人教A版高中數(shù)學(xué)(必修第一冊(cè))同步講義第31講 拓展二：函數(shù)與方程的綜合應(yīng)用（含解析）

第09講拓展二：函數(shù)與方程的綜合應(yīng)用題型01根據(jù)零點(diǎn)求參數(shù)【典例1】（2023春·江蘇宿遷·高一統(tǒng)考期中）函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)，則實(shí)數(shù)m的值為（

）A．9 B．12 C．0或9 D．0或12【答案】C【詳解】因?yàn)镾KIPIF1<0，令SKIPIF1<0，得到SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，得到SKIPIF1<0，滿足題意，當(dāng)SKIPIF1<0時(shí)，因?yàn)楹瘮?shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)，故SKIPIF1<0，得到SKIPIF1<0，綜上，SKIPIF1<0或SKIPIF1<0.故選：C.【典例2】（2023·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0恰有SKIPIF1<0個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因?yàn)镾KIPIF1<0時(shí)至多有一個(gè)零點(diǎn)，單調(diào)函數(shù)SKIPIF1<0至多一個(gè)零點(diǎn)，而函數(shù)SKIPIF1<0恰有SKIPIF1<0個(gè)零點(diǎn)，所以需滿足SKIPIF1<0有1個(gè)零點(diǎn)，SKIPIF1<0有1個(gè)零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，故選：D【典例3】（2023秋·四川雅安·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0若SKIPIF1<0恰有2個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是.【答案】SKIPIF1<0或SKIPIF1<0【詳解】又SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0；由SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0恰有2個(gè)零點(diǎn)，所以若SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0不是函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0；若SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0不是函數(shù)SKIPIF1<0的零點(diǎn)，則SKIPIF1<0，若SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的零點(diǎn)，SKIPIF1<0不是函數(shù)SKIPIF1<0的零點(diǎn)，則不存在這樣的SKIPIF1<0.綜上所述：實(shí)數(shù)a的取值范圍是SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0.【變式1】（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0有零點(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【答案】C【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0有零點(diǎn)，SKIPIF1<0與SKIPIF1<0有交點(diǎn)，SKIPIF1<0，即SKIPIF1<0，故選：C【變式2】（2023春·新疆昌吉·高二統(tǒng)考期末）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是．【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，符合題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，要使得函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，在方程SKIPIF1<0有兩個(gè)小于SKIPIF1<0的實(shí)根，設(shè)SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上與SKIPIF1<0軸有兩個(gè)交點(diǎn)，則滿足SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.【變式3】（2023·高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0，若1是此函數(shù)的零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的值是．【答案】0【詳解】因?yàn)?是此函數(shù)的零點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0題型02求函數(shù)的零點(diǎn)（方程的根）的個(gè)數(shù)【典例1】（2023·全國(guó)·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)的充分條件是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)楹瘮?shù)SKIPIF1<0恒過(guò)點(diǎn)SKIPIF1<0，所以函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)SKIPIF1<0函數(shù)SKIPIF1<0沒(méi)有零點(diǎn)SKIPIF1<0函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0無(wú)交點(diǎn)，數(shù)形結(jié)合可得，SKIPIF1<0或SKIPIF1<0即函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)的充要條件是SKIPIF1<0或SKIPIF1<0,只有選項(xiàng)SKIPIF1<0是函數(shù)有且只有一個(gè)零點(diǎn)的充分條件,

故選：A【典例2】（2023·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0，則方程SKIPIF1<0的實(shí)根個(gè)數(shù)為（

）A．3 B．4 C．5 D．6【答案】A【詳解】由SKIPIF1<0，則可作出函數(shù)SKIPIF1<0的圖象如下：由方程SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以方程SKIPIF1<0的實(shí)根個(gè)數(shù)為3．故選：A．【典例3】（2023秋·上海浦東新·高一?？计谀┮阎瘮?shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，判斷SKIPIF1<0在SKIPIF1<0上的單調(diào)性并證明；(2)討論函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)．【答案】(1)單調(diào)遞減，證明見(jiàn)解析(2)答案見(jiàn)解析【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0該函數(shù)為單調(diào)遞減函數(shù)，證明如下：在區(qū)間SKIPIF1<0上任取SKIPIF1<0，且SKIPIF1<0則SKIPIF1<0SKIPIF1<0因?yàn)镾KIPIF1<0，故SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0則函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減.即證.（2）SKIPIF1<0SKIPIF1<0，等價(jià)于SKIPIF1<0即等價(jià)于SKIPIF1<0的根的個(gè)數(shù)，令SKIPIF1<0，則其函數(shù)圖像如下所示：由圖可知：當(dāng)SKIPIF1<0或SKIPIF1<0或SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0有兩個(gè)交點(diǎn)；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0只有一個(gè)交點(diǎn)；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0有三個(gè)交點(diǎn).故：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有1個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有2個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有3個(gè)零點(diǎn).【典例4】（2023春·高一平湖市當(dāng)湖高級(jí)中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0）.(1)若SKIPIF1<0且方程SKIPIF1<0有解，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)若SKIPIF1<0是偶函數(shù)，討論函數(shù)SKIPIF1<0的零點(diǎn)情況.【答案】(1)SKIPIF1<0(2)當(dāng)SKIPIF1<0時(shí)函數(shù)無(wú)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)函數(shù)有一個(gè)零點(diǎn).【詳解】（1）因?yàn)榉匠蘏KIPIF1<0有解，所以方程SKIPIF1<0有解，即SKIPIF1<0的值域與方程SKIPIF1<0的值域相同.SKIPIF1<0SKIPIF1<0所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0；（2）因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)SKIPIF1<0滿足題意.函數(shù)SKIPIF1<0的零點(diǎn)情況等價(jià)于SKIPIF1<0的解的情況，即SKIPIF1<0，討論SKIPIF1<0的解的情況，令SKIPIF1<0，則SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)方程SKIPIF1<0無(wú)解，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0開(kāi)口向上，且恒過(guò)定點(diǎn)SKIPIF1<0，則SKIPIF1<0只有一解，此時(shí)方程SKIPIF1<0只有1解，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0開(kāi)口向下，且恒過(guò)定點(diǎn)SKIPIF1<0，且函數(shù)的對(duì)稱軸SKIPIF1<0，則方程（*）無(wú)解，綜上所述：當(dāng)SKIPIF1<0時(shí)函數(shù)無(wú)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)函數(shù)有一個(gè)零點(diǎn).【變式1】（2023·山東濰坊·統(tǒng)考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個(gè)數(shù)是（

）A．3 B．4 C．5 D．6【答案】A【詳解】求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個(gè)數(shù)，轉(zhuǎn)化為方程SKIPIF1<0在區(qū)間SKIPIF1<0上的根的個(gè)數(shù).由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個(gè)數(shù)為3.故選：A.【變式2】（2023秋·內(nèi)蒙古烏蘭察布·高一?？计谀┖瘮?shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是（

）．A．3個(gè) B．2個(gè) C．1個(gè) D．0個(gè)【答案】C【詳解】分別做出函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0的圖像，如上圖所示，由圖像可知，兩個(gè)函數(shù)的交點(diǎn)個(gè)數(shù)是SKIPIF1<0，所以函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是SKIPIF1<0.故選:C【變式3】（2023春·安徽·高一安徽省舒城中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0是偶函數(shù).(1)求實(shí)數(shù)SKIPIF1<0的值；(2)求方程SKIPIF1<0的實(shí)根的個(gè)數(shù)；(3)若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有且只有一個(gè)公共點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)1(3)SKIPIF1<0【詳解】（1）因?yàn)楹瘮?shù)SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.因?yàn)閷?duì)定義域內(nèi)的任意SKIPIF1<0上式恒成立，所以SKIPIF1<0.（2）由（1）可知SKIPIF1<0的解析式為SKIPIF1<0.所以SKIPIF1<0.因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0.所以方程SKIPIF1<0的實(shí)根的個(gè)數(shù)為1.（3）由題可知SKIPIF1<0.由SKIPIF1<0，可得SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0.所以SKIPIF1<0可化為SKIPIF1<0.令函數(shù)SKIPIF1<0.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，舍去.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的圖象開(kāi)口向上，因?yàn)镾KIPIF1<0，所以SKIPIF1<0一定存在唯一的正根，符合題意.當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0的圖象開(kāi)口向下，因?yàn)镾KIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0.又SKIPIF1<0，所以對(duì)稱軸SKIPIF1<0，所以SKIPIF1<0（舍去）或SKIPIF1<0.所以SKIPIF1<0.綜上，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.【變式4】（2023秋·貴州黔東南·高一統(tǒng)考期末）已知函數(shù)SKIPIF1<0為偶函數(shù)．(1)求實(shí)數(shù)SKIPIF1<0的值；(2)判斷函數(shù)SKIPIF1<0零點(diǎn)的個(gè)數(shù)．【答案】(1)SKIPIF1<0(2)一個(gè)零點(diǎn)【詳解】（1）因?yàn)楹瘮?shù)SKIPIF1<0SKIPIF1<0為偶函數(shù)，所以SKIPIF1<0即SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.（2）由（1）知SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)的個(gè)數(shù)等價(jià)于方程SKIPIF1<0根的個(gè)數(shù)，即方程SKIPIF1<0根的個(gè)數(shù)，令SKIPIF1<0，則轉(zhuǎn)化為方程SKIPIF1<0根的個(gè)數(shù)，而方程SKIPIF1<0的根為SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0有一個(gè)零點(diǎn)．【點(diǎn)睛】利用函數(shù)的奇偶性求參數(shù)，關(guān)鍵點(diǎn)在于利用好函數(shù)的奇偶性，即函數(shù)SKIPIF1<0是奇函數(shù)時(shí)，有SKIPIF1<0，函數(shù)SKIPIF1<0是偶函數(shù)時(shí)，有SKIPIF1<0.題型03函數(shù)與方程的綜合應(yīng)用【典例1】（2023秋·湖北·高一湖北省黃梅縣第一中學(xué)校聯(lián)考期末）已知SKIPIF1<0為偶函數(shù)．(1)求SKIPIF1<0的值；(2)解不等式SKIPIF1<0；(3)若關(guān)于SKIPIF1<0的方程SKIPIF1<0有4個(gè)不相等的實(shí)根，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0．【詳解】（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，SKIPIF1<0恒成立，即SKIPIF1<0恒成立，而SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0．（2）SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增且SKIPIF1<0，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增且SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，值域?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0的值域?yàn)镾KIPIF1<0．SKIPIF1<0，SKIPIF1<0不等式SKIPIF1<0的解集為SKIPIF1<0．（3）令SKIPIF1<0，則原方程可化為SKIPIF1<0，由（2）知SKIPIF1<0且方程SKIPIF1<0僅有一根，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩個(gè)不相等的實(shí)根，SKIPIF1<0關(guān)于SKIPIF1<0的方程SKIPIF1<0有4個(gè)不相等的實(shí)根，SKIPIF1<0關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0上有2個(gè)不相等的實(shí)根，記SKIPIF1<0，則SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0．【典例2】（2023春·江蘇南京·高二南京市中華中學(xué)?？计谀┮阎瘮?shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0的值域；(2)令SKIPIF1<0，則SKIPIF1<0，已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0有零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0，SKIPIF1<0(2)SKIPIF1<0【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0，由二次函數(shù)的性質(zhì)可知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0為增函數(shù)，所以函數(shù)的最大值為SKIPIF1<0，函數(shù)的最小值為SKIPIF1<0，則函數(shù)的值域?yàn)镾KIPIF1<0．（2）SKIPIF1<0，令SKIPIF1<0，由于SKIPIF1<0，則SKIPIF1<0，則問(wèn)題等價(jià)為SKIPIF1<0在SKIPIF1<0上有零點(diǎn)，即SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，則由對(duì)勾函數(shù)的性質(zhì)可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．【典例3】（2023秋·云南昆明·高一統(tǒng)考期末）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0定義域內(nèi)的一個(gè)子集，若存在SKIPIF1<0，使得SKIPIF1<0成立，則稱SKIPIF1<0是SKIPIF1<0的一個(gè)“不動(dòng)點(diǎn)”，也稱SKIPIF1<0在區(qū)間SKIPIF1<0上存在不動(dòng)點(diǎn)，例如SKIPIF1<0的“不動(dòng)點(diǎn)”滿足SKIPIF1<0，即SKIPIF1<0的“不動(dòng)點(diǎn)”是SKIPIF1<0.設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的不動(dòng)點(diǎn)；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上不存在不動(dòng)點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）根據(jù)題目給出的“不動(dòng)點(diǎn)”的定義，可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上的不動(dòng)點(diǎn)為SKIPIF1<0.（2）根據(jù)已知，得SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)解，所以SKIPIF1<0在SKIPIF1<0上無(wú)解，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)解，所以SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)解，設(shè)SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0所以SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，又因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上恒成立，所以SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，所以SKIPIF1<0，則SKIPIF1<0綜上，實(shí)數(shù)a的取值范圍是SKIPIF1<0.【變式1】（2023春·江蘇泰州·高二泰州中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0SKIPIF1<0是偶函數(shù).(1)求SKIPIF1<0的值；(2)若方程SKIPIF1<0有解，求SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）由已知可得，SKIPIF1<0SKIPIF1<0.因?yàn)镾KIPIF1<0為R上的偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0恒成立，所以，SKIPIF1<0，解得SKIPIF1<0.（2）由（1）知，SKIPIF1<0SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.因?yàn)榉匠蘏KIPIF1<0有解，即SKIPIF1<0有解，所以SKIPIF1<0.【變式2】（2023春·湖南株洲·高一株洲二中校考開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0是偶函數(shù).(1)求SKIPIF1<0的值；(2)若方程SKIPIF1<0有兩個(gè)不等的實(shí)數(shù)解，求實(shí)數(shù)m的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，又因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0，有SKIPIF1<0，即SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<0對(duì)SKIPIF1<0恒成立，即SKIPIF1<0對(duì)SKIPIF1<0恒成立，因?yàn)镾KIP