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

18/18解密函數(shù)零點相關(guān)問題一、方法綜述新課標下的高考越來越注重對學(xué)生的綜合素質(zhì)的考察，函數(shù)的零點問題便是一個考察學(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)零點問題，其形式逐漸多樣化，但都與函數(shù)、導(dǎo)數(shù)知識密不可分．根據(jù)函數(shù)零點的定義：對于函數(shù)SKIPIF1<0，把使SKIPIF1<0成立的實數(shù)SKIPIF1<0叫做函數(shù)SKIPIF1<0的零點．即：方程SKIPIF1<0有實數(shù)根SKIPIF1<0函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸有交點的橫坐標SKIPIF1<0函數(shù)SKIPIF1<0有零點．圍繞三者之間的關(guān)系，在高考數(shù)學(xué)中函數(shù)零點的題型主要①函數(shù)的零點的分布；②函數(shù)的零點的個數(shù)問題；③利用導(dǎo)數(shù)結(jié)合圖像的變動將兩個函數(shù)的圖像的交點問題轉(zhuǎn)化成函數(shù)的零點的個數(shù)問題．二、解題策略類型一：函數(shù)零點的分布問題例1．【2020·河南高考模擬】已知單調(diào)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，對于定義域內(nèi)任意SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的零點所在的區(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，因為SKIPIF1<0，所以零點所在的區(qū)間為（3，4）.【解題秘籍】判斷函數(shù)零點所在區(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)的零點所在區(qū)間為SKIPIF1<0．學(xué)科$網(wǎng)類型二函數(shù)零點的個數(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的零點的個數(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，方程xf（x）＝1不成立，即x≠0，則等價為f（x）＝SKIPIF1<0，當(dāng)2＜x≤4時，0＜x﹣2≤2，此時f（x）＝SKIPIF1<0f（x﹣2）＝SKIPIF1<0（1﹣|x﹣2﹣1|）＝SKIPIF1<0﹣SKIPIF1<0|x﹣3|，當(dāng)4＜x≤6時，2＜x﹣2≤4，此時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，由圖象知兩個函數(shù)圖象有3個交點，即函數(shù)g（x）的零點個數(shù)為3個，故選：B．【點睛】函數(shù)零點的求解與判斷方法：(1)直接求零點：令f(x)＝0，如果能求出解，則有幾個解就有幾個零點．(2)零點存在性定理：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)有多少個零點．(3)利用圖象交點的個數(shù)：將函數(shù)變形為兩個函數(shù)的差，畫兩個函數(shù)的圖象，看其交點的橫坐標有幾個不同的值，就有幾個不同的零點．【舉一反三】【2020·安徽高考模擬】已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有兩個零點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，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，所以當(dāng)SKIPIF1<0時，SKIPIF1<0的最小值為SKIPIF1<0.又在SKIPIF1<0上，SKIPIF1<0的圖像如圖所示：因為SKIPIF1<0有兩個不同的零點，所以方程SKIPIF1<0有兩個不同的解即直線SKIPIF1<0與SKIPIF1<0有兩個不同交點且交點的橫坐標分別為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恰有三個不相等的實數(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恰有三個不相等的實數(shù)解，即方程SKIPIF1<0恰有三個不相等的實數(shù)解，即SKIPIF1<0與SKIPIF1<0有三個不同的交點.令SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)單調(diào)遞增；且當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，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SKIPIF1<0的取值范圍是SKIPIF1<0.本題選擇C選項.【舉一反三】【2020·江蘇高考模擬】已知函數(shù)SKIPIF1<0有且僅有三個零點，并且這三個零點構(gòu)成等差數(shù)列，則實數(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個根為x1，x2，x3，且x1＜x2＜x3，當(dāng)x＞﹣a時，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的兩個解，即x1，x2是x2+（2a+3）x+4＝0的兩個解．得到x1+x2＝﹣（2a+3），x1x2＝4，又由設(shè)f（x）＝0的3個根為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，f（x）＝0最多只有兩個解，不滿足題意；綜上所述，aSKIPIF1<0或﹣1SKIPIF1<0．三、強化訓(xùn)練1．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有SKIPIF1<0個零點，則實數(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個零點即直線SKIPIF1<0與函數(shù)SKIPIF1<0有SKIPIF1<0個交點，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個單位，即直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像有SKIPIF1<0個交點，因為SKIPIF1<0，滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，因為直線SKIPIF1<0過點SKIPIF1<0，所以只需滿足直線SKIPIF1<0與SKIPIF1<0剛好有除點SKIPIF1<0外的另一個交點即可，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，如圖，將直線SKIPIF1<0繞原點逆時針旋轉(zhuǎn)，顯然SKIPIF1<0與SKIPIF1<0只有一個交點，故實數(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上恒有兩個零點，則實數(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上恒有兩個零點，即函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個交點，易知當(dāng)SKIPIF1<0時，滿足題意；當(dāng)SKIPIF1<0時，有三個交點，不滿足題意；當(dāng)SKIPIF1<0時，考慮SKIPIF1<0與SKIPIF1<0相切時，設(shè)切點坐標為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時，有兩個交點，滿足題意；當(dāng)SKIPIF1<0時，有四個交點，不滿足題意；當(dāng)SKIPIF1<0時，無交點，不滿足題意綜上，實數(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SKIPIF1<0?SKIPIF1<0，使得SKIPIF1<0?SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，則實數(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，其關(guān)于SKIPIF1<0對稱的點的坐標為SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對稱的函數(shù)為SKIPIF1<0.由于SKIPIF1<0與SKIPIF1<0的圖象上分別存在點SKIPIF1<0?SKIPIF1<0，使得SKIPIF1<0?SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，故函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0圖象在區(qū)間SKIPIF1<0有交點，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個實根，則SKIPIF1<0D．若函數(shù)SKIPIF1<0在SKIPIF1<0上有個零點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，對于A中，當(dāng)SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時，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錯誤；對于B中，因為SKIPIF1<0時，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錯誤；對于C中，直線SKIPIF1<0恒過定點SKIPIF1<0，函數(shù)SKIPIF1<0的圖象和函數(shù)SKIPIF1<0的圖象有三個交點，當(dāng)SKIPIF1<0，設(shè)SKIPIF1<0與SKIPIF1<0相切于點SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0，根據(jù)對稱性可知，當(dāng)SKIPIF1<0與SKIPIF1<0相切時，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的圖象有三個交點時，SKIPIF1<0，所以C正確．對于D中，又由函數(shù)SKIPIF1<0在SKIPIF1<0上有個零點SKIPIF1<0，故直線SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上由6個交點，不妨設(shè)SKIPIF1<0，由圖象可知SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，所以SKIPIF1<0，所以D錯誤.故選：C.5．SKIPIF1<0為實數(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點與SKIPIF1<0的圖像上某點關(guān)于SKIPIF1<0軸對稱，則實數(shù)SKIPIF1<0的取值范圍為___________.【答案】SKIPIF1<0【解析】設(shè)SKIPIF1<0，點SKIPIF1<0關(guān)于SKIPIF1<0軸對稱，由題意可知SKIPIF1<0在SKIPIF1<0有一個解，故SKIPIF1<0在SKIPIF1<0有一個解設(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只有一個交點，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為實數(shù))有兩個不同的零點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有兩個不同零點，等價于方程SKIPIF1<0有兩個不等實根.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上有兩個不同的實根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，由零點存在性定理知，SKIPIF1<0在SKIPIF1<0上有唯一零點，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個零點，則實數(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，SKIPIF1<0.所以，函數(shù)SKIPIF1<0的定義域為SKIPIF1<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個零點，則直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有兩個交點，設(shè)交點的橫坐標分別為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，即當(dāng)SKIPIF1<0時，直線SKIPIF1<0與函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象有兩個交點，綜上所述，實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.8．已知函數(shù)SKIPIF1<0給出下列四個結(jié)論：①存在實數(shù)SKIPIF1<0，使函數(shù)SKIPIF1<0為奇函數(shù)；②對任意實數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0既無最大值也無最小值；③對任意實數(shù)SKIPIF1<0和SKIPIF1<0，函數(shù)SKIPIF1<0總存在零點；④對于任意給定的正實數(shù)SKIPIF1<0，總存在實數(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時函數(shù)SKIPIF1<0的圖象，對于①：當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0圖象如圖SKIPIF1<0關(guān)于原點對稱，所以存在SKIPIF1<0使得函數(shù)SKIPIF1<0為奇函數(shù)，故①正確；對于②：由三個圖知當(dāng)SKIPIF1<0時，SKIPIF1<0