高一上學(xué)期期末數(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選項中，只有一項是符合題目要求的.1.命題“SKIPIF1<0”的否定是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)全稱命題的否定形式書寫即可判斷.【詳解】利用全稱量詞命題否定是存在量詞命題，所以命題“SKIPIF1<0”的否定為：“SKIPIF1<0”，故選：SKIPIF1<0.2.已知集合SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】利用一元二次不等式的解法和指數(shù)函數(shù)的單調(diào)性求出集合SKIPIF1<0，然后利用集合的運(yùn)算即可求解.【詳解】集合SKIPIF1<0，集合SKIPIF1<0，則SKIPIF1<0，由并集的運(yùn)算可知：SKIPIF1<0，故選：A3.已知函數(shù)SKIPIF1<0，角SKIPIF1<0終邊經(jīng)過SKIPIF1<0與SKIPIF1<0圖象的交點(diǎn)，則SKIPIF1<0()A.1 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】根據(jù)冪函數(shù)的性質(zhì)求出兩函數(shù)圖象的交點(diǎn)坐標(biāo)，結(jié)合任意角的三角函數(shù)的定義即可求解.【詳解】因為冪函數(shù)SKIPIF1<0和SKIPIF1<0圖象的交點(diǎn)為SKIPIF1<0，所以角SKIPIF1<0的終邊經(jīng)過交點(diǎn)SKIPIF1<0，所以SKIPIF1<0.故選：A.4.“SKIPIF1<0”是“SKIPIF1<0”的()A.充分必要條件 B.充分條件C.必要條件 D.既不充分又不必要條件【答案】C【解析】【分析】根據(jù)SKIPIF1<0可得到SKIPIF1<0或SKIPIF1<0，進(jìn)而利用充分條件和必要條件的判斷即可求解.【詳解】由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，所以充分性不成立；由SKIPIF1<0可推出SKIPIF1<0成立，所以必要性成立，結(jié)合選項可知：“SKIPIF1<0”是“SKIPIF1<0”的必要條件，故選：SKIPIF1<0.5.設(shè)SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)指數(shù)函數(shù)的單調(diào)性可得SKIPIF1<0，根據(jù)對數(shù)運(yùn)算性質(zhì)和對數(shù)函數(shù)的單調(diào)性可得SKIPIF1<0，即可求解.【詳解】由題意知，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故選：D.6.拱券是教堂建筑的主要素材之一，常見的拱券包括半圓拱?等邊哥特拱?弓形拱?馬蹄拱?二心內(nèi)心拱?四心拱?土耳其拱?波斯拱等.如圖，分別以點(diǎn)A和B為圓心，以線段AB為半徑作圓弧，交于點(diǎn)C，等邊哥特拱是由線段AB，SKIPIF1<0，SKIPIF1<0所圍成的圖形.若SKIPIF1<0，則該拱券的面積是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】求出扇形SKIPIF1<0的面積和三角形SKIPIF1<0的面積即得解.【詳解】解：設(shè)SKIPIF1<0的長為SKIPIF1<0.所以扇形SKIPIF1<0的面積為SKIPIF1<0.SKIPIF1<0的面積為SKIPIF1<0.所以該拱券的面積為SKIPIF1<0.故選：D7.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<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】首先根據(jù)不等式的解集，利用韋達(dá)定理得到SKIPIF1<0的關(guān)系，再代入求解不等式的解集.【詳解】由條件可知，SKIPIF1<0的兩個實數(shù)根是SKIPIF1<0和SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，所以不等式的解集為SKIPIF1<0.故選：A8.若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)僅有1個零點(diǎn)，則SKIPIF1<0的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】求出函數(shù)的零點(diǎn)，即對稱點(diǎn)的橫坐標(biāo)，列出3個相鄰的對稱點(diǎn)，由SKIPIF1<0在SKIPIF1<0內(nèi)僅有一個零點(diǎn)可得SKIPIF1<0，解之即可.【詳解】由題意知，令SKIPIF1<0，解得SKIPIF1<0，得函數(shù)SKIPIF1<03個相鄰的對稱點(diǎn)分別為SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)僅有一個零點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，得SKIPIF1<0.故選：C.二?多選題：本題共4小題，每小題5分，共20分.在每小題給出的選項中，有多項符合題目要求.全部選對的得5分，部分選對的得2分，有選铓的得0分.9.已知SKIPIF1<0都是正數(shù)，且SKIPIF1<0，則()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【解析】【分析】根據(jù)不等式的性質(zhì)判斷選項SKIPIF1<0，利用作差法判斷選項SKIPIF1<0.【詳解】對于SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，故選項SKIPIF1<0正確；對于SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，則無法判斷SKIPIF1<0的符號，故選項SKIPIF1<0錯誤；對于SKIPIF1<0，因為SKIPIF1<0都是正數(shù)，且SKIPIF1<0，所以SKIPIF1<0，故選項SKIPIF1<0正確；對于SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0都是正數(shù)，且SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0，則SKIPIF1<0，故選項SKIPIF1<0正確，故選：SKIPIF1<0.10.若函數(shù)SKIPIF1<0在一個周期內(nèi)的圖象如圖所示，則()A.SKIPIF1<0最小正周期為SKIPIF1<0B.SKIPIF1<0的增區(qū)間是SKIPIF1<0C.SKIPIF1<0D.將SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0倍(縱坐標(biāo)不變)得到SKIPIF1<0的圖象【答案】ABD【解析】【分析】結(jié)合圖象根據(jù)正弦函數(shù)的圖象和性質(zhì)逐項進(jìn)行分析即可求解.【詳解】由圖象可知：SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，又因為函數(shù)圖象過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，又因為SKIPIF1<0，所以SKIPIF1<0，則函數(shù)解析式為：SKIPIF1<0.對于SKIPIF1<0，函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0，故選項SKIPIF1<0正確；對于SKIPIF1<0，因為SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，所以函數(shù)SKIPIF1<0的增區(qū)間是SKIPIF1<0，故選項SKIPIF1<0正確；對于SKIPIF1<0，因為函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，故選項SKIPIF1<0錯誤；對于SKIPIF1<0，將SKIPIF1<0的圖象上所有點(diǎn)的橫坐標(biāo)變?yōu)樵瓉淼腟KIPIF1<0倍(縱坐標(biāo)不變)得到SKIPIF1<0，故選項SKIPIF1<0正確，故選：SKIPIF1<0.11.已知函數(shù)SKIPIF1<0，則下列命題正確的是()A.函數(shù)SKIPIF1<0是奇函數(shù)B.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在零點(diǎn)C.當(dāng)SKIPIF1<0時，SKIPIF1<0D.若SKIPIF1<0，則SKIPIF1<0【答案】BC【解析】【分析】根據(jù)函數(shù)的奇偶性判斷A；根據(jù)零點(diǎn)的存在性定理判斷B；結(jié)合圖形，根據(jù)函數(shù)的單調(diào)性判斷C；根據(jù)賦值法判斷D.【詳解】A：函數(shù)SKIPIF1<0的定義域為R，關(guān)于原點(diǎn)對稱，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0為非奇非偶函數(shù)，故A錯誤；B：SKIPIF1<0，有SKIPIF1<0，又函數(shù)SKIPIF1<0是連續(xù)的，由零點(diǎn)的存在性定理，得函數(shù)SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)，故B正確；C：如圖，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0，且在R上單調(diào)遞減，且SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，故C正確；D：SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故D錯誤.故選：BC.12.懸鏈線是平面曲線，是柔性鏈條或纜索兩端固定在兩根支柱頂部，中間自然下垂所形成的外形.在工程中有廣泛的應(yīng)用，例如縣索橋?雙曲拱橋?架空電纜都用到了懸鏈線的原理.當(dāng)微積分尚未出現(xiàn)的伽利略時期，伽利略猜測這種形狀是拋物線.直到1691年萊布尼茲和伯努利利用微積分推導(dǎo)出懸鏈線的方程是SKIPIF1<0，其中SKIPIF1<0為有關(guān)參數(shù).這樣，數(shù)學(xué)上又多了一對與SKIPIF1<0有關(guān)的著名函數(shù)——雙曲函數(shù)：雙曲正弦函數(shù)SKIPIF1<0和雙曲余弦函數(shù)SKIPIF1<0.則()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)新定義，直接運(yùn)算SKIPIF1<0即可判斷A，根據(jù)SKIPIF1<0即可判斷B，結(jié)合同底數(shù)冪的乘法法則，利用作差法即可判斷CD.【詳解】A：SKIPIF1<0SKIPIF1<0，故A錯誤；B：SKIPIF1<0，故B正確；C：SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，故C正確；D：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，故D正確.故選：BCD.三?填空題：本題共4小題，每小題5分，共20分.13.函數(shù)SKIPIF1<0定義域為__________.【答案】SKIPIF1<0【解析】【分析】根據(jù)對數(shù)函數(shù)與分式、根式的定義域求解即可.【詳解】由題意，SKIPIF1<0，解得SKIPIF1<0，故函數(shù)的定義域為SKIPIF1<0.故答案為：SKIPIF1<0.14.已知SKIPIF1<0，則SKIPIF1<0的值為__________.【答案】SKIPIF1<0【解析】【分析】根據(jù)角SKIPIF1<0與SKIPIF1<0互補(bǔ)，角SKIPIF1<0與SKIPIF1<0的關(guān)系，再結(jié)合誘導(dǎo)公式即可求解.【詳解】由題意可知：SKIPIF1<0，則SKIPIF1<0，又因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.15.已知正數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為__________.【答案】SKIPIF1<0##SKIPIF1<0【解析】【分析】首先將條件變形為SKIPIF1<0，再利用“1”的妙用，結(jié)合基本不等式求SKIPIF1<0的最小值.【詳解】因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時等號成立，所以SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<016.已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0的解集是__________.【答案】SKIPIF1<0【解析】【分析】利用奇偶性求出函數(shù)SKIPIF1<0的解析式SKIPIF1<0，分類討論即可求解.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，因為函數(shù)SKIPIF1<0是定義在R上的奇函數(shù)，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，要解不等式SKIPIF1<0，只需SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，綜上，不等式的解集為SKIPIF1<0.故答案為：SKIPIF1<0.四?解答題：本題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.已知集合SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0；(2)若SKIPIF1<0，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【解析】【分析】(1)先化簡集合SKIPIF1<0，再利用集合的并集運(yùn)算即可得解；(2)先由條件得到SKIPIF1<0，再對SKIPIF1<0與SKIPIF1<0分兩種情況討論得解.【小問1詳解】因為當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0.【小問2詳解】因為SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，滿足SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0；綜上，實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.18.已知SKIPIF1<0，且SKIPIF1<0.求下列各式的值：(1)SKIPIF1<0：(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)角的范圍和同角三角函數(shù)的基本關(guān)系得出SKIPIF1<0，進(jìn)一步得到SKIPIF1<0，將式子弦化切即可求解；(2)利用誘導(dǎo)公式將式子化簡為SKIPIF1<0，結(jié)合(1)即可求解.【小問1詳解】因為SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.【小問2詳解】SKIPIF1<0.19.已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的值域；(2)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)利用換元法注意新元的范圍及二次函數(shù)的性質(zhì)即可求解；(2)根據(jù)對數(shù)的運(yùn)算性質(zhì)及對數(shù)不等式的解法，將不等式恒成立的問題轉(zhuǎn)化為求函數(shù)的最值問題，結(jié)合基本不等式即可求解.【小問1詳解】令SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0，由二次函數(shù)的性質(zhì)知，對稱軸為SKIPIF1<0，開口向上，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0取得最小值為SKIPIF1<0，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0取得最大值為SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0.【小問2詳解】因為函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.因為SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立等價于SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0，SKIPIF1<0即可.因為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時取等號，所以當(dāng)SKIPIF1<0時，SKIPIF1<0的最小值為SKIPIF1<0,即SKIPIF1<0，故實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.20.“硬科技”是以人工智能?航空航天?生物技術(shù)?光電芯片?信息技術(shù)?新材料?新能源?智能制造等為代表的高精尖科技，屬于由科技創(chuàng)新構(gòu)成的物理世界，是需要長期研發(fā)投入?持續(xù)積累才能形成的原創(chuàng)技術(shù)，具有極高技術(shù)門檻和技術(shù)壁壘，難以被復(fù)制和模仿?最近十年，我國的一大批自主創(chuàng)新的企業(yè)都在打造自己的科技品牌，某高科技企業(yè)自主研發(fā)了一款具有自主知識產(chǎn)權(quán)的高級設(shè)備，并從2023年起全面發(fā)售.經(jīng)測算，生產(chǎn)該高級設(shè)備每年需投入固定成本1000萬元，每生產(chǎn)x百臺高級設(shè)備需要另投成本SKIPIF1<0萬元，且SKIPIF1<0每百臺高級設(shè)備售價為160萬元，假設(shè)每年生產(chǎn)的高級設(shè)備能夠全部售出，且高級設(shè)備年產(chǎn)展最大為10000臺.(1)求企業(yè)獲得年利潤SKIPIF1<0(萬元)關(guān)于年產(chǎn)量SKIPIF1<0(百臺)的函數(shù)關(guān)系式；(2)當(dāng)年產(chǎn)量為多少時，企業(yè)所獲年利潤最大？并求最大年利潤.【答案】(1)SKIPIF1<0；(2)當(dāng)年產(chǎn)量為30百臺時公司獲利最大，且最大利潤為800萬元.【解析】【分析】(1)根據(jù)利潤、成本、收入之間的關(guān)系分類討論即可；(2)當(dāng)SKIPIF1<0時，結(jié)合二次函數(shù)的性質(zhì)求出函數(shù)的最大值；當(dāng)SKIPIF1<0時，利用基本不等式求出函數(shù)的最大值，再比大小，即可求解.【小問1詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0；【小問2詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0(萬元).當(dāng)SKIPIF1<0時，SKIPIF1<0(萬元)，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時，等號成立.因為SKIPIF1<0，所以當(dāng)年產(chǎn)量為30百臺時，公司獲利最大，且最大利潤為800萬元.21.已知函數(shù)SKIPIF1<0的圖象與x軸的兩個相鄰交點(diǎn)之間的距離為SKIPIF1<0，直線SKIPIF1<0是SKIPIF1<0的圖象的一條對稱軸.(1)求函數(shù)SKIPIF1<0的解析式；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個零點(diǎn)SKIPIF1<0，請直接寫出SKIPIF1<0的取值范圍，并求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0；SKIPIF1<0【解析】【分析】(1)根據(jù)函數(shù)的圖象性質(zhì)，求解函數(shù)的解析式；(2)首先求函數(shù)SKIPIF1<0，將函數(shù)的零點(diǎn)轉(zhuǎn)化為函數(shù)圖象的交點(diǎn)問題，利用數(shù)形結(jié)合求參數(shù)的取值范圍，得到零點(diǎn)的關(guān)系，即可求解.【小問1詳解】由條件可知，周期SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0；【小問2詳解】SKIPIF1<0，當(dāng)SKIPIF1<0，設(shè)SKIPIF1<0，由條件轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0，在SKIPIF1<0上的圖象恰有3個不同的交點(diǎn)，作出SKIPIF1<0與SKIPIF1<0的圖象，如圖所示，由圖可知，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0.22.對于兩個定義域相同的函數(shù)SKIPIF1<0和SKIPIF1<0，若存在實數(shù)SKIPIF1<0，使SKIPIF1<0，則稱函數(shù)SKIPIF1<0是由“基函數(shù)SKIPIF1<0和SKIPIF1<0”生成的.(1)若SKIPIF1<0是由“基函數(shù)SKIPIF1<0和SKIPIF1<0”生成的，求實數(shù)SKIPIF1<0的值；(2)試?yán)谩盎瘮?shù)SKIPIF1<0和SKIPIF1<0”生成一個函數(shù)SKIPIF1<0，使之滿足SKIPIF1<0為偶函數(shù)，且SKIPIF1<0.①求函數(shù)SKIPIF1<0的解析式；②已知SKIPIF1<0，對于區(qū)間SKIPIF1<0上的任意值SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0恒成立，求實數(shù)SKIPIF1<0的最小值.(注：SKIPIF1<0.)【答案】(1)SKIPIF1<0；(2)①SKIPIF1<0；②SKIPIF1<0.【解析】【分析】(1)根據(jù)題意，可得SKIPIF1<0，化簡，利用對應(yīng)項的系數(shù)相等即可求解；①設(shè)SKIPIF1<0，根據(jù)函數(shù)SKIPI