湖南師xx中學(xué)2022-2023學(xué)年高一上學(xué)期期末模擬數(shù)學(xué)試題（一）（含解析）

高一上學(xué)期期中數(shù)學(xué)試題一、單選題(本大題共8小題，共40.0分.在每小題列出的選項(xiàng)中，選出符合題目的一項(xiàng))1.已知集合SKIPIF1<0，則SKIPIF1<0中元素的個(gè)數(shù)為()A.2 B.3 C.4 D.5【答案】C【解析】【分析】化簡(jiǎn)集合SKIPIF1<0，根據(jù)交集概念求出SKIPIF1<0，從而可得答案.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0,因?yàn)镾KIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0滿(mǎn)足SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0中元素的個(gè)數(shù)為SKIPIF1<0.故選：C2.已知實(shí)數(shù)a，b，c滿(mǎn)足SKIPIF1<0，則下列不等式一定成立的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】利用作差法逐項(xiàng)判斷可得答案.【詳解】因?yàn)閍，b，c滿(mǎn)足SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，對(duì)于A，SKIPIF1<0，所以SKIPIF1<0，故A錯(cuò)誤；對(duì)于B，SKIPIF1<0，所以SKIPIF1<0，故B錯(cuò)誤；對(duì)于C，SKIPIF1<0，所以SKIPIF1<0，故C錯(cuò)誤；對(duì)于D，SKIPIF1<0，所以SKIPIF1<0，故D正確；故選：D.3.已知定義域?yàn)镽的函數(shù)f(x)不是偶函數(shù)，則下列命題一定為真命題的是()A.?x∈R，f(－x)≠f(x)B.?x∈R，f(－x)≠－f(x)C.?x0∈R，f(－x0)≠f(x0)D.?x0∈R，f(－x0)≠－f(x0)【答案】C【解析】【分析】利用偶函數(shù)的定義和全稱(chēng)命題的否定分析判斷解答.【詳解】∵定義域?yàn)镽的函數(shù)f(x)不是偶函數(shù)，∴?x∈R，f(－x)＝f(x)為假命題，∴?x0∈R，f(－x0)≠f(x0)為真命題.故選C【點(diǎn)睛】本題主要考查偶函數(shù)的定義和全稱(chēng)命題的否定，意在考查學(xué)生對(duì)該知識(shí)的理解掌握水平，屬于基礎(chǔ)題.4.若正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，則A.SKIPIF1<0有最大值4 B.SKIPIF1<0有最小值SKIPIF1<0C.SKIPIF1<0有最大值SKIPIF1<0 D.SKIPIF1<0有最小值SKIPIF1<0【答案】C【解析】【詳解】試題分析：因?yàn)檎龑?shí)數(shù)，滿(mǎn)足，所以SKIPIF1<0，故SKIPIF1<0有最小值4，故A不正確；由基本不等式可得SKIPIF1<0，故有最大值SKIPIF1<0，故B不正確；由于SKIPIF1<0，故SKIPIF1<0由最大值為SKIPIF1<0，故C正確；SKIPIF1<0，故SKIPIF1<0由最小值SKIPIF1<0，故D不正確．考點(diǎn)：基本不等式5.已知SKIPIF1<0是第三象限角，且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)誘導(dǎo)公式及同角三角函數(shù)關(guān)系與二倍角公式即可得解.【詳解】由已知得SKIPIF1<0，SKIPIF1<0，則原式SKIPIF1<0.故選：D6.已知SKIPIF1<0，則“存在SKIPIF1<0使得SKIPIF1<0”是“SKIPIF1<0”的()．A.充分而不必要條件 B.必要而不充分條件C.充分必要條件 D.既不充分也不必要條件【答案】C【解析】【分析】根據(jù)充分條件，必要條件的定義，以及誘導(dǎo)公式分類(lèi)討論即可判斷.【詳解】(1)當(dāng)存SKIPIF1<0使得SKIPIF1<0時(shí)，若SKIPIF1<0為偶數(shù)，則SKIPIF1<0；若SKIPIF1<0為奇數(shù)，則SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，亦即存在SKIPIF1<0使得SKIPIF1<0．所以，“存在SKIPIF1<0使得SKIPIF1<0”是“SKIPIF1<0”的充要條件.故選：C.【點(diǎn)睛】本題主要考查充分條件，必要條件的定義的應(yīng)用，誘導(dǎo)公式的應(yīng)用，涉及分類(lèi)討論思想的應(yīng)用，屬于基礎(chǔ)題.7.為了保障交通安全，某地根據(jù)《道路交通安全法》規(guī)定：汽車(chē)駕駛員血液中的酒精含量不得超過(guò)SKIPIF1<0．據(jù)儀器監(jiān)測(cè)，某駕駛員喝了二兩白酒后，血液中的酒精含量迅速上升到SKIPIF1<0，在停止喝酒后，血液中每小時(shí)末的酒精含量都比上一個(gè)小時(shí)末減少25%．那么此人在開(kāi)車(chē)前至少要休息()(參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0)A.4．1小時(shí) B.4．2小時(shí) C.4．3小時(shí) D.4．4小時(shí)【答案】B【解析】【分析】由題意知經(jīng)過(guò)SKIPIF1<0小時(shí)，血液中的酒精含量為SKIPIF1<0，則SKIPIF1<0，解不等式SKIPIF1<0即可.【詳解】設(shè)經(jīng)過(guò)SKIPIF1<0小時(shí)，血液中的酒精含量為SKIPIF1<0，則SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0.因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以開(kāi)車(chē)前至少要休息4.2小時(shí)，故選：B．【點(diǎn)晴】關(guān)鍵點(diǎn)點(diǎn)晴：實(shí)際問(wèn)題，關(guān)鍵是讀懂題意抽象出具體函數(shù).8.已知定義在R上的函數(shù)SKIPIF1<0對(duì)于任意的x都滿(mǎn)足SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若函數(shù)SKIPIF1<0至少有6個(gè)零點(diǎn)，則a的取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】函數(shù)的根轉(zhuǎn)化為兩個(gè)新函數(shù)圖像的焦點(diǎn)問(wèn)題，再對(duì)對(duì)數(shù)函數(shù)的SKIPIF1<0進(jìn)行分類(lèi)討論即可.【詳解】由SKIPIF1<0知SKIPIF1<0是周期為2的周期函數(shù)，函數(shù)SKIPIF1<0至少有6個(gè)零點(diǎn)等價(jià)于函數(shù)SKIPIF1<0SKIPIF1<0與SKIPIF1<0的圖象至少有6個(gè)交點(diǎn)，①當(dāng)SKIPIF1<0時(shí),畫(huà)出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象如下圖所示,根據(jù)圖象可得SKIPIF1<0,即SKIPIF1<0.

②當(dāng)SKIPIF1<0時(shí),畫(huà)出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象如下圖所示,根據(jù)圖象可得SKIPIF1<0,即SKIPIF1<0SKIPIF1<0.綜上所述,SKIPIF1<0的取值范圍是SKIPIF1<0.

故選：A二、多選題(本大題共4小題，共20.0分.在每小題有多項(xiàng)符合題目要求)9.下列計(jì)算結(jié)果為有理數(shù)的是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)特殊角的三角函數(shù)判斷A，根據(jù)對(duì)數(shù)的運(yùn)算性質(zhì)與換底公式判斷BCD.【詳解】SKIPIF1<0，不是有理數(shù)，故A錯(cuò)誤；SKIPIF1<0，是有理數(shù)，故B正確；SKIPIF1<0，是有理數(shù)，故C正確；SKIPIF1<0，是有理數(shù)，故D正確.故選:BCD10.已知函數(shù)SKIPIF1<0在一個(gè)周期內(nèi)的圖象如圖所示，其中圖象最高點(diǎn)、最低點(diǎn)的橫坐標(biāo)分別為SKIPIF1<0、SKIPIF1<0，圖象在SKIPIF1<0軸上的截距為SKIPIF1<0．則下列結(jié)論正確的是()A.SKIPIF1<0的最小正周期為SKIPIF1<0B.SKIPIF1<0的最大值為2C.SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D.SKIPIF1<0為偶函數(shù)【答案】BC【解析】【分析】由周期求SKIPIF1<0，由五點(diǎn)法作圖求出SKIPIF1<0的值，由特殊點(diǎn)的坐標(biāo)求出A，再利用三角函數(shù)的圖象和性質(zhì)，得出結(jié)論．【詳解】由圖知，SKIPIF1<0的最小正周期SKIPIF1<0，則SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞增.因?yàn)镾KIPIF1<0，則SKIPIF1<0不是偶函數(shù)，故選：BC．【點(diǎn)睛】本題主要考查三角函數(shù)的圖象與性質(zhì)，解題的關(guān)鍵是會(huì)根據(jù)圖象求解析式.11.已知函數(shù)SKIPIF1<0，則下列結(jié)論正確的是()A.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0無(wú)零點(diǎn)B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0只有一個(gè)零點(diǎn)C.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)零點(diǎn)D.若SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0【答案】ABD【解析】【分析】判斷函數(shù)零點(diǎn)轉(zhuǎn)化為判斷方程SKIPIF1<0的根，再轉(zhuǎn)化為考察直線SKIPIF1<0和拋物線SKIPIF1<0的位置關(guān)系即可求解.【詳解】令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0．考察直線SKIPIF1<0和拋物線SKIPIF1<0的位置關(guān)系，由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0無(wú)零點(diǎn)；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0只有一個(gè)零點(diǎn)，當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)零點(diǎn)；若SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0是方程SKIPIF1<0的兩根，由韋達(dá)定理，得SKIPIF1<0，故選：ABD12.若a，b，c∈R，且ab＋bc＋ca＝1，則下列不等式成立的是()A.a2＋b2＋c2≥1 B.a＋b＋c≤SKIPIF1<0C.SKIPIF1<0+SKIPIF1<0+SKIPIF1<0≤2SKIPIF1<0 D.(a＋b＋c)2≥3【答案】AD【解析】【分析】先利用均值不等式得到a2＋b2＋c2≥ab＋bc＋ca，確定A正確，進(jìn)而推出BD選項(xiàng)正誤，再利用特殊值驗(yàn)證C項(xiàng)錯(cuò)誤即可.【詳解】由均值不等式知a2＋b2≥2ab，b2＋c2≥2bc，a2＋c2≥2ac，于是a2＋b2＋c2≥ab＋bc＋ca＝1，故A正確；而(a＋b＋c)2＝a2＋b2＋c2＋2(ab＋bc＋ca)≥3(ab＋bc＋ca)＝3，故D項(xiàng)正確，B項(xiàng)錯(cuò)誤；令a＝b＝c＝SKIPIF1<0，則ab＋bc＋ca＝1，但SKIPIF1<0＝3SKIPIF1<0＞2SKIPIF1<0，故C項(xiàng)錯(cuò)誤．故選：AD.三、填空題(本大題共4小題，共20.0分)13.已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0______【答案】SKIPIF1<0【解析】【分析】解不等式求出集合SKIPIF1<0,根據(jù)集合的交集運(yùn)算即可求得答案.【詳解】由題意解不等式SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，解不等式SKIPIF1<0，即SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.14.設(shè)SKIPIF1<0是第二象限角，SKIPIF1<0為其終邊上一點(diǎn)，且SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0##SKIPIF1<0【解析】【分析】根據(jù)三角函數(shù)定義，求得SKIPIF1<0，以及SKIPIF1<0，再結(jié)合正切的倍角公式，即可求得結(jié)果.【詳解】根據(jù)題意，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0是第二象限角，故SKIPIF1<0；則SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.15.在等式SKIPIF1<0的等號(hào)右側(cè)兩個(gè)分?jǐn)?shù)的分母方塊處，各填上一個(gè)正整數(shù)，并且使這兩個(gè)正整數(shù)的和最小，則這兩個(gè)數(shù)的積為_(kāi)_____.【答案】SKIPIF1<0【解析】【分析】將題意轉(zhuǎn)化為SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0最小時(shí)SKIPIF1<0的值，再根據(jù)基本不等式求解即可.【詳解】由題意，即SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0最小時(shí)SKIPIF1<0的值.因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，此時(shí)SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<016.對(duì)于函數(shù)SKIPIF1<0，若在其定義域內(nèi)存在兩個(gè)實(shí)數(shù)SKIPIF1<0，使當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值域也是SKIPIF1<0，則稱(chēng)函數(shù)SKIPIF1<0為“保值”函數(shù)，區(qū)間SKIPIF1<0稱(chēng)為函數(shù)SKIPIF1<0的“等域區(qū)間”.(1)請(qǐng)寫(xiě)出一個(gè)滿(mǎn)足條件的“保值”函數(shù)：______(2)若函數(shù)SKIPIF1<0是“保值”函數(shù)，則實(shí)數(shù)k的取值范圍是______.【答案】①.SKIPIF1<0②.SKIPIF1<0【解析】【分析】(1)單調(diào)函數(shù)，定義域與值域一樣，固然想到SKIPIF1<0(2)根據(jù)判斷SKIPIF1<0的單調(diào)性，轉(zhuǎn)化為關(guān)于SKIPIF1<0的方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根.【詳解】(1)由題意得方程SKIPIF1<0至少有兩個(gè)根，設(shè)函數(shù)SKIPIF1<0(2)因?yàn)镾KIPIF1<0是增函數(shù)，若SKIPIF1<0是“保值”函數(shù)，則存在實(shí)數(shù)SKIPIF1<0，使SKIPIF1<0即SKIPIF1<0所以SKIPIF1<0是關(guān)于SKIPIF1<0的方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根，從而方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根.令SKIPIF1<0，則SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0SKIPIF1<0，根據(jù)二次函數(shù)的圖象可知，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)不同的交點(diǎn)，即方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.四、解答題(本大題共6小題，共70.0分.解答應(yīng)寫(xiě)出文字說(shuō)明，證明過(guò)程或演算步驟)17.已知函數(shù)SKIPIF1<0，集合SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的最大值；(2)記集合SKIPIF1<0，若SKIPIF1<0是SKIPIF1<0的必要條件，求實(shí)數(shù)a的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)先求出對(duì)稱(chēng)軸，再討論區(qū)間與對(duì)稱(chēng)軸的關(guān)系，即可求解．(2)先得到SKIPIF1<0，再分SKIPIF1<0和SKIPIF1<0，分別列出不等式組，求解即可．【小問(wèn)1詳解】SKIPIF1<0，SKIPIF1<0，對(duì)稱(chēng)軸為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0,SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0,SKIPIF1<0上單調(diào)遞減，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0,SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，綜上所述：SKIPIF1<0．【小問(wèn)2詳解】SKIPIF1<0是SKIPIF1<0的充分條件，SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，②當(dāng)SKIPIF1<0時(shí)，由題意得方程SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0,SKIPIF1<0上有兩個(gè)實(shí)根，令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜上所述：實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．18.已知函數(shù)SKIPIF1<0.(1)求證：SKIPIF1<0是奇函數(shù)；(2)求證：SKIPIF1<0；(3)若SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0，SKIPIF1<0的值．【答案】(1)證明見(jiàn)解析(2)證明見(jiàn)解析(3)SKIPIF1<0，SKIPIF1<0【解析】【分析】(1)由函數(shù)解析式可得SKIPIF1<0，求得函數(shù)的定義域關(guān)于原點(diǎn)對(duì)稱(chēng)．再根據(jù)SKIPIF1<0，可得SKIPIF1<0是奇函數(shù)．(2)根據(jù)對(duì)數(shù)的運(yùn)算法則分別求得SKIPIF1<0，SKIPIF1<0，可得要證的等式成立．(3)由條件利用(2)的結(jié)論可得SKIPIF1<0，SKIPIF1<0，由此求得SKIPIF1<0和SKIPIF1<0的值．【小問(wèn)1詳解】解：由函數(shù)SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱(chēng)．再根據(jù)SKIPIF1<0，可得SKIPIF1<0是奇函數(shù)．【小問(wèn)2詳解】證明：SKIPIF1<0，而SKIPIF1<0，SKIPIF1<0成立．小問(wèn)3詳解】解：若SKIPIF1<0，SKIPIF1<0，則由(2)可得SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0．19.設(shè)SKIPIF1<0．(1)求使不等式SKIPIF1<0成立的SKIPIF1<0的取值集合；(2)先將SKIPIF1<0圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來(lái)的2倍，縱坐標(biāo)不變；再向右平移SKIPIF1<0個(gè)單位；最后向下平移SKIPIF1<0個(gè)單位得到函數(shù)SKIPIF1<0的圖象．若不等式SKIPIF1<0在SKIPIF1<0上恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0．【解析】【分析】(1)利用降冪公式和輔助角公式可得SKIPIF1<0，因此SKIPIF1<0等價(jià)于SKIPIF1<0，利用正弦函數(shù)的性質(zhì)可求不等式的解集.(2)根據(jù)圖象變換可得SKIPIF1<0，從而原不等式可化為SKIPIF1<0在SKIPIF1<0，換元后利用二次函數(shù)的性質(zhì)可求SKIPIF1<0的取值范圍.【詳解】解：SKIPIF1<0.(1)SKIPIF1<0即：SKIPIF1<0SKIPIF1<0，所以原不等式的解集為：SKIPIF1<0.(2)SKIPIF1<0將圖象上所有點(diǎn)的橫坐標(biāo)伸長(zhǎng)為原來(lái)的2倍，縱坐標(biāo)不變,得SKIPIF1<0；再向右平移SKIPIF1<0個(gè)單位，得SKIPIF1<0；最后向下平移SKIPIF1<0個(gè)單位得到函數(shù)SKIPIF1<0，∴SKIPIF1<0.設(shè)SKIPIF1<0，由SKIPIF1<0可得：SKIPIF1<0，則原不等式等價(jià)于：SKIPIF1<0在SKIPIF1<0上恒成立；設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0遞增，在SKIPIF1<0遞減，所以SKIPIF1<0，所以SKIPIF1<0．【點(diǎn)睛】形如SKIPIF1<0的函數(shù)，可以利用降冪公式和輔助角公式將其化為SKIPIF1<0的形式，再根據(jù)正弦函數(shù)的性質(zhì)求與SKIPIF1<0相關(guān)的不等式或方程的求解問(wèn)題．另外，含SKIPIF1<0的二次式的恒成立問(wèn)題，常通過(guò)換元轉(zhuǎn)化為一元二次不等式在相應(yīng)范圍上的恒成立問(wèn)題.20.如圖所示，有一塊扇形鋼板OPQ，面積是SKIPIF1<0平方米，其所在圓的半徑為1米，(1)求扇形圓心角的大?。?2)現(xiàn)在鋼板OPQ上裁下一塊平行四邊形鋼板ABOC，要求使裁下的鋼板面積最大.試問(wèn)如何確定A的位置，才能使裁下的鋼板符合要求？最大面積為多少？【答案】(1)SKIPIF1<0(2)當(dāng)SKIPIF1<0是SKIPIF1<0的中點(diǎn)時(shí)，裁下的鋼板符合要求，最大面積為SKIPIF1<0平方米【解析】【分析】(1)利用扇形面積公式列方程，從而求得扇形圓心角的大小.(2)連接SKIPIF1<0，設(shè)SKIPIF1<0，將裁下的鋼板的面積用SKIPIF1<0來(lái)表示，結(jié)合三角函數(shù)的性質(zhì)求得面積的最大值以及此時(shí)SKIPIF1<0點(diǎn)的位置.【小問(wèn)1詳解】依題意，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，即扇形圓心角的大小為SKIPIF1<0.【小問(wèn)2詳解】連接SKIPIF1<0，設(shè)SKIPIF1<0，過(guò)SKIPIF1<0作SKIPIF1<0，垂足為SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，所以SKIPIF1<0，設(shè)四邊形SKIPIF1<0的面積為SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0，由于SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值為SKIPIF1<0(平方米).所以當(dāng)SKIPIF1<0是SKIPIF1<0的中點(diǎn)時(shí)，裁下的鋼板符合要求，最大面積為SKIPIF1<0平方米.21.某產(chǎn)品近日開(kāi)始上市，通過(guò)市場(chǎng)調(diào)查，得到該產(chǎn)品每1件的市場(chǎng)價(jià)SKIPIF1<0單位：元SKIPIF1<0與上市時(shí)間SKIPIF1<0單位：天SKIPIF1<0的數(shù)據(jù)如下：上市時(shí)間x天41036市場(chǎng)價(jià)y元905190(1)根據(jù)上表數(shù)據(jù)，從下列函數(shù)中選取一個(gè)恰當(dāng)?shù)暮瘮?shù)描述該產(chǎn)品的市場(chǎng)價(jià)y與上市時(shí)間x的變化關(guān)系，并簡(jiǎn)要說(shuō)明你選取的理由；①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0(2)利用你選取的函數(shù)，求該產(chǎn)品市場(chǎng)價(jià)最低時(shí)的上市天數(shù)以及最低的價(jià)格；(3)設(shè)你所選取的函數(shù)為SKIPIF1<0，若對(duì)任意實(shí)數(shù)k，關(guān)于x的方程SKIPIF1<0恒有兩個(gè)相異實(shí)數(shù)根，求實(shí)數(shù)m的取值范圍.【答案】(1)SKIPIF1<0(2)該產(chǎn)品上市20天時(shí)市場(chǎng)價(jià)最低，最低的價(jià)格為26元；(3)SKIPIF1<0【解析】【分析】(1)隨著時(shí)間SKIPIF1<0的增加，SKIPIF1<0的值先減后增，結(jié)合函數(shù)的單調(diào)性即可得出結(jié)論；(2)把點(diǎn)SKIPIF1<0代入SKIPIF1<0中，求出函數(shù)的解析式，利用配方法，即可求出該產(chǎn)品市場(chǎng)價(jià)最低時(shí)的上市天數(shù)以及最低的價(jià)格；(3)由(2)結(jié)合題意可得SKIPIF1<0有兩個(gè)相異的實(shí)根，然后由SKIPIF1<0可求出實(shí)數(shù)m的取值范圍.【小問(wèn)1詳解】因?yàn)殡S著時(shí)間SKIPIF1<0的增加，SKIPIF1<0的值先減后增，而所給的函數(shù)中SKIPIF1<0和SKIPIF1<0都是單調(diào)函數(shù)，不滿(mǎn)足題意，所以選擇SKIPIF1<0【小問(wèn)2詳解】把點(diǎn)SKIPIF1<0代入SKIPIF1<0中，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值26，所以當(dāng)該產(chǎn)品上市20天時(shí)市場(chǎng)價(jià)最低，最低的價(jià)格為26元；【小問(wèn)3詳解】由(2)可知SKIPIF1<0，所以由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，因?yàn)榉匠逃袃蓚€(gè)相異實(shí)數(shù)根，所以SKIPIF1<0，所以SKIPIF1<0，因?yàn)閷?duì)任意實(shí)數(shù)k，上式恒成立，所以SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)m的取值范圍為SKIPIF1<0.22.已知函數(shù)SKIPIF1<0(SKIPIF1<0).(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的最小值；(2)若函數(shù)SKIPIF1<0存在兩個(gè)不同的零點(diǎn)SKIPIF1<0與SKIPIF1<0，求SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)由題意可知SKIPIF1<0，對(duì)自變量SKIPIF1<0進(jìn)行分類(lèi)討論，將函數(shù)SKIPIF1<0寫(xiě)成分段函數(shù)形式利用函數(shù)單調(diào)性即可求得函數(shù)SKIPIF1<0的最小值；(2)對(duì)參數(shù)SKIPIF1<0的取值進(jìn)行分類(lèi)討論，利用韋達(dá)定理寫(xiě)出SKIPIF1<0關(guān)于SKIPIF1<0的表達(dá)式，再利用換元法構(gòu)造函數(shù)根據(jù)函數(shù)單調(diào)性即可求得其取值范圍.【小問(wèn)1詳解】解法一：若SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0.故SKIPIF1<