新高考一輪復(fù)習(xí)核心考點(diǎn)講與練考點(diǎn)03 函數(shù)及其性質(zhì)原卷版

考點(diǎn)03函數(shù)及其性質(zhì)（核心考點(diǎn)講與練）1.函數(shù)的概念設(shè)A，B是兩個(gè)非空數(shù)集，如果按照確定的法則f，對(duì)A中的任意數(shù)x，都有唯一確定的數(shù)y與它對(duì)應(yīng)，那么就稱f：A→B為從集合A到集合B的一個(gè)函數(shù)，記作y＝f(x)，x∈A.2.函數(shù)的定義域、值域(1)函數(shù)y＝f(x)自變量取值的范圍(數(shù)集A)叫做這個(gè)函數(shù)的定義域；所有函數(shù)值構(gòu)成的集合{y|y＝f(x)，x∈A}叫做這個(gè)函數(shù)的值域.(2)如果兩個(gè)函數(shù)的定義域相同，并且對(duì)應(yīng)法則完全一致，則這兩個(gè)函數(shù)為相等函數(shù).3.函數(shù)的表示法表示函數(shù)的常用方法有解析法、圖象法和列表法.4.分段函數(shù)(1)在函數(shù)的定義域內(nèi)，對(duì)于自變量x的不同取值區(qū)間，有著不同的對(duì)應(yīng)法則，這種函數(shù)稱為分段函數(shù).(2)分段函數(shù)是一個(gè)函數(shù)，分段函數(shù)的定義域是各段定義域的并集，值域是各段值域的并集.5.函數(shù)的單調(diào)性(1)單調(diào)函數(shù)的定義增函數(shù)減函數(shù)定義設(shè)函數(shù)y＝f(x)的定義域?yàn)锳，區(qū)間M?A，如果取區(qū)間M中任意兩個(gè)值x1，x2，改變量Δx＝x2－x1>0，則當(dāng)Δy＝f(x2)－f(x1)>0時(shí)，就稱函數(shù)y＝f(x)在區(qū)間M上是增函數(shù)Δy＝f(x2)－f(x1)<0時(shí)，就稱函數(shù)y＝f(x)在區(qū)間M上是減函數(shù)圖象描述自左向右看圖象是上升的自左向右看圖象是下降的(2)如果一個(gè)函數(shù)在某個(gè)區(qū)間M上是增函數(shù)或是減函數(shù)，就說(shuō)這個(gè)函數(shù)在這個(gè)區(qū)間M上具有單調(diào)性，區(qū)間M稱為單調(diào)區(qū)間.6.函數(shù)的最值前提設(shè)函數(shù)y＝f(x)的定義域?yàn)镮，如果存在實(shí)數(shù)M滿足條件(1)對(duì)于任意x∈I，都有f(x)≤M；(2)存在x0∈I，使得f(x0)＝M(3)對(duì)于任意x∈I，都有f(x)≥M；(4)存在x0∈I，使得f(x0)＝M結(jié)論M為最大值M為最小值7.函數(shù)的奇偶性奇偶性定義圖象特點(diǎn)奇函數(shù)設(shè)函數(shù)y＝f(x)的定義域?yàn)镈，如果對(duì)D內(nèi)的任意一個(gè)x，都有－x∈D，且f(－x)＝－f(x)，則這個(gè)函數(shù)叫做奇函數(shù)關(guān)于原點(diǎn)對(duì)稱偶函數(shù)設(shè)函數(shù)y＝g(x)的定義域?yàn)镈，如果對(duì)D內(nèi)的任意一個(gè)x，都有－x∈D，且g(－x)＝g(x)，則這個(gè)函數(shù)叫做偶函數(shù)關(guān)于y軸對(duì)稱8.函數(shù)的周期性(1)周期函數(shù)：對(duì)于函數(shù)y＝f(x)，如果存在一個(gè)非零常數(shù)T，使得當(dāng)x取定義域內(nèi)的任何值時(shí)，都有f(x＋T)＝f(x)，那么就稱函數(shù)y＝f(x)為周期函數(shù)，稱T為這個(gè)函數(shù)的周期.(2)最小正周期：如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù)，那么這個(gè)最小正數(shù)就叫做f(x)的最小正周期.1.利用基本不等式求最值時(shí)，要注意其必須滿足的三個(gè)條件：（1）“一正”就是各項(xiàng)必須為正數(shù)；（2）“二定”就是要求和的最小值，必須把構(gòu)成和的二項(xiàng)之積轉(zhuǎn)化成定值；要求積的最大值，則必須把構(gòu)成積的因式的和轉(zhuǎn)化成定值；（3）“三相等”是利用基本不等式求最值時(shí)，必須驗(yàn)證等號(hào)成立的條件，若不能取等號(hào)則這個(gè)定值就不是所求的最值，這也是最容易發(fā)生錯(cuò)誤的地方.2.已知函數(shù)零點(diǎn)(方程根)的個(gè)數(shù)求參數(shù)值(取值范圍)常用的方法：（1）直接法：直接求解方程得到方程的根，再通過(guò)解不等式確定參數(shù)范圍；（2）分離參數(shù)法：先將參數(shù)分離，轉(zhuǎn)化成求函數(shù)的值域問(wèn)題加以解決；（3）數(shù)形結(jié)合法：先對(duì)解析式變形，進(jìn)而構(gòu)造兩個(gè)函數(shù)，然后在同一平面直角坐標(biāo)系中畫出函數(shù)的圖象，利用數(shù)形結(jié)合的方法求解.3.函數(shù)的對(duì)稱性與單調(diào)性，指數(shù)式、對(duì)數(shù)式的大小比較．比較指數(shù)式大小時(shí)，常常化為同底數(shù)的冪，利用指數(shù)函數(shù)性質(zhì)比較，或化為同指數(shù)的冪，利用冪函數(shù)性質(zhì)比較，比較對(duì)數(shù)式大小，常?；癁橥讛?shù)的對(duì)數(shù)，利用對(duì)數(shù)函數(shù)性質(zhì)比較，如果不能化為同底數(shù)或同指數(shù)，或不同類型的數(shù)常常借助中間值如0或1比較大?。?.導(dǎo)數(shù)是研究函數(shù)的單調(diào)性、極值(最值)最有效的工具，而函數(shù)是高中數(shù)學(xué)中重要的知識(shí)點(diǎn)，對(duì)導(dǎo)數(shù)的應(yīng)用的考查主要從以下幾個(gè)角度進(jìn)行：（1）考查導(dǎo)數(shù)的幾何意義，往往與解析幾何、微積分相聯(lián)系．（2）利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間，判斷單調(diào)性；已知單調(diào)性，求參數(shù)．（3）利用導(dǎo)數(shù)求函數(shù)的最值(極值)，解決生活中的優(yōu)化問(wèn)題．（4）考查數(shù)形結(jié)合思想的應(yīng)用．函數(shù)及其表示一、單選題1．（2022·天津市第四十七中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若在區(qū)間SKIPIF1<0上存在SKIPIF1<0個(gè)不同的數(shù)SKIPIF1<0，使得SKIPIF1<0成立，則SKIPIF1<0的取值集合是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·天津市第四十七中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·河北·模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·浙江嘉興·二模）已知函數(shù)SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的解析式可能是（

）（SKIPIF1<0是自然對(duì)數(shù)的底數(shù)）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·天津·耀華中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）在區(qū)間SKIPIF1<0上為單調(diào)函數(shù)，若函數(shù)SKIPIF1<0有三個(gè)不同的零點(diǎn)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2022·浙江省義烏中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則圖象為下圖的函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2021·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0（m，n為整數(shù)），值域是SKIPIF1<0，則滿足條件的整數(shù)對(duì)SKIPIF1<0的個(gè)數(shù)是（

）A．2 B．3 C．4 D．58．（2020·南開(kāi)中學(xué)模擬預(yù)測(cè)）下列各組函數(shù)中，表示同一函數(shù)的是（

）A．SKIPIF1<0與SKIPIF1<0 B．SKIPIF1<0與SKIPIF1<0C．SKIPIF1<0與SKIPIF1<0 D．SKIPIF1<0與SKIPIF1<09．（2020·廣東中山·模擬預(yù)測(cè)）下列各組表示同一函數(shù)的是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<010.（2020·全國(guó)·一模）網(wǎng)購(gòu)女鞋時(shí)，常常會(huì)看到一張女鞋尺碼對(duì)照表如下，第一行是我們習(xí)慣稱呼的“鞋號(hào)”（單位：.號(hào)），第二行是腳長(zhǎng)（單位：SKIPIF1<0），請(qǐng)根據(jù)表中數(shù)據(jù)，思考：他們家正好有一款“32號(hào)”的女鞋在搞打折，那么適合購(gòu)買這款鞋的腳長(zhǎng)的取值范圍是（

）鞋碼3536373839腳長(zhǎng)225230235240245A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題11．（2022·江蘇南通·模擬預(yù)測(cè)）已知定義在R上的函數(shù)SKIPIF1<0的圖象連續(xù)不間斷，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0是SKIPIF1<0的兩個(gè)零點(diǎn)，且SKIPIF1<0，則SKIPIF1<012．（2022·江蘇·沭陽(yáng)如東中學(xué)模擬預(yù)測(cè)）華人數(shù)學(xué)家李天巖和美國(guó)數(shù)學(xué)家約克給出了“混沌”的數(shù)學(xué)定義，由此發(fā)展的混沌理論在生物學(xué)?經(jīng)濟(jì)學(xué)和社會(huì)學(xué)領(lǐng)域都有重要作用.在混沌理論中，函數(shù)的周期點(diǎn)是一個(gè)關(guān)鍵概念，定義如下：設(shè)SKIPIF1<0是定義在R上的函數(shù)，對(duì)于SKIPIF1<0R，令SKIPIF1<0，若存在正整數(shù)k使得SKIPIF1<0，且當(dāng)0<j<k時(shí)，SKIPIF1<0，則稱SKIPIF1<0是SKIPIF1<0的一個(gè)周期為k的周期點(diǎn).若SKIPIF1<0，下列各值是SKIPIF1<0周期為2的周期點(diǎn)的有（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．113．（2022·江蘇江蘇·一模）下列函數(shù)中，最大值是1的函數(shù)有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<014．（2021·江西·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則下列敘述正確的是（

）A．SKIPIF1<0的值域?yàn)镾KIPIF1<0 B．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0 D．若SKIPIF1<0，則SKIPIF1<0的最小值為-315．（2021·江西·模擬預(yù)測(cè)）下列各組函數(shù)中表示同一個(gè)函數(shù)的是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<016．（2021·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0和SKIPIF1<0滿足（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題17．（2022·福建·三模）寫出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)SKIPIF1<0________.①定義域?yàn)镾KIPIF1<0；②值域?yàn)镾KIPIF1<0；③對(duì)任意SKIPIF1<0且SKIPIF1<0，均有SKIPIF1<0.18．（2022·山東淄博·一模）以模型SKIPIF1<0去擬合一組數(shù)據(jù)時(shí)，設(shè)SKIPIF1<0，將其變換后得到線性回歸方程SKIPIF1<0，則SKIPIF1<0______．19．（2022·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0，則SKIPIF1<0______．20．（2022·重慶·模擬預(yù)測(cè)）已知定義在R上的函數(shù)SKIPIF1<0不是常值函數(shù)，且同時(shí)滿足：①SKIPIF1<0；②對(duì)任意SKIPIF1<0，均存在SKIPIF1<0使得SKIPIF1<0成立；則函數(shù)SKIPIF1<0______．（寫出一個(gè)符合條件的答案即可）四、雙空題21．（2022·浙江嘉興·二模）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，且滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0___________，SKIPIF1<0___________.22．（2022·江蘇·金陵中學(xué)二模）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最小正周期為_(kāi)__________；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值域?yàn)開(kāi)__________.23．（2022·山東濟(jì)南·一模）已知函數(shù)SKIPIF1<0，對(duì)任意非零實(shí)數(shù)x，均滿足SKIPIF1<0.則SKIPIF1<0的值為_(kāi)__________；函數(shù)SKIPIF1<0的最小值為_(kāi)__________.函數(shù)的基本性質(zhì)一、函數(shù)的單調(diào)性一、單選題1．（2022·天津·耀華中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則下述關(guān)系式正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·廣東廣州·二模）下列函數(shù)中，既是偶函數(shù)又在SKIPIF1<0上單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·遼寧撫順·一模）已知函數(shù)SKIPIF1<0對(duì)任意SKIPIF1<0都有SKIPIF1<0，若SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，且對(duì)任意的，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，都有SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國(guó)·哈師大附中模擬預(yù)測(cè)（理））已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題5．（2022·廣東茂名·二模）若對(duì)任意的SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，則m的值可能是（

）（注SKIPIF1<0…為自然對(duì)數(shù)的底數(shù)）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．16．（2022·福建泉州·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為非零常數(shù)，則下列說(shuō)法正確的是（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0的值域?yàn)镾KIPIF1<0D．當(dāng)SKIPIF1<0，且SKIPIF1<0時(shí)，若將函數(shù)SKIPIF1<0與SKIPIF1<0的圖象在SKIPIF1<0的SKIPIF1<0個(gè)交點(diǎn)記為SKIPIF1<0，則SKIPIF1<07．（2022·廣東廣東·一模）下列四個(gè)函數(shù)中，以SKIPIF1<0為周期且在SKIPIF1<0上單調(diào)遞增的偶函數(shù)有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題8．（2022·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若對(duì)任意的正數(shù)SKIPIF1<0，滿足SKIPIF1<0SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)________.9．（2022·山東·濟(jì)南市歷城第二中學(xué)模擬預(yù)測(cè)）函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)SKIPIF1<0的范圍是_______.10．（2022·北京豐臺(tái)·一模）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，能說(shuō)明“若函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增“為假命題的一個(gè)函數(shù)是__________.四、解答題11．（2022·江蘇江蘇·一模）已知實(shí)數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，SKIPIF1<0是自然對(duì)數(shù)的底數(shù).(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)求證：SKIPIF1<0存在極值點(diǎn)SKIPIF1<0，并求SKIPIF1<0的最小值.12．（2022·山東青島·一模）已知函數(shù)SKIPIF1<0．(1)若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，求SKIPIF1<0的值．13．（2022·湖北·黃岡中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0(1)若SKIPIF1<0，且SKIPIF1<0，試比較SKIPIF1<0與SKIPIF1<0的大小關(guān)系，并說(shuō)明理由；(2)若SKIPIF1<0，且SKIPIF1<0，證明：（i）SKIPIF1<0；（ii）SKIPIF1<0.（參考數(shù)據(jù)：SKIPIF1<0）二、函數(shù)的最值一、單選題1．（2021·湖南·模擬預(yù)測(cè)）已知奇函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，且在區(qū)間SKIPIF1<0上的最大值為9，最小值為-6，則SKIPIF1<0的值為（

）A．3 B．1 C．-1 D．-32．（2022·浙江嘉興·二模）設(shè)a，SKIPIF1<0，若SKIPIF1<0時(shí)，恒有SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·浙江省義烏中學(xué)模擬預(yù)測(cè)）設(shè)SKIPIF1<0，則有（

）A．存在SKIPIF1<0成立 B．任意SKIPIF1<0恒成立C．任意SKIPIF1<0恒成立 D．存在SKIPIF1<0成立4．（2022·重慶·二模）已知SKIPIF1<0，若SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則實(shí)數(shù)m的最大值為（

）A．2 B．4 C．SKIPIF1<0 D．SKIPIF1<05．（2022·天津?qū)嶒?yàn)中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若?SKIPIF1<0∈R，使得SKIPIF1<0成立，則實(shí)數(shù)m的取值范圍為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題6．（2022·湖北·一模）已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0在（0，+∞）上單調(diào)遞減C．SKIPIF1<0是周期函數(shù) D．SKIPIF1<0≥-1恒成立7．（2022·福建漳州·一模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的定義域?yàn)镾KIPIF1<0 B．SKIPIF1<0是偶函數(shù)C．函數(shù)SKIPIF1<0的零點(diǎn)為0 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0三、填空題8．（2022·浙江省諸暨市第二高級(jí)中學(xué)模擬預(yù)測(cè)）已知平面單位向量SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0.設(shè)SKIPIF1<0，SKIPIF1<0，向量SKIPIF1<0，SKIPIF1<0的夾角為SKIPIF1<0，則SKIPIF1<0的最大值是_______________.9．（2022·湖北·一模）已知函數(shù)SKIPIF1<0（x>0），若SKIPIF1<0的最大值為SKIPIF1<0，則正實(shí)數(shù)a=___________.10．（2020·山東臨沂·二模）若SKIPIF1<0，SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.11．（2021·北京延慶·模擬預(yù)測(cè)）同學(xué)們，你們是否注意到：自然下垂的鐵鏈；空曠的田野上，兩根電線桿之間的電線；峽谷的上空，橫跨深澗的觀光索道的鋼索．這些現(xiàn)象中都有相似的曲線形態(tài)．事實(shí)上，這些曲線在數(shù)學(xué)上常常被稱為懸鏈線．懸鏈線的相關(guān)理論在工程、航海、光學(xué)等方面有廣泛的應(yīng)用．在恰當(dāng)?shù)淖鴺?biāo)系中，這類函數(shù)的表達(dá)式可以為SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0是非零常數(shù)，無(wú)理數(shù)SKIPIF1<0…），對(duì)于函數(shù)SKIPIF1<0以下結(jié)論正確的是______．①如果SKIPIF1<0，那么函數(shù)SKIPIF1<0為奇函數(shù)；②如果SKIPIF1<0，那么SKIPIF1<0為單調(diào)函數(shù)；③如果SKIPIF1<0，那么函數(shù)SKIPIF1<0沒(méi)有零點(diǎn)；④如果SKIPIF1<0那么函數(shù)SKIPIF1<0的最小值為2．12．（2021·浙江浙江·二模）設(shè)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0的最大值是SKIPIF1<0，則SKIPIF1<0___________.四、解答題13．（2022·海南·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0在區(qū)間-π2,0上的最大值和最小值；(2)設(shè)SKIPIF1<0，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求實(shí)數(shù)a的取值范圍.14．（2021·江西·模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)求SKIPIF1<0的解析式；(2)若SKIPIF1<0，使得SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍.三、函數(shù)的奇偶性一、單選題1．（2022·廣東茂名·二模）已知SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·湖南湘潭·三模）函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．3．（2022·廣東茂名·二模）已知函數(shù)SKIPIF1<0，SKIPIF1<0分別是定義在R上的偶函數(shù)和奇函數(shù)，且SKIPIF1<0，若函數(shù)SKIPIF1<0有唯一零點(diǎn)，則正實(shí)數(shù)SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．24．（2021·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0為定義在R上的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．﹣2022 B．2022 C．SKIPIF1<0 D．SKIPIF1<05．（2022·遼寧·大連二十四中模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0且SKIPIF1<0，則有（

）A．SKIPIF1<0可能是奇函數(shù)，也可能是偶函數(shù) B．SKIPIF1<0C．SKIPIF1<0時(shí)，SKIPIF1<0 D．SKIPIF1<06．（2022·江蘇南通·模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0為奇函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<07．（2022·湖北·二模）已知函數(shù)SKIPIF1<0，則使不等式SKIPIF1<0成立的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．（2022·天津三中二模）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镈，若對(duì)任意的SKIPIF1<0，且SKIPIF1<0，恒有SKIPIF1<0，則稱函數(shù)SKIPIF1<0具有對(duì)稱性，其中點(diǎn)SKIPIF1<0為函數(shù)SKIPIF1<0的對(duì)稱中心，研究函數(shù)SKIPIF1<0的對(duì)稱中心，求SKIPIF1<0（

）A．2022 B．4043 C．4044 D．8086二、多選題9．（2022·江蘇泰州·模擬預(yù)測(cè)）已知定義在SKIPIF1<0上的單調(diào)遞增的函數(shù)SKIPIF1<0滿足：任意SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0，則（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0B．任意SKIPIF1<0，SKIPIF1<0C．存在非零實(shí)數(shù)SKIPIF1<0，使得任意SKIPIF1<0，SKIPIF1<0D．存在非零實(shí)數(shù)SKIPIF1<0，使得任意SKIPIF1<0，SKIPIF1<0三、填空題10．（2022·廣東深圳·二模）已知函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0___________．11．（2022·山東菏澤·一模）已知奇函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)，且SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，都有SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)_____．四、函數(shù)的周期性一、單選題1．（2022·山東濟(jì)寧·一模）定義在R上的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．0 B．1 C．－1 D．20222．（2022·福建·模擬預(yù)測(cè)）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．1 B．0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·江蘇江蘇·二模）已知SKIPIF1<0是定義域?yàn)镽的偶函數(shù)，f(5.5)＝2，g(x)＝(x－1)SKIPIF1<0.若g(x＋1)是偶函數(shù)，則SKIPIF1<0＝(

)A．－3 B．－2 C．2 D．3二、多選題4．（2022·河北·模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0（SKIPIF1<0）是周期為2的奇函數(shù)．則下列選項(xiàng)一定正確的是（

）A．函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱B．2是函數(shù)SKIPIF1<0的一個(gè)周期C．SKIPIF1<0D．SKIPIF1<05．（2022·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)周期B．SKIPIF1<0是函數(shù)SKIPIF1<0的一條對(duì)稱軸C．函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增6．（2021·廣東肇慶·模擬預(yù)測(cè)）已知定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0對(duì)任意的SKIPIF1<0滿足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0且SKIPIF1<0，則下列結(jié)論正確的有（

）A．SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．若SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0D．若方程SKIPIF1<0在SKIPIF1<0上有SKIPIF1<0個(gè)不同的實(shí)數(shù)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0三、填空題7．（2022·廣東茂名·二模）請(qǐng)寫出一個(gè)函數(shù)SKIPIF1<0_______，使之同時(shí)具有以下性質(zhì)：①圖象關(guān)于y軸對(duì)稱；②SKIPIF1<0，SKIPIF1<0．8．（2022·重慶·二模）已知定義域?yàn)镽的函數(shù)SKIPIF1<0滿足SKIPIF1<0且SKIPIF1<0，則函數(shù)SKIPIF1<0的解析式可以是______.五、函數(shù)的對(duì)稱性一、單選題1．（2022·廣東佛山·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度后得到函數(shù)SKIPIF1<0的圖象，則函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，且對(duì)任意的實(shí)數(shù)x，SKIPIF1<0恒成立.若存在實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0（SKIPIF1<0），使得SKIPIF1<0成立，則n的最大值為（

）A．25 B．26 C．28 D．313．（2021·浙江·模擬預(yù)測(cè)）已知定義在SKIPIF1<0上的圖象連續(xù)的函數(shù)SKIPIF1<0的導(dǎo)數(shù)是SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·廣東·梅州市梅江區(qū)嘉應(yīng)中學(xué)模擬預(yù)測(cè)）已知定義在R上的函數(shù)SKIPIF1<0滿足：對(duì)任意SKIPIF1<0，都有SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（其中SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)）．設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題5．（2022·江蘇·新沂市第一中學(xué)模擬預(yù)測(cè)）已知三次函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)（1，0）對(duì)稱，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0有3個(gè)零點(diǎn)C．SKIPIF1<0的對(duì)稱中心是SKIPIF1<0 D．SKIPIF1<0三、填空題6．（2022·重慶·模擬預(yù)測(cè)）寫出一個(gè)同時(shí)滿足下列三個(gè)條件的函數(shù)SKIPIF1<0的解析式__________.①SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0；②SKIPIF1<0；③SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.7．（2022·廣東廣州·二模）函數(shù)SKIPIF1<0的所有零點(diǎn)之和為_(kāi)_________．一、單選題1．（2020·山東·高考真題）已知函數(shù)SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則該函數(shù)在SKIPIF1<0上的圖像大致是（

）A． B．C． D．2．（2020·山東·高考真題）已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，若對(duì)于任意兩個(gè)不相等的實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，總有SKIPIF1<0成立，則函數(shù)SKIPIF1<0一定是（

）A．奇函數(shù) B．偶函數(shù) C．增函數(shù) D．減函數(shù)3．（2020·山東·高考真題）函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·全國(guó)·高考真題（理））設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0為奇函數(shù)，SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·全國(guó)·高考真題（文））設(shè)SKIPIF1<0是定義域?yàn)镽的奇函數(shù)，且SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2021·全國(guó)·高考真題（理））設(shè)函數(shù)SKIPIF1<0，則下列函數(shù)中為奇函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題7．（2021·全國(guó)·高考真題）已知函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0______.三、解答題8．（2021·全國(guó)·高考真題（文））已知函數(shù)SKIPIF1<0．（1）畫出SKIPIF1<0和SKIPIF1<0的圖像；（2）若SKIPIF1<0，求a的取值范圍．一、單選題1.（2022·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則a的最小值為（）A.1 B.2 C.3 D.42.（2022·北京·北師大實(shí)驗(yàn)中學(xué)模擬預(yù)測(cè)）下列函數(shù)中，既是偶函數(shù)又在SKIPIF1<0單調(diào)遞增的函數(shù)是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.（2022·全國(guó)·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0在SKIPIF1<0上的大致圖象為（）A. B.C. D.4.（2022·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0（）A.e B.1 C.SKIPIF1<0 D.SKIPIF1<05.（2022·全國(guó)·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的圖象大致為（）A. B.C. D.6.（2022·全國(guó)·模擬預(yù)測(cè)）已知定義域?yàn)镽的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<07.（2022·全國(guó)·模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值與最小值之和不小于SKIPIF1<0，則實(shí)數(shù)a的取值范圍為（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<08.（2022·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0為R上的奇函數(shù)，SKIPIF1<0，若SKIPIF1<0且SKIPIF1<0，都有SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<09.（2022·廣東汕頭·一模）定義在R上的偶函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0至少有8個(gè)實(shí)數(shù)解，則實(shí)數(shù)m的取值范圍是（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0