新高考數(shù)學(xué)三輪沖刺 北京卷押題練習(xí) 第15題 分段函數(shù)與函數(shù)的零點(diǎn) （原卷版）

分段函數(shù)與函數(shù)的零點(diǎn)核心考點(diǎn)考情統(tǒng)計(jì)考向預(yù)測(cè)備考策略分段函數(shù)的性質(zhì)2023·北京卷T15可以預(yù)測(cè)2024年新高考命題方向?qū)⒗^續(xù)分段函數(shù)的綜合問題或函數(shù)的零點(diǎn)作為壓軸題展開命題．分段函數(shù)的綜合問題或函數(shù)的零點(diǎn)中填空題較難，縱觀近幾年的新高考試題，分別考查分段函數(shù)的性質(zhì)、函數(shù)的零點(diǎn)，同時(shí)備考也需強(qiáng)化函數(shù)的性質(zhì)和數(shù)形結(jié)合的應(yīng)用，也是高考沖刺復(fù)習(xí)的重點(diǎn)復(fù)習(xí)內(nèi)容。分段函數(shù)開放題2022·北京卷T14函數(shù)零點(diǎn)2021·北京卷T151．（2023·北京卷T15）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，則SKIPIF1<0；④設(shè)SKIPIF1<0．若SKIPIF1<0存在最小值，則a的取值范圍是SKIPIF1<0．其中所有正確結(jié)論的序號(hào)是．2．（2022·北京卷T14）設(shè)函數(shù)SKIPIF1<0若SKIPIF1<0存在最小值，則a的一個(gè)取值為；a的最大值為．3．（2021·北京卷T15）已知函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①若SKIPIF1<0，SKIPIF1<0恰有2個(gè)零點(diǎn)；②存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有1個(gè)零點(diǎn)；③存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)；④存在正數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)．其中所有正確結(jié)論的序號(hào)是．1.分段函數(shù)(1)若函數(shù)在其定義域的不同子集上，因?qū)?yīng)關(guān)系不同而分別用幾個(gè)不同的式子來表示，這種函數(shù)稱為分段函數(shù).分段函數(shù)表示的是一個(gè)函數(shù).(2)分段函數(shù)的定義域等于各段函數(shù)的定義域的并集，其值域等于各段函數(shù)的值域的并集.2.根據(jù)分段函數(shù)解析式求函數(shù)值，首先確定自變量的值屬于哪個(gè)區(qū)間，其次選定相應(yīng)的解析式代入求解.3.已知函數(shù)值或函數(shù)的取值范圍求自變量的值或范圍時(shí)，應(yīng)根據(jù)每一段的解析式分別求解，但要注意檢驗(yàn)所求自變量的值或范圍是否符合相應(yīng)段的自變量的取值范圍.4.判斷函數(shù)零點(diǎn)個(gè)數(shù)的方法：(1)利用零點(diǎn)存在定理判斷.(2)代數(shù)法：求方程f(x)＝0的實(shí)數(shù)根.(3)幾何法：對(duì)于不易求根的方程，將它與函數(shù)y＝f(x)的圖象聯(lián)系起來，利用函數(shù)的性質(zhì)找出零點(diǎn)或利用兩個(gè)函數(shù)圖象的交點(diǎn)求解.在利用函數(shù)性質(zhì)時(shí)，可用求導(dǎo)的方法判斷函數(shù)的單調(diào)性.5.利用函數(shù)零點(diǎn)的情況求參數(shù)值(或取值范圍)的三種方法1．已知函數(shù)SKIPIF1<0，若SKIPIF1<0存在最小值，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．定義運(yùn)算SKIPIF1<0則函數(shù)SKIPIF1<0的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．已知函數(shù)SKIPIF1<0，若存在非零實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）.A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．已知函數(shù)SKIPIF1<0,其中SKIPIF1<0，則下列結(jié)論中一定正確的是（

）A．函數(shù)SKIPIF1<0一定存在最大值 B．函數(shù)SKIPIF1<0一定存在最小值C．函數(shù)SKIPIF1<0一定不存在最大值 D．函數(shù)SKIPIF1<0一定不存在最小值6．已知函數(shù)SKIPIF1<0，其中SKIPIF1<0，且SKIPIF1<0．給出下列三個(gè)結(jié)論：①函數(shù)SKIPIF1<0是單調(diào)函數(shù)；②當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱；③當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0根的個(gè)數(shù)可能是1或2．其中所有正確結(jié)論的序號(hào)是（

）A．①② B．①③ C．②③ D．①②③7．函數(shù)SKIPIF1<0，定義SKIPIF1<0，則SKIPIF1<0滿足（

）A．只有最小值，沒有最大值 B．既有最大值，又有最小值C．只有最大值，沒有最小值 D．既無最大值，也無最小值8．已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．已知函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0的方程SKIPIF1<0.有四個(gè)不同的實(shí)數(shù)解SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．關(guān)于函數(shù)SKIPIF1<0，給出下列結(jié)論：①SKIPIF1<0是偶函數(shù)且在在SKIPIF1<0上單調(diào)遞減；②方程SKIPIF1<0一定有實(shí)數(shù)解；③如果方程SKIPIF1<0（SKIPIF1<0為常數(shù)）有解，則解的個(gè)數(shù)一定是偶數(shù).則正確結(jié)論的個(gè)數(shù)（

）A．3 B．2 C．1 D．011．已知函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①若SKIPIF1<0，則函數(shù)SKIPIF1<0至少有一個(gè)零點(diǎn)；②存在實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，使得函數(shù)SKIPIF1<0無零點(diǎn)；③若SKIPIF1<0，則不存在實(shí)數(shù)SKIPIF1<0，使得函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)；②對(duì)任意實(shí)數(shù)SKIPIF1<0，總存在實(shí)數(shù)SKIPIF1<0使得函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn).其中所有正確結(jié)論的個(gè)數(shù)是（

）A．4 B．3 C．2 D．112．關(guān)于函數(shù)SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，給出下列四個(gè)結(jié)論：甲：6是該函數(shù)的零點(diǎn)；乙：4是該函數(shù)的零點(diǎn)；丙：該函數(shù)的零點(diǎn)之積為0；丁：方程SKIPIF1<0有兩個(gè)根.若上述四個(gè)結(jié)論中有且只有一個(gè)結(jié)論錯(cuò)誤，則該錯(cuò)誤結(jié)論是（

）A．甲 B．乙 C．丙 D．丁13．設(shè)函數(shù)SKIPIF1<0①若SKIPIF1<0，則SKIPIF1<0的最小值為.②若SKIPIF1<0有最小值，則實(shí)數(shù)SKIPIF1<0的取值范圍是.14．已知函數(shù)SKIPIF1<0，（1）若SKIPIF1<0，則SKIPIF1<0的最大值是；（2）若SKIPIF1<0存在最大值，則SKIPIF1<0的取值范圍為.15．函數(shù)SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0的值為；若SKIPIF1<0有兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是.16．設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0.①若SKIPIF1<0，則SKIPIF1<0；②若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，則a的取值范圍是.17．已知函數(shù)SKIPIF1<0，那么SKIPIF1<0；當(dāng)方程SKIPIF1<0有且僅有3個(gè)不同的根時(shí)，實(shí)數(shù)SKIPIF1<0的取值范圍是.18．已知函數(shù)SKIPIF1<0，關(guān)于x的方程SKIPIF1<0有3個(gè)不同的解，則m的取值范圍是.19．已知函數(shù)SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值域?yàn)?；②若關(guān)于SKIPIF1<0的方程SKIPIF1<0