新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專題03 函數(shù)的圖象與應(yīng)用（練）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專題03函數(shù)的圖象與應(yīng)用（練）【對點(diǎn)演練】一、單選題1．（2022·北京海淀·高三期中）在同一個坐標(biāo)系中，函數(shù)SKIPIF1<0與SKIPIF1<0且SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】A【分析】根據(jù)同底的指數(shù)函數(shù)和對數(shù)函數(shù)圖象關(guān)于SKIPIF1<0對稱可確定結(jié)果.【詳解】由指數(shù)函數(shù)和對數(shù)函數(shù)性質(zhì)可知：SKIPIF1<0與SKIPIF1<0圖象關(guān)于SKIPIF1<0對稱，由選項(xiàng)中圖象對稱關(guān)系可知A正確.故選：A.2．（2022·海南·模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖象如圖所示，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由函數(shù)圖象可確定SKIPIF1<0大小關(guān)系，結(jié)合指數(shù)函數(shù)單調(diào)性可得結(jié)果.【詳解】由圖象可知：SKIPIF1<0，SKIPIF1<0.故選：C.3．（2022·天津市建華中學(xué)高三階段練習(xí)）若函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸有公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)指數(shù)函數(shù)性質(zhì)可求得SKIPIF1<0的值域，由此可構(gòu)造不等式求得結(jié)果.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0與SKIPIF1<0軸有公共點(diǎn)，SKIPIF1<0，解得：SKIPIF1<0.故選：D.4．（2022·廣東·廣州六中高三階段練習(xí)）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】D【分析】舉例SKIPIF1<0，求導(dǎo)分析函數(shù)的單調(diào)性再判斷即可.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞增，SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，又SKIPIF1<0時SKIPIF1<0，而SKIPIF1<0時SKIPIF1<0，所以D圖象可能；故選：D5．（2022·四川省鄰水縣第二中學(xué)高三階段練習(xí)（理））定義運(yùn)算SKIPIF1<0，則函數(shù)SKIPIF1<0的圖像是（）A． B．C． D．【答案】A【分析】結(jié)合函數(shù)新定義與指數(shù)函數(shù)圖像求解即可.【詳解】解：因?yàn)檫\(yùn)算SKIPIF1<0，所以，SKIPIF1<0，所以，根據(jù)指數(shù)函數(shù)圖像可知A選項(xiàng)滿足題意.故選：A6．（2022·陜西·寶雞市金臺區(qū)教育體育局教研室高三階段練習(xí)（文））函數(shù)SKIPIF1<0的部分圖象如圖所示，則SKIPIF1<0的解析式可能為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】通過函數(shù)奇偶性的定義對選項(xiàng)逐個進(jìn)行判斷，再取圖象上的特殊點(diǎn)進(jìn)行排除即可.【詳解】由圖可知，SKIPIF1<0在SKIPIF1<0上的圖象關(guān)于SKIPIF1<0軸對稱，所以SKIPIF1<0在SKIPIF1<0上為偶函數(shù)，故應(yīng)先判斷各選項(xiàng)中函數(shù)SKIPIF1<0的奇偶性.對A，SKIPIF1<0，SKIPIF1<0為偶函數(shù)，故A選項(xiàng)的函數(shù)SKIPIF1<0為其定義域內(nèi)的偶函數(shù).同理：對C、D選項(xiàng)的SKIPIF1<0均為其定義域內(nèi)的偶函數(shù)，只有SKIPIF1<0選項(xiàng)的SKIPIF1<0為其定義域內(nèi)的奇函數(shù)，從而排除選項(xiàng)B.又SKIPIF1<0，對A選項(xiàng)：SKIPIF1<0，所以排除A.而由圖可知SKIPIF1<0，對C選項(xiàng)：SKIPIF1<0，SKIPIF1<0，故排除C.故選：D.二、多選題7．（2022·全國·高三專題練習(xí)）已知實(shí)數(shù)SKIPIF1<0滿足等式SKIPIF1<0，則下列可能成立的關(guān)系式為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【分析】在同一坐標(biāo)系內(nèi)分別畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖像，結(jié)合圖像即可判斷.【詳解】由題意，在同一坐標(biāo)系內(nèi)分別畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖像，如圖所示，由圖像知，當(dāng)SKIPIF1<0時，SKIPIF1<0，故選項(xiàng)A正確；做出直線SKIPIF1<0，當(dāng)SKIPIF1<0時，若SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)B正確；當(dāng)SKIPIF1<0時，若SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)C正確；當(dāng)SKIPIF1<0時，易得SKIPIF1<0，則SKIPIF1<0，故選項(xiàng)D錯誤.故選：ABC.8．（2022·江蘇·句容碧桂園學(xué)校高三期中）已知函數(shù)SKIPIF1<0,則下列結(jié)論中正確的是（

）A．SKIPIF1<0在(0,1)單調(diào)遞增B．SKIPIF1<0在(1,2)單調(diào)遞減C．SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對稱D．SKIPIF1<0的圖像關(guān)于點(diǎn)(0,1)對稱【答案】ABC【分析】先求定義域,用對數(shù)運(yùn)算性質(zhì)化為對數(shù)型復(fù)合函數(shù),根據(jù)復(fù)合函數(shù)的單調(diào)性判斷A,B的正誤;再根據(jù)SKIPIF1<0和SKIPIF1<0的關(guān)系判斷函數(shù)的對稱性.【詳解】解：由題意知,SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0,由復(fù)合函數(shù)的單調(diào)性知,函數(shù)SKIPIF1<0在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減,所以A,B正確;∵函數(shù)SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對稱,所以C正確,D錯誤．故選:ABC.三、填空題9.（2022·廣東·深圳市福田區(qū)福田中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0的最大值是________．【答案】SKIPIF1<0##SKIPIF1<0【分析】分別求得SKIPIF1<0和SKIPIF1<0時對應(yīng)的自變量SKIPIF1<0的值，結(jié)合SKIPIF1<0的圖象可確定SKIPIF1<0的取值范圍，由此可得結(jié)果.【詳解】令SKIPIF1<0，解得：SKIPIF1<0；令SKIPIF1<0，解得：SKIPIF1<0；SKIPIF1<0圖象如下圖所示，由圖象可知：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0.10．（2022·黑龍江·鐵人中學(xué)高三開學(xué)考試）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0時，SKIPIF1<0，若方程SKIPIF1<0恰有3個根，則實(shí)數(shù)SKIPIF1<0的取值范圍是_________.【答案】SKIPIF1<0【分析】根據(jù)題意可知，函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象有3個交點(diǎn)，作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象,數(shù)形結(jié)合即可求出．【詳解】依題可知，函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象有3個交點(diǎn),根據(jù)題意,可畫出SKIPIF1<0和SKIPIF1<0的圖象,

由圖可知:SKIPIF1<0解得SKIPIF1<0．故答案為：SKIPIF1<0．【沖刺提升】一、單選題1．（2022·遼寧·東北育才學(xué)校高三階段練習(xí)）函數(shù)SKIPIF1<0的大致圖象為（

）A． B． C． D．【答案】D【分析】根據(jù)函數(shù)的奇偶性可排除A,C,根據(jù)特殊點(diǎn)處的函數(shù)值可排除B,進(jìn)而可求解.【詳解】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對稱，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0是定義域內(nèi)的偶函數(shù)，故可排除A,C，又SKIPIF1<0，故可排除B，故選：D2．（2022·河南安陽·高三階段練習(xí)（理））如圖是某個函數(shù)SKIPIF1<0的圖象的一部分，則該函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)選項(xiàng)判斷函數(shù)的奇偶性并計(jì)算SKIPIF1<0的值，根據(jù)SKIPIF1<0的圖象即可求解.【詳解】對于A,SKIPIF1<0,為偶函數(shù)，且SKIPIF1<0，對于B，SKIPIF1<0，為奇函數(shù)，且SKIPIF1<0對于C,SKIPIF1<0，為偶函數(shù)，且SKIPIF1<0，對于D,SKIPIF1<0,為奇函數(shù)，且SKIPIF1<0，由SKIPIF1<0的圖象可知：SKIPIF1<0的圖象關(guān)于原點(diǎn)對稱且過SKIPIF1<0，故選：B3．（2022·北京市房山區(qū)良鄉(xiāng)中學(xué)高三期中）已知函數(shù)SKIPIF1<0，則下列命題錯誤的是（

）A．該函數(shù)圖象關(guān)于點(diǎn)SKIPIF1<0對稱；B．該函數(shù)的圖象關(guān)于直線SKIPIF1<0對稱；C．該函數(shù)在定義域內(nèi)單調(diào)遞減；D．將該函數(shù)圖象向左平移一個單位，再向下平移一個單位后與函數(shù)SKIPIF1<0的圖象重合．【答案】C【分析】依題意可得SKIPIF1<0，再根據(jù)函數(shù)的平移變換及反比例函數(shù)SKIPIF1<0的性質(zhì)判斷即可.【詳解】解：SKIPIF1<0把SKIPIF1<0向右，向上分別平移1個單位即可得到SKIPIF1<0的圖象，因?yàn)镾KIPIF1<0為奇函數(shù)，關(guān)于SKIPIF1<0對稱，所以SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對稱，故A正確；則將SKIPIF1<0的圖象向左平移一個單位，再向下平移一個單位得到SKIPIF1<0，故D正確由于函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0對稱，根據(jù)函數(shù)的圖象的平移可知函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0對稱，故B正確SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，但在整個定義域內(nèi)不具備單調(diào)性，故C錯誤故選：C．4．（2022·福建寧德·高三期中）函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．【答案】A【分析】根據(jù)特殊點(diǎn)的函數(shù)值、函數(shù)的奇偶性求得正確答案.【詳解】SKIPIF1<0，排除C選項(xiàng).SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0是偶函數(shù)，排除D選項(xiàng).SKIPIF1<0，所以B選項(xiàng)錯誤.故A選項(xiàng)正確.故選：A5．（2022·北京通州·高三期中）已知函數(shù)SKIPIF1<0設(shè)SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個零點(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】D【分析】先把函數(shù)零點(diǎn)問題轉(zhuǎn)化成兩個函數(shù)圖象有交點(diǎn)問題，再畫出圖象，結(jié)合導(dǎo)函數(shù)求出兩個函數(shù)有一個交點(diǎn)時實(shí)數(shù)SKIPIF1<0的值，再結(jié)合圖象分析有兩個交點(diǎn)時實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0有兩個零點(diǎn)，所以函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象有兩個不同的交點(diǎn).函數(shù)SKIPIF1<0恒過定點(diǎn)SKIPIF1<0,SKIPIF1<0，如圖所示，兩個函數(shù)圖象已經(jīng)有一個交點(diǎn)SKIPIF1<0.SKIPIF1<0時，SKIPIF1<0，其導(dǎo)函數(shù)SKIPIF1<0，當(dāng)直線SKIPIF1<0與函數(shù)SKIPIF1<0相切時，只有一個交點(diǎn)SKIPIF1<0，此時SKIPIF1<0，解得SKIPIF1<0，則當(dāng)SKIPIF1<0時，有兩個交點(diǎn).SKIPIF1<0時，SKIPIF1<0，其導(dǎo)函數(shù)SKIPIF1<0，當(dāng)直線SKIPIF1<0與函數(shù)SKIPIF1<0相切時，只有一個交點(diǎn)SKIPIF1<0，此時SKIPIF1<0，解得SKIPIF1<0，則當(dāng)SKIPIF1<0時，有兩個交點(diǎn).綜上，要使函數(shù)SKIPIF1<0有兩個零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D.6．（2022·河北保定·高三階段練習(xí)）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0.若對任意SKIPIF1<0，都有SKIPIF1<0，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)已知條件求出當(dāng)SKIPIF1<0，SKIPIF1<0時，函數(shù)SKIPIF1<0的解析式，做出函數(shù)圖象，結(jié)合圖象可求SKIPIF1<0的范圍.【詳解】因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0.所以當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，又SKIPIF1<0，且對任意SKIPIF1<0，都有SKIPIF1<0，所以SKIPIF1<0，作出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象，要使SKIPIF1<0，則需SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故選：B.二、多選題7．（2022·河北滄州·高三階段練習(xí)）函數(shù)SKIPIF1<0的大致圖象可能是（

）A． B．C． D．【答案】ABD【分析】先根據(jù)當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，排除C，再舉出適當(dāng)?shù)腟KIPIF1<0的值，分別得到ABD三個圖象.【詳解】由題意知SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的大致圖象不可能為C，而當(dāng)SKIPIF1<0為其他值時，A，B，D均有可能出現(xiàn)，不妨設(shè)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，此時A選項(xiàng)符合要求；當(dāng)SKIPIF1<0時，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，故函數(shù)SKIPIF1<0為奇函數(shù)，所以B選項(xiàng)符合要求，當(dāng)SKIPIF1<0時，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，故函數(shù)SKIPIF1<0為偶函數(shù)，所以D選項(xiàng)符合要求.故選：ABD8．（2022·江蘇省灌南高級中學(xué)高三階段練習(xí)）已知奇函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，若對SKIPIF1<0，有SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0，則下列結(jié)論中正確的是（

）A．SKIPIF1<0B．函數(shù)SKIPIF1<0是周期函數(shù)，且周期為2C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個數(shù)是7個D．對SKIPIF1<0，SKIPIF1<0【答案】BCD【分析】通過賦值法可以判斷A選項(xiàng)；根據(jù)函數(shù)的周期性判斷B選項(xiàng)；由對稱性及函數(shù)圖像即可判斷C、D選項(xiàng)．【詳解】解：由SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，SKIPIF1<0．∵SKIPIF1<0為奇函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以選項(xiàng)A錯誤，選項(xiàng)B正確；函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的零點(diǎn)個數(shù)等價為SKIPIF1<0的左右兩函數(shù)的交點(diǎn)個數(shù)，分別作出SKIPIF1<0與SKIPIF1<0的圖像如下所示：由圖像易知有7個交點(diǎn)，故選項(xiàng)C正確；對于選項(xiàng)D，對SKIPIF1<0，由對稱性可知：SKIPIF1<0關(guān)于SKIPIF1<0對稱，所以SKIPIF1<0，又SKIPIF1<0大于0，SKIPIF1<0，SKIPIF1<0小于0，SKIPIF1<0，所以SKIPIF1<0，所以選項(xiàng)D正確．故選：BCD．三、填空題9．（2022·上海師范大學(xué)附屬嘉定高級中學(xué)高三期中）已知SKIPIF1<0，若函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)a的取值范圍是___________.【答案】SKIPIF1<0【分析】分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0討論，結(jié)合二次函數(shù)的圖像及性質(zhì)即可得解.【詳解】因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，不符合題意；若SKIPIF1<0時，則SKIPIF1<0，符合題意，故SKIPIF1<0成立；當(dāng)SKIPIF1<0，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜上，SKIPIF1<0.故答案為：SKIPIF1<0.10．（2023·江蘇南京·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，則滿足SKIPIF1<0的x的取值范圍是________．【答案】SKIPIF1<0.【分析】結(jié)合函數(shù)圖象，利用復(fù)合函數(shù)的單調(diào)性解不等式.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0，由SKIPIF1<0有：SKIPIF1<0且SKIPIF1<0，因?yàn)镾KIPIF1<0，大致圖象如圖，①當(dāng)SKIPIF1<0且SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，顯然滿足SKIPIF1<0；②當(dāng)SKIPIF1<0時，根據(jù)復(fù)合函數(shù)的單調(diào)性法則同增異減可得，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，根據(jù)復(fù)合函數(shù)的單調(diào)性法則同增異減可得，SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，所以根據(jù)函數(shù)SKIPIF1<0的單調(diào)性有：由SKIPIF1<0,解得：SKIPIF1<0或SKIPIF1<0.綜上，滿足SKIPIF1<0的SKIPIF1<0取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.11．（2022·黑龍江·尚志市尚志中學(xué)高三階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有四個實(shí)數(shù)解SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【分析】方程SKIPIF1<0的解，即函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0交點(diǎn)的橫坐標(biāo)，可畫出函數(shù)圖象，結(jié)合二次函數(shù)的對稱性和對數(shù)函數(shù)的性質(zhì)求解.【詳解】函數(shù)SKIPIF1<0圖象如圖所示：∵關(guān)于SKIPIF1<0的方程SKIPIF1<0有四個實(shí)數(shù)解，∴函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有四個交點(diǎn)，交點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0與SKIPIF1<0關(guān)于SKIPIF1<0的對稱軸SKIPIF1<0對稱，∴SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.令SKIPIF1<0，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0