江蘇省南京市2022-2023學(xué)年高一上學(xué)期期中聯(lián)考數(shù)學(xué)試題（含解析）

高一上學(xué)期期中數(shù)學(xué)試題一?選擇題：本題共8小題，每小題5分，共40分.在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的.1.已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】確定集合A中元素，根據(jù)集合的交集運(yùn)算即可求得答案.【詳解】由題意得集合SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，故選:C.2.命題“SKIPIF1<0，SKIPIF1<0”的否定是()A.SKIPIF1<0，SKIPIF1<0 B.SKIPIF1<0，SKIPIF1<0C.SKIPIF1<0，SKIPIF1<0 D.SKIPIF1<0，SKIPIF1<0【答案】B【解析】【分析】根據(jù)含有一個(gè)量詞的命題的否定，即可確定答案.【詳解】命題“SKIPIF1<0，SKIPIF1<0”為特稱命題，其否定為全稱命題：SKIPIF1<0，SKIPIF1<0，故選：B.3.函數(shù)SKIPIF1<0的定義域?yàn)?)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)函數(shù)的解析式有意義，列出不等式組，即可求解.【詳解】由題意，函數(shù)SKIPIF1<0有意義，則滿足SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故選：B4.設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0=()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)對數(shù)的運(yùn)算，化簡SKIPIF1<0為SKIPIF1<0，即可得答案.【詳解】由題意知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故選：D5.已知SKIPIF1<0均為實(shí)數(shù)，且SKIPIF1<0，則下列結(jié)論正確的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)不等式的性質(zhì)可判斷A,D;舉反例可判斷B,C,即得答案.【詳解】由題意SKIPIF1<0均為實(shí)數(shù)，且SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，A錯(cuò)誤；取SKIPIF1<0，滿足條件，但是SKIPIF1<0，B，C錯(cuò)誤；由SKIPIF1<0知，SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，D正確，故選：D.6.已知函數(shù)SKIPIF1<0，則SKIPIF1<0的大致圖象是A. B.C. D.【答案】B【解析】【分析】利用特殊值SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，排除錯(cuò)誤選項(xiàng).【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除C，故選B.【點(diǎn)睛】從函數(shù)解析式結(jié)合選項(xiàng)，發(fā)現(xiàn)零點(diǎn)、單調(diào)性、奇偶性、過特殊點(diǎn)等性質(zhì)，是求解函數(shù)圖象問題的常見方法.7.已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集是SKIPIF1<0，則不等式SKIPIF1<0的解集是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)不等式的解集確定SKIPIF1<0是方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根，且SKIPIF1<0，進(jìn)而得SKIPIF1<0，化簡SKIPIF1<0為SKIPIF1<0，即可求得答案.【詳解】由題意關(guān)于SKIPIF1<0不等式SKIPIF1<0的解集是SKIPIF1<0，可知SKIPIF1<0是方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根，且SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，即不等式SKIPIF1<0解集是SKIPIF1<0，故選：C.8.已知SKIPIF1<0是奇函數(shù)，且在SKIPIF1<0上是增函數(shù)，又SKIPIF1<0，則SKIPIF1<0的解集為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)函數(shù)SKIPIF1<0的奇偶性以及在SKIPIF1<0上的單調(diào)性確定函數(shù)值的正負(fù)情況，結(jié)合SKIPIF1<0可得相應(yīng)不等式組，即可求得答案.【詳解】因?yàn)槎x在R上的奇函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上也是單調(diào)遞增，且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0,即SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0的解集為SKIPIF1<0，故選：A.【點(diǎn)睛】本題考查了函數(shù)的奇偶性以及單調(diào)性的綜合應(yīng)用，考查抽象不等式的解法，解答時(shí)要明確函數(shù)的對稱性質(zhì)，進(jìn)而判斷函數(shù)值的正負(fù)情況，解答的關(guān)鍵時(shí)根據(jù)不等式結(jié)合函數(shù)值情況得到相應(yīng)不等式組，求得結(jié)果.二?選擇題：本題共4小題，每小題5分，共20分.在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求.全部選對的得5分，部分選對的得2分，有選錯(cuò)的得0分.9.下列各組函數(shù)中是同一個(gè)函數(shù)的是()A.SKIPIF1<0與SKIPIF1<0B.SKIPIF1<0與SKIPIF1<0C.SKIPIF1<0與SKIPIF1<0D.SKIPIF1<0與SKIPIF1<0【答案】AC【解析】【分析】根據(jù)函數(shù)的定義,只需對應(yīng)關(guān)系和定義域一致,即可判斷為同一個(gè)函數(shù).【詳解】關(guān)于選項(xiàng)A,因?yàn)閷?yīng)關(guān)系和定義域一致,所以A是同一個(gè)函數(shù);關(guān)于選項(xiàng)B,因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0定義域?yàn)镾KIPIF1<0R,定義域不一致,所以B不是同一個(gè)函數(shù);關(guān)于選項(xiàng)C,因?yàn)閷?yīng)關(guān)系和定義域一致,所以C是同一個(gè)函數(shù);關(guān)于選項(xiàng)D,因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0,可得SKIPIF1<0,SKIPIF1<0定義域?yàn)镾KIPIF1<0,定義域不一致,所以D不是同一個(gè)函數(shù).故選:AC10.下列命題中正確的是()A.若SKIPIF1<0，則SKIPIF1<0 B.若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ABD【解析】【分析】將SKIPIF1<0時(shí)，SKIPIF1<0化為SKIPIF1<0，利用均值不等式可判斷A;利用SKIPIF1<0，利用均值不等式可判斷B；將SKIPIF1<0化為SKIPIF1<0，利用均值不等式可判斷C；利用SKIPIF1<0，結(jié)合均值不等式判斷D.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故A正確；若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)取等號(hào)，B正確；由SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不成立，故SKIPIF1<0等號(hào)取不到，C錯(cuò)誤；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，D正確，故選：SKIPIF1<0.11.已知命題p：函數(shù)SKIPIF1<0有零點(diǎn)，命題SKIPIF1<0，SKIPIF1<0.若p，q全為真命題，則實(shí)數(shù)a的取值可以是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【解析】【分析】分別求出p，q為真命題時(shí)a的取值范圍，取交集確定a的范圍，結(jié)合各選項(xiàng)即可確定答案.【詳解】命題p：函數(shù)SKIPIF1<0有零點(diǎn)為真命題時(shí)，若SKIPIF1<0，則SKIPIF1<0，函數(shù)無零點(diǎn)，不合題意；故SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0；命題SKIPIF1<0，SKIPIF1<0為真命題時(shí)，由于SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，若p，q全為真命題，則SKIPIF1<0或SKIPIF1<0，結(jié)合各選項(xiàng)可知實(shí)數(shù)a的取值可以是SKIPIF1<0，故選：SKIPIF1<0.12.已知函數(shù)SKIPIF1<0是偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，關(guān)于SKIPIF1<0的方程SKIPIF1<0的根，下列說法正確的有()A.當(dāng)SKIPIF1<0時(shí)，方程有4個(gè)不等實(shí)根B.當(dāng)SKIPIF1<0時(shí)，方程有6個(gè)不等實(shí)根C.當(dāng)SKIPIF1<0時(shí)，方程有4個(gè)不等實(shí)根D.當(dāng)SKIPIF1<0時(shí)，方程有6個(gè)不等實(shí)根【答案】BC【解析】【分析】結(jié)合函數(shù)奇偶性以及SKIPIF1<0時(shí)解析式，作出函數(shù)圖象，將關(guān)于SKIPIF1<0的方程SKIPIF1<0的根的問題轉(zhuǎn)化為函數(shù)圖象的交點(diǎn)問題，數(shù)形結(jié)合，求得答案.【詳解】由題意函數(shù)SKIPIF1<0是偶函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可作出函數(shù)SKIPIF1<0的圖象如圖示：則關(guān)于SKIPIF1<0的方程SKIPIF1<0的根，即轉(zhuǎn)化為函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0的交點(diǎn)問題，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0與SKIPIF1<0的圖象有三個(gè)交點(diǎn)，方程有3個(gè)不等實(shí)根，A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0的圖象有6個(gè)交點(diǎn)，方程有6個(gè)不等實(shí)根，B正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0的圖象有4個(gè)交點(diǎn)，方程有4個(gè)不等實(shí)根，C正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0的圖象有4個(gè)或2個(gè)或0個(gè)交點(diǎn)，方程有有4個(gè)或2個(gè)或0個(gè)實(shí)根，D錯(cuò)誤；故選：BC.【點(diǎn)睛】本題考查了函數(shù)的奇偶性的以及分段函數(shù)的應(yīng)用，考查了方程的根的個(gè)數(shù)的確定，解答時(shí)要注意函數(shù)圖象的應(yīng)用以及數(shù)形結(jié)合的思想方法，解答的關(guān)鍵是將方程的根的問題轉(zhuǎn)化為函數(shù)圖象的交點(diǎn)問題.三?填空題：本題共4小題，每小題5分，共20分.13.已知函數(shù)SKIPIF1<0，則SKIPIF1<0___________.【答案】0【解析】【分析】令SKIPIF1<0代入函數(shù)解析式中，可得答案.【詳解】由函數(shù)SKIPIF1<0可知，令SKIPIF1<0，則得SKIPIF1<0，故答案為：0.14.若“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【解析】【分析】根據(jù)題意可知SKIPIF1<0SKIPIF1<0，由此可求得m的范圍.【詳解】由題意“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件，則SKIPIF1<0SKIPIF1<0，故SKIPIF1<0，故答案為：SKIPIF1<0.15.已知SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為___________.【答案】3【解析】【分析】由條件SKIPIF1<0得SKIPIF1<0.后利用基本不等式可得答案.【詳解】由題SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0.又SKIPIF1<0.則SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).故答案為:SKIPIF1<016.函數(shù)SKIPIF1<0.(1)若SKIPIF1<0,則SKIPIF1<0___________；(2)若SKIPIF1<0是SKIPIF1<0上的減函數(shù),則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】①.1②.SKIPIF1<0##SKIPIF1<0【解析】【分析】(1)將SKIPIF1<0進(jìn)行分類討論,代入對應(yīng)的解析式求解即可;(2)解不等式組SKIPIF1<0即可得到結(jié)果;【詳解】(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0即SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0根據(jù)單調(diào)性的性質(zhì)函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù),且SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上無解.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0即SKIPIF1<0,解得SKIPIF1<0,滿足SKIPIF1<0.綜上SKIPIF1<0.(2)SKIPIF1<0SKIPIF1<0是SKIPIF1<0上的減函數(shù),SKIPIF1<0SKIPIF1<0,解得SKIPIF1<0.故答案為:(1)1;(2)SKIPIF1<0(或SKIPIF1<0).四?解答題：本大題共6小題，共70分.解答應(yīng)寫出文字說明?證明過程或演算步驟.17.已知全集SKIPIF1<0，集合SKIPIF1<0，集合SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0；(2)若SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0或SKIPIF1<0；(2)SKIPIF1<0或SKIPIF1<0.【解析】【分析】(1)確定集合A，B，求出集合B的補(bǔ)集，根據(jù)集合的并集運(yùn)算，即可求得答案.(2)求出集合A的補(bǔ)集，根據(jù)SKIPIF1<0，列出相應(yīng)不等式，求得答案.【小問1詳解】集合SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0SKIPIF1<0或SKIPIF1<0；【小問2詳解】由題意可知SKIPIF1<0或SKIPIF1<0，SKIPIF1<0,由SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.18.化簡求值：(1)SKIPIF1<0；(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)分?jǐn)?shù)指數(shù)冪以及根式的運(yùn)算法則，化簡求值，可得答案;(2)根據(jù)對數(shù)的運(yùn)算法則化簡求值，可得答案.【小問1詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0;【小問2詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0.19.已知二次函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0.(1)求SKIPIF1<0的解析式；(2)解關(guān)于SKIPIF1<0的不等式SKIPIF1<0.【答案】(1)SKIPIF1<0(2)答案見解析【解析】【分析】(1)設(shè)二次函數(shù)的解析式為SKIPIF1<0(SKIPIF1<0)，根據(jù)題意利用待定系數(shù)法求出a、b、c即可；(2)將原不等式化為SKIPIF1<0，分類討論，結(jié)合一元二次不等式的解法求出不等式當(dāng)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0時(shí)的解集即可.【小問1詳解】設(shè)SKIPIF1<0，SKIPIF1<0由SKIPIF1<0，得SKIPIF1<0SKIPIF1<0SKIPIF1<0又SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0【小問2詳解】由已知，SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，原不等式即為：SKIPIF1<0，解得SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，解得SKIPIF1<0綜上，當(dāng)SKIPIF1<0時(shí)，不等式的解集為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式的解集為：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式的解集為：SKIPIF1<0.20.已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)求SKIPIF1<0的值；(2)求函數(shù)SKIPIF1<0的解析式；(3)判斷函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性，并證明.【答案】(1)5;(2)SKIPIF1<0;(3)減函數(shù)，證明見解析.【解析】【分析】(1)根據(jù)函數(shù)SKIPIF1<0時(shí)的解析式結(jié)合其奇偶性，可求得SKIPIF1<0的值，繼而求得SKIPIF1<0的值；(2)由函數(shù)SKIPIF1<0時(shí)的解析式結(jié)合其奇偶性，可求得SKIPIF1<0時(shí)的解析式，由奇函數(shù)定義確定SKIPIF1<0，即可確定函數(shù)解析式；(3)利用函數(shù)單調(diào)性的定義可證明函數(shù)在SKIPIF1<0的單調(diào)性.【小問1詳解】由題意當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；【小問2詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，又因?yàn)楹瘮?shù)SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0，故SKIPIF1<0；【小問3詳解】由(2)可得SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上為減函數(shù)；證明如下：設(shè)SKIPIF1<0，則SKIPIF1<0，又由SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù).21.2022年8月17日，為進(jìn)一步捍衛(wèi)國家主權(quán)和領(lǐng)土完整，中國人民解放軍東部戰(zhàn)區(qū)繼續(xù)開展圍繞某島的軍事演習(xí)，海陸空三軍聯(lián)手展開全域作戰(zhàn)演練，各類現(xiàn)役主力裝備悉數(shù)登場，其中解放軍長航時(shí)無人機(jī)遠(yuǎn)海作戰(zhàn)能力再一次強(qiáng)力震懾住了敵對勢力.例如兩型偵察干擾無人機(jī)可以在遙控設(shè)備或自備程序控制操縱的情況下執(zhí)行任務(wù)，進(jìn)行對敵方通訊設(shè)施的電磁壓制和干擾，甚至壓制敵方的防空系統(tǒng).為了檢驗(yàn)實(shí)戰(zhàn)效果，某作戰(zhàn)部門對某處戰(zhàn)場實(shí)施“電磁干擾”實(shí)驗(yàn)，據(jù)測定，該處的“干擾指數(shù)”與無人機(jī)干擾源的強(qiáng)度和距離的比值成正比，比例系數(shù)為常數(shù)SKIPIF1<0.現(xiàn)已知相距36SKIPIF1<0的SKIPIF1<0兩處配置兩架無人機(jī)干擾源，其對敵干擾的強(qiáng)度分別為SKIPIF1<0和SKIPIF1<0，線段SKIPIF1<0上任意一點(diǎn)SKIPIF1<0處的干擾指數(shù)SKIPIF1<0等于兩機(jī)對該處的干擾指數(shù)之和，設(shè)SKIPIF1<0.(1)試將SKIPIF1<0表示為SKIPIF1<0的函數(shù)，并求出定義域；(2)當(dāng)SKIPIF1<0時(shí)，試確定“干擾指數(shù)”最小時(shí)SKIPIF1<0所處的位置.【答案】(1)SKIPIF1<0，定義域?yàn)镾KIPIF1<0(2)“干擾指數(shù)”最小時(shí)SKIPIF1<0所處位置在距離A點(diǎn)SKIPIF1<0處【解析】【分析】(1)根據(jù)題意即可求出SKIPIF1<0，繼而根據(jù)問題實(shí)際意義求得函數(shù)定義域；(2)將SKIPIF1<0變?yōu)镾KIPIF1<0，利用基本不等式即可求得答案.小問1詳解】由題意，點(diǎn)SKIPIF1<0受A干擾指數(shù)為SKIPIF1<0，點(diǎn)SKIPIF1<0受SKIPIF1<0干擾指數(shù)為SKIPIF1<0，其中SKIPIF1<0，從而點(diǎn)SKIPIF1<0處干擾指數(shù)：SKIPIF1<0，又SKIPIF1<0，故定義域?yàn)镾KIPIF1<0.【小問2詳解】當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)，等號(hào)成立.故“干擾指數(shù)”最小時(shí)SKIPIF1<0所處位置在距離A點(diǎn)SKIPIF1<0處.22已知二次函數(shù)SKIPIF1<0.(1)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0對SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍；(2)已知函數(shù)SKIPIF1<0，若對SKIPIF1<0，SKIPIF1<0，使不等式SKIPIF1<0成立，求SKIPIF1<0