高考數(shù)學(xué)三輪沖刺壓軸小題03 解密函數(shù)零點(diǎn)相關(guān)問題 (解析版)

18/18解密函數(shù)零點(diǎn)相關(guān)問題一、方法綜述新課標(biāo)下的高考越來越注重對(duì)學(xué)生的綜合素質(zhì)的考察，函數(shù)的零點(diǎn)問題便是一個(gè)考察學(xué)生綜合素質(zhì)的很好途徑，它主要涉及到基本初等函數(shù)的圖象，滲透著轉(zhuǎn)化、化歸、數(shù)形結(jié)合、函數(shù)與方程等思想方法，在培養(yǎng)思維的靈活性、創(chuàng)造性等方面起到了積極的作用．近幾年的數(shù)學(xué)高考中頻頻出現(xiàn)零點(diǎn)問題，其形式逐漸多樣化，但都與函數(shù)、導(dǎo)數(shù)知識(shí)密不可分．根據(jù)函數(shù)零點(diǎn)的定義：對(duì)于函數(shù)SKIPIF1<0，把使SKIPIF1<0成立的實(shí)數(shù)SKIPIF1<0叫做函數(shù)SKIPIF1<0的零點(diǎn)．即：方程SKIPIF1<0有實(shí)數(shù)根SKIPIF1<0函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸有交點(diǎn)的橫坐標(biāo)SKIPIF1<0函數(shù)SKIPIF1<0有零點(diǎn)．圍繞三者之間的關(guān)系，在高考數(shù)學(xué)中函數(shù)零點(diǎn)的題型主要①函數(shù)的零點(diǎn)的分布；②函數(shù)的零點(diǎn)的個(gè)數(shù)問題；③利用導(dǎo)數(shù)結(jié)合圖像的變動(dòng)將兩個(gè)函數(shù)的圖像的交點(diǎn)問題轉(zhuǎn)化成函數(shù)的零點(diǎn)的個(gè)數(shù)問題．二、解題策略類型一：函數(shù)零點(diǎn)的分布問題例1．【2020·河南高考模擬】已知單調(diào)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，對(duì)于定義域內(nèi)任意SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】根據(jù)題意，對(duì)任意的SKIPIF1<0，都有SKIPIF1<0，又由SKIPIF1<0是定義在SKIPIF1<0上的單調(diào)函數(shù)，則SKIPIF1<0為定值，設(shè)SKIPIF1<0，則SKIPIF1<0，又由SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，所以零點(diǎn)所在的區(qū)間為（3，4）.【解題秘籍】判斷函數(shù)零點(diǎn)所在區(qū)間有三種常用方法：①直接法，解方程判斷；②定理法；③圖象法．【舉一反三】函數(shù)f(x)＝lnx＋x－eq\f(1,2)，則函數(shù)的零點(diǎn)所在區(qū)間是()A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．(1，2)【答案】C【解析】函數(shù)f(x)＝lnx＋x－eq\f(1,2)的圖象在(0，＋∞)上連續(xù)，且SKIPIF1<0＝lneq\f(3,4)＋eq\f(3,4)－eq\f(1,2)＝lneq\f(3,4)＋eq\f(1,4)＜0，f(1)＝ln1＋1－eq\f(1,2)＝eq\f(1,2)＞0，故f(x)的零點(diǎn)所在區(qū)間為SKIPIF1<0．學(xué)科$網(wǎng)類型二函數(shù)零點(diǎn)的個(gè)數(shù)問題例2．【2020·陜西高考模擬】已知函數(shù)SKIPIF1<0，則函數(shù)g（x）＝xf（x）﹣1的零點(diǎn)的個(gè)數(shù)為（）A．2 B．3 C．4 D．5【答案】B【解析】由g（x）＝xf（x）﹣1＝0得xf（x）＝1，當(dāng)x＝0時(shí)，方程xf（x）＝1不成立，即x≠0，則等價(jià)為f（x）＝SKIPIF1<0，當(dāng)2＜x≤4時(shí)，0＜x﹣2≤2，此時(shí)f（x）＝SKIPIF1<0f（x﹣2）＝SKIPIF1<0（1﹣|x﹣2﹣1|）＝SKIPIF1<0﹣SKIPIF1<0|x﹣3|，當(dāng)4＜x≤6時(shí)，2＜x﹣2≤4，此時(shí)f（x）＝SKIPIF1<0f（x﹣2）＝SKIPIF1<0[SKIPIF1<0﹣SKIPIF1<0|x﹣2﹣3|]＝SKIPIF1<0﹣SKIPIF1<0|x﹣5|，作出f（x）的圖象如圖，則f（1）＝1，f（3）＝SKIPIF1<0f（1）＝SKIPIF1<0，f（5）＝SKIPIF1<0f（3）＝SKIPIF1<0，設(shè)h（x）＝SKIPIF1<0，則h（1）＝1，h（3）＝SKIPIF1<0，h（5）＝SKIPIF1<0＞f（5），作出h（x）的圖象，由圖象知兩個(gè)函數(shù)圖象有3個(gè)交點(diǎn)，即函數(shù)g（x）的零點(diǎn)個(gè)數(shù)為3個(gè)，故選：B．【點(diǎn)睛】函數(shù)零點(diǎn)的求解與判斷方法：(1)直接求零點(diǎn)：令f(x)＝0，如果能求出解，則有幾個(gè)解就有幾個(gè)零點(diǎn)．(2)零點(diǎn)存在性定理：利用定理不僅要函數(shù)在區(qū)間[a，b]上是連續(xù)不斷的曲線，且f(a)·f(b)＜0，還必須結(jié)合函數(shù)的圖象與性質(zhì)(如單調(diào)性、奇偶性)才能確定函數(shù)有多少個(gè)零點(diǎn)．(3)利用圖象交點(diǎn)的個(gè)數(shù)：將函數(shù)變形為兩個(gè)函數(shù)的差，畫兩個(gè)函數(shù)的圖象，看其交點(diǎn)的橫坐標(biāo)有幾個(gè)不同的值，就有幾個(gè)不同的零點(diǎn)．【舉一反三】【2020·安徽高考模擬】已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0或SKIPIF1<0【答案】D【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為SKIPIF1<0.又在SKIPIF1<0上，SKIPIF1<0的圖像如圖所示：因?yàn)镾KIPIF1<0有兩個(gè)不同的零點(diǎn)，所以方程SKIPIF1<0有兩個(gè)不同的解即直線SKIPIF1<0與SKIPIF1<0有兩個(gè)不同交點(diǎn)且交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0.綜上，選D.類型三已知函數(shù)零點(diǎn)求參數(shù)例3．【2020·天津高考模擬】已知函數(shù)SKIPIF1<0，SKIPIF1<0若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)解,則SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)解，即方程SKIPIF1<0恰有三個(gè)不相等的實(shí)數(shù)解，即SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn).令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增；且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，據(jù)此繪制函數(shù)SKIPIF1<0的圖像如圖所示，結(jié)合函數(shù)圖像可知，滿足題意時(shí)SKIPIF1<0的取值范圍是SKIPIF1<0.本題選擇C選項(xiàng).【舉一反三】【2020·江蘇高考模擬】已知函數(shù)SKIPIF1<0有且僅有三個(gè)零點(diǎn)，并且這三個(gè)零點(diǎn)構(gòu)成等差數(shù)列，則實(shí)數(shù)a的值為_______．【答案】SKIPIF1<0或SKIPIF1<0【解析】函數(shù)SKIPIF1<00，得|x+a|SKIPIF1<0a＝3，設(shè)g（x）＝|x+a|SKIPIF1<0a，h（x）＝3，則函數(shù)g（x）SKIPIF1<0，不妨設(shè)f（x）＝0的3個(gè)根為x1，x2，x3，且x1＜x2＜x3，當(dāng)x＞﹣a時(shí)，由f（x）＝0，得g（x）＝3，即xSKIPIF1<03，得x2﹣3x﹣4＝0，得（x+1）（x﹣4）＝0，解得x＝﹣1，或x＝4；若①﹣a≤﹣1，即a≥1，此時(shí)x2＝﹣1，x3＝4，由等差數(shù)列的性質(zhì)可得x1＝﹣6，由f（﹣6）＝0，即g（﹣6）＝3得6SKIPIF1<02a＝3，解得aSKIPIF1<0，滿足f（x）＝0在（﹣∞，﹣a]上有一解．若②﹣1＜﹣a≤4，即﹣4≤a＜1，則f（x）＝0在（﹣∞，﹣a]上有兩個(gè)不同的解，不妨設(shè)x1，x2，其中x3＝4，所以有x1，x2是﹣xSKIPIF1<02a＝3的兩個(gè)解，即x1，x2是x2+（2a+3）x+4＝0的兩個(gè)解．得到x1+x2＝﹣（2a+3），x1x2＝4，又由設(shè)f（x）＝0的3個(gè)根為x1，x2，x3成差數(shù)列，且x1＜x2＜x3，得到2x2＝x1+4，解得：a＝﹣1SKIPIF1<0（舍去）或a＝﹣1SKIPIF1<0．③﹣a＞4，即a＜﹣4時(shí)，f（x）＝0最多只有兩個(gè)解，不滿足題意；綜上所述，aSKIPIF1<0或﹣1SKIPIF1<0．三、強(qiáng)化訓(xùn)練1．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有SKIPIF1<0個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【來源】四川省成都市南開為明學(xué)校2020-2021學(xué)年高三上學(xué)期第二次調(diào)研考試數(shù)學(xué)（理）試題【答案】A【解析】令SKIPIF1<0，則SKIPIF1<0，則函數(shù)SKIPIF1<0有SKIPIF1<0個(gè)零點(diǎn)即直線SKIPIF1<0與函數(shù)SKIPIF1<0有SKIPIF1<0個(gè)交點(diǎn)，將直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像分別沿SKIPIF1<0軸的正方向上移SKIPIF1<0個(gè)單位，即直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像有SKIPIF1<0個(gè)交點(diǎn)，因?yàn)镾KIPIF1<0，滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，因?yàn)橹本€SKIPIF1<0過點(diǎn)SKIPIF1<0，所以只需滿足直線SKIPIF1<0與SKIPIF1<0剛好有除點(diǎn)SKIPIF1<0外的另一個(gè)交點(diǎn)即可，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，如圖，將直線SKIPIF1<0繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)，顯然SKIPIF1<0與SKIPIF1<0只有一個(gè)交點(diǎn)，故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0，故選：A.2．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0在SKIPIF1<0上恒有兩個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0【來源】百師聯(lián)盟2020-2021學(xué)年高三上學(xué)期一輪復(fù)習(xí)聯(lián)考（四）全國卷I文科數(shù)學(xué)試題【答案】B【解析】作出SKIPIF1<0和SKIPIF1<0，如圖所示，要使函數(shù)SKIPIF1<0在SKIPIF1<0上恒有兩個(gè)零點(diǎn)，即函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，易知當(dāng)SKIPIF1<0時(shí)，滿足題意；當(dāng)SKIPIF1<0時(shí)，有三個(gè)交點(diǎn)，不滿足題意；當(dāng)SKIPIF1<0時(shí)，考慮SKIPIF1<0與SKIPIF1<0相切時(shí)，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，有兩個(gè)交點(diǎn)，滿足題意；當(dāng)SKIPIF1<0時(shí)，有四個(gè)交點(diǎn)，不滿足題意；當(dāng)SKIPIF1<0時(shí)，無交點(diǎn)，不滿足題意綜上，實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0或SKIPIF1<0，故選SKIPIF1<0.3．已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0與SKIPIF1<0的圖象上分別存在點(diǎn)SKIPIF1<0?SKIPIF1<0，使得SKIPIF1<0?SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【來源】四川省內(nèi)江市高中2020-2021學(xué)年高三上學(xué)期第一次模擬考試數(shù)學(xué)理科試題【答案】C【解析】設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的圖象上的任意一點(diǎn)，其關(guān)于SKIPIF1<0對(duì)稱的點(diǎn)的坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱的函數(shù)為SKIPIF1<0.由于SKIPIF1<0與SKIPIF1<0的圖象上分別存在點(diǎn)SKIPIF1<0?SKIPIF1<0，使得SKIPIF1<0?SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，故函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0圖象在區(qū)間SKIPIF1<0有交點(diǎn)，所以方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故選：C.4．已知函數(shù)SKIPIF1<0，以下結(jié)論正確的是（）A．SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)B．SKIPIF1<0C．若方程SKIPIF1<0恰有SKIPIF1<0個(gè)實(shí)根，則SKIPIF1<0D．若函數(shù)SKIPIF1<0在SKIPIF1<0上有個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0【來源】四川省師范大學(xué)附屬中學(xué)2020-2021學(xué)年高三上學(xué)期期中數(shù)學(xué)（理）試題【答案】C【解析】由題意，作出函數(shù)SKIPIF1<0的圖象，如圖所示，對(duì)于A中，當(dāng)SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為周期為SKIPIF1<0的函數(shù)，作出SKIPIF1<0在區(qū)間SKIPIF1<0的函數(shù)，可知SKIPIF1<0在區(qū)間SKIPIF1<0上先增后減，所以A錯(cuò)誤；對(duì)于B中，因?yàn)镾KIPIF1<0時(shí)，函數(shù)SKIPIF1<0為周期為SKIPIF1<0的函數(shù)，又由SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以B錯(cuò)誤；對(duì)于C中，直線SKIPIF1<0恒過定點(diǎn)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象和函數(shù)SKIPIF1<0的圖象有三個(gè)交點(diǎn)，當(dāng)SKIPIF1<0，設(shè)SKIPIF1<0與SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0，根據(jù)對(duì)稱性可知，當(dāng)SKIPIF1<0與SKIPIF1<0相切時(shí)，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，綜上可得，當(dāng)函數(shù)SKIPIF1<0的圖象和函數(shù)SKIPIF1<0的圖象有三個(gè)交點(diǎn)時(shí)，SKIPIF1<0，所以C正確．對(duì)于D中，又由函數(shù)SKIPIF1<0在SKIPIF1<0上有個(gè)零點(diǎn)SKIPIF1<0，故直線SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上由6個(gè)交點(diǎn)，不妨設(shè)SKIPIF1<0，由圖象可知SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，所以SKIPIF1<0，所以D錯(cuò)誤.故選：C.5．SKIPIF1<0為實(shí)數(shù)，SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù).SKIPIF1<0，若SKIPIF1<0的圖像上恰好存在一個(gè)點(diǎn)與SKIPIF1<0的圖像上某點(diǎn)關(guān)于SKIPIF1<0軸對(duì)稱，則實(shí)數(shù)SKIPIF1<0的取值范圍為___________.【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0，點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱，由題意可知SKIPIF1<0在SKIPIF1<0有一個(gè)解，故SKIPIF1<0在SKIPIF1<0有一個(gè)解設(shè)SKIPIF1<0，SKIPIF1<0寫成分段函數(shù)形式即為SKIPIF1<0作出函數(shù)圖象可知SKIPIF1<0與SKIPIF1<0，SKIPIF1<0只有一個(gè)交點(diǎn)，由圖象可知，SKIPIF1<0的取值范圍為SKIPIF1<0或SKIPIF1<0故答案為：SKIPIF1<06．已知SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0(SKIPIF1<0為實(shí)數(shù))有兩個(gè)不同的零點(diǎn)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為___________.【答案】SKIPIF1<0【解析】SKIPIF1<0，求導(dǎo)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.函數(shù)SKIPIF1<0有兩個(gè)不同零點(diǎn)，等價(jià)于方程SKIPIF1<0有兩個(gè)不等實(shí)根.設(shè)SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，作出函數(shù)SKIPIF1<0的圖像，則問題轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上有兩個(gè)不同的實(shí)根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0則SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，由零點(diǎn)存在性定理知，SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0.7．已知函數(shù)SKIPIF1<0存在SKIPIF1<0個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是__________.【來源】江西宜春市2021屆高三上學(xué)期數(shù)學(xué)（理）期末試題【答案】SKIPIF1<0【解析】令SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0極大值SKIPIF1<0所以，函數(shù)SKIPIF1<0在SKIPIF1<0處取得最大值，即SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.所以，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0，由于SKIPIF1<0，解得SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0極大值SKIPIF1<0所以，函數(shù)SKIPIF1<0在SKIPIF1<0處取得最大值，即SKIPIF1<0，若使得函數(shù)SKIPIF1<0存在SKIPIF1<0個(gè)零點(diǎn)，則直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有兩個(gè)交點(diǎn)，設(shè)交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0、SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖如下圖所示：由圖象可知，SKIPIF1<0.作出函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象如下圖所示：由圖象可知，當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象有兩個(gè)交點(diǎn)，綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.8．已知函數(shù)SKIPIF1<0給出下列四個(gè)結(jié)論：①存在實(shí)數(shù)SKIPIF1<0，使函數(shù)SKIPIF1<0為奇函數(shù)；②對(duì)任意實(shí)數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0既無最大值也無最小值；③對(duì)任意實(shí)數(shù)SKIPIF1<0和SKIPIF1<0，函數(shù)SKIPIF1<0總存在零點(diǎn)；④對(duì)于任意給定的正實(shí)數(shù)SKIPIF1<0，總存在實(shí)數(shù)SKIPIF1<0，使函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減.其中所有正確結(jié)論的序號(hào)是______________.【來源】中國人民大學(xué)附屬中學(xué)2021屆高三3月開學(xué)檢測(cè)數(shù)學(xué)試題【答案】①②③④【解析】如上圖分別為SKIPIF1<0，SKIPIF1<0和SKIPIF1<0時(shí)函數(shù)SKIPIF1<0的圖象，對(duì)于①：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0圖象如圖SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，所以存在SKIPIF1<0使得函數(shù)SKIPIF1<0為奇函數(shù)，故①正確；對(duì)于②：由三個(gè)圖知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0