(cè)得實(shí)際聲壓分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD

【解析】【分析】本題考查了對(duì)數(shù)函數(shù)的實(shí)際應(yīng)用，屬于中檔題.利用公式聲壓級(jí)公式結(jié)合每種汽車聲壓級(jí)范圍計(jì)算即可逐項(xiàng)判斷.【解答】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以A正確SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以B錯(cuò)誤SKIPIF1<0SKIPIF1<0，SKIPIF1<0，所以C正確SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以D正確．故選ACD4.（2023·新課標(biāo)=1\*ROMANI卷第11題）（多選）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)【答案】ABC

【解析】【分析】本題主要考查抽象函數(shù)的奇偶性、函數(shù)的極值點(diǎn)，屬中檔題.通過(guò)賦值法，可判斷ABC選項(xiàng).對(duì)于D選項(xiàng)可設(shè)常函數(shù)SKIPIF1<0

，進(jìn)行排除.【解答】解：選項(xiàng)A，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故A正確;選項(xiàng)B，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故B正確;選項(xiàng)C，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，再令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，故C正確;選項(xiàng)D，不妨設(shè)SKIPIF1<0為常函數(shù)，且滿足原題SKIPIF1<0，而常函數(shù)沒(méi)有極值點(diǎn)，故D錯(cuò)誤.故選：SKIPIF1<0【2022年真題】5.（2022·新高考I卷第12題）（多選）已知函數(shù)SKIPIF1<0及其導(dǎo)函數(shù)SKIPIF1<0的定義域?yàn)镽，記SKIPIF1<0若SKIPIF1<0，SKIPIF1<0均為偶函數(shù)，則(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】BC

【解析】【分析】本題主要考查導(dǎo)函數(shù)與原函數(shù)的關(guān)系，函數(shù)的對(duì)稱性及奇偶性，屬于難題.利用函數(shù)的奇偶性及周期性，導(dǎo)函數(shù)與原函數(shù)的關(guān)系逐項(xiàng)分析即可.【解答】解：由SKIPIF1<0為偶函數(shù)可知SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，

由SKIPIF1<0為偶函數(shù)可知SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，

結(jié)合SKIPIF1<0，根據(jù)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱可知SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，

根據(jù)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱可知：SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，

綜上，函數(shù)SKIPIF1<0與SKIPIF1<0均是周期為2的周期函數(shù)，所以有SKIPIF1<0，所以A不正確;

SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，所以C正確.

SKIPIF1<0，SKIPIF1<0，所以B正確;

又SKIPIF1<0，所以SKIPIF1<0，所以D不正確.6.（2022·新高考II卷第8題）若函數(shù)SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.0 D.1【答案】A

【解析】【分析】本題考查函數(shù)性質(zhì)的應(yīng)用，涉及函數(shù)的周期與賦值法的應(yīng)用?！窘獯稹拷猓毫頢KIPIF1<0得SKIPIF1<0

故SKIPIF1<0，SKIPIF1<0，

消去SKIPIF1<0和SKIPIF1<0得到SKIPIF1<0，故SKIPIF1<0周期為SKIPIF1<0

令SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，

SKIPIF1<0，

SKIPIF1<0，

SKIPIF1<0，

SKIPIF1<0，

SKIPIF1<0，

故SKIPIF1<0

SKIPIF1<0

即【2021年真題】7.（2021·新高考I卷第13題）已知函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0__________.【答案】1

【解析】【分析】本題考查函數(shù)的奇偶性，屬于基礎(chǔ)題.

利用SKIPIF1<0即可求出a的值.【解答】解：SKIPIF1<0函數(shù)SKIPIF1<0是偶函數(shù)；

SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0，解得SKIPIF1<0，故答案為SKIPIF1<08.（2021·新高考II卷第7題）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列判斷正確的是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C

【解析】【分析】本題考查了對(duì)數(shù)的單調(diào)性與大小比較，合理轉(zhuǎn)化是關(guān)鍵.

利用對(duì)數(shù)函數(shù)的單調(diào)性可比較a、b與c的大小關(guān)系，由此可得出結(jié)論.【解答】解：SKIPIF1<0，

即SKIPIF1<0故選SKIPIF1<09.（2021·新高考II卷第8題）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，則

(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B

【解析】【分析】本題是對(duì)函數(shù)奇偶性和周期性的綜合考查.

推導(dǎo)出函數(shù)是以4為周期的周期函數(shù)，由已知條件得出SKIPIF1<0，結(jié)合已知條件可得出結(jié)論.【解答】解：因?yàn)楹瘮?shù)為偶函數(shù)，則SKIPIF1<0，可得SKIPIF1<0，因?yàn)楹瘮?shù)為奇函數(shù)，則SKIPIF1<0，所以SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，故函數(shù)是以4為周期的周期函數(shù)，因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，故SKIPIF1<0，其它三個(gè)選項(xiàng)未知.故選SKIPIF1<010.（2021·新高考II卷第14題）寫出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)SKIPIF1<0：_________.①SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；③SKIPIF1<0是奇函數(shù)．【答案】SKIPIF1<0答案不唯一，SKIPIF1<0均滿足SKIPIF1<0

【解析】【分析】本題是開(kāi)放性問(wèn)題，合理分析所給條件找出合適的函數(shù)是關(guān)鍵，屬于中檔題.

根據(jù)冪函數(shù)的性質(zhì)可得所求的【解答】解：取SKIPIF1<0，則SKIPIF1<0，滿足①，SKIPIF1<0，SKIPIF1<0時(shí)有，滿足②，SKIPIF1<0的定義域?yàn)镽，又SKIPIF1<0，故是奇函數(shù)，滿足③.故答案為：SKIPIF1<0答案不唯一，SKIPIF1<0均滿足SKIPIF1<0【2020年真題】11.（2020·新高考I卷第6題）基本再生數(shù)SKIPIF1<0與世代間隔T是新冠肺炎流行病學(xué)基本參數(shù).基本再生數(shù)指一個(gè)感染者傳染的平均人數(shù)，世代間隔指兩代間傳染所需的平均時(shí)間.在新冠肺炎疫情初始階段，可以用指數(shù)模型：SKIPIF1<0描述累計(jì)感染病例數(shù)SKIPIF1<0隨時(shí)間SKIPIF1<0單位：天SKIPIF1<0的變化規(guī)律，指數(shù)增長(zhǎng)率r與SKIPIF1<0，T近似滿足SKIPIF1<0有學(xué)者基于已有數(shù)據(jù)估計(jì)出SKIPIF1<0，SKIPIF1<0據(jù)此，在新冠肺炎疫情初始階段，累計(jì)感染病例數(shù)增加1倍需要的時(shí)間約為SKIPIF1<0(

)A.SKIPIF1<0天 B.SKIPIF1<0天 C.SKIPIF1<0天 D.SKIPIF1<0天【答案】B

【解析】【分析】本題結(jié)合實(shí)際問(wèn)題考查指數(shù)對(duì)數(shù)化簡(jiǎn)求值，屬于中檔題.

根據(jù)題意，先將SKIPIF1<0，SKIPIF1<0代入SKIPIF1<0，求得r，再由題意即可求解.【解答】解：將SKIPIF1<0，SKIPIF1<0代入SKIPIF1<0，

得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，

當(dāng)增加1倍時(shí)，，

所需時(shí)間為故選SKIPIF1<012.（2020·新高考I卷、II卷第8題）若定義在R上的奇函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，則滿足SKIPIF1<0的x的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0

C.SKIPIF1<0 D.SKIPIF1<0【答案】D

【解析】【分析】本題考查函數(shù)奇偶性和單調(diào)性的應(yīng)用，考查運(yùn)算求解及邏輯推理能力，屬于一般題.

根據(jù)題意，不等式SKIPIF1<0可化為SKIPIF1<0

或SKIPIF1<0，從而利用奇函數(shù)性質(zhì)及函數(shù)的單調(diào)性求解即可.【解答】解：根據(jù)題意，不等式SKIPIF1<0可化為SKIPIF1<0

或SKIPIF1<0，由奇函數(shù)性質(zhì)得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以或，

解得SKIPIF1<0或SKIPIF1<0滿足SKIPIF1<0的x的取值范圍是SKIPIF1<0故選SKIPIF1<013.（2020·新高考II卷第7題）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則a的取值范圍是(

)A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D

【解析】【分析】本題考查復(fù)合函數(shù)單調(diào)性的求法，屬于中檔題．

由對(duì)數(shù)式的真數(shù)大于0求得函數(shù)的定義域，令SKIPIF1<0，由外層函數(shù)SKIPIF1<0是其定義域內(nèi)的增函數(shù)，結(jié)合復(fù)合函數(shù)的單調(diào)性可知，要使函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，需內(nèi)層函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增且恒大于0，轉(zhuǎn)化為SKIPIF1<0，即可得到a的范圍．【解答】解：由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0令SKIPIF1<0，SKIPIF1<0外層函數(shù)SKIPIF1<0是其定義域內(nèi)的增函數(shù)，SKIPIF1<0要使函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則需內(nèi)層函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增且恒大于0，則SKIPIF1<0，即SKIPIF1<0SKIPIF1<0的取值范圍是SKIPIF1<0故選：SKIPIF1<014.（2020·新高考I卷第12題）（多選）信息熵是信息論中的一個(gè)重要概念.設(shè)隨機(jī)變量X所有可能的取值為1，2，SKIPIF1<0，n，且SKIPIF1<0，SKIPIF1<0，定義X的信息熵SKIPIF1<0(

)A.若SKIPIF1<0，則SKIPIF1<0

B.若SKIPIF1<0，則SKIPIF1<0隨著SKIPIF1<0的增大而增大

C.若SKIPIF1<0=SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0隨著n的增大而增大

D.若SKIPIF1<0，隨機(jī)變量Y的所有可能取值為1，2，SKIPIF1<0，m，且SKIPIF1<0SKIPIF1<0+SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1