河北省xx中學(xué)2022-2023學(xué)年高一上學(xué)期綜合素質(zhì)檢測(cè)二數(shù)學(xué)試題（含解析）

高一上學(xué)期期中數(shù)學(xué)試題第I卷(選擇題共60分)一?單項(xiàng)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.下列函數(shù)中，其定義域和值域分別與函數(shù)SKIPIF1<0的定義域和值域相同的是()A.y=x B.y=lnx C.y=SKIPIF1<0 D.y=SKIPIF1<0【答案】D【解析】【分析】分別求出各個(gè)函數(shù)的定義域和值域，比較后可得答案．【詳解】解：函數(shù)SKIPIF1<0的定義域和值域均為SKIPIF1<0，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，不滿足要求；函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，不滿足要求；函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0，不滿足要求；函數(shù)SKIPIF1<0的定義域和值域均為SKIPIF1<0，滿足要求；故選：SKIPIF1<0．【點(diǎn)睛】本題考查的知識(shí)點(diǎn)是函數(shù)的定義域和值域，熟練掌握各種基本初等函數(shù)的定義域和值域，是解答的關(guān)鍵．2.已知SKIPIF1<0，則A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【詳解】因?yàn)镾KIPIF1<0，且冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以b<a<c.故選A.點(diǎn)睛：本題主要考查冪函數(shù)的單調(diào)性及比較大小問(wèn)題，解答比較大小問(wèn)題，常見(jiàn)思路有兩個(gè)：一是判斷出各個(gè)數(shù)值所在區(qū)間(一般是看三個(gè)區(qū)間SKIPIF1<0)；二是利用函數(shù)的單調(diào)性直接解答；數(shù)值比較多的比大小問(wèn)題也可以?xún)煞N方法綜合應(yīng)用；三是借助于中間變量比較大小.3.已知SKIPIF1<0，則SKIPIF1<0的值是A.SKIPIF1<0 B.SKIPIF1<0CSKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】由題意結(jié)合根式的運(yùn)算法則整理計(jì)算即可求得最終結(jié)果.【詳解】由題意知SKIPIF1<0，SKIPIF1<0，由于SKIPIF1<0，故SKIPIF1<0，則原式SKIPIF1<0.故選B.【點(diǎn)睛】本題主要考查根式的運(yùn)算法則及其應(yīng)用，屬于中等題.4.區(qū)塊鏈作為一種新型的技術(shù)，已經(jīng)被應(yīng)用于許多領(lǐng)域.在區(qū)塊鏈技術(shù)中，某個(gè)密碼的長(zhǎng)度設(shè)定為512B，則密碼一共有SKIPIF1<0種可能，為了破解該密碼，最壞的情況需要進(jìn)行SKIPIF1<0次運(yùn)算.現(xiàn)在有一臺(tái)計(jì)算機(jī)，每秒能進(jìn)行SKIPIF1<0次運(yùn)算，那么在最壞的情況下，這臺(tái)計(jì)算機(jī)破譯該密碼所需時(shí)間大約為()(參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)題意所求時(shí)間為SKIPIF1<0，利用對(duì)數(shù)的運(yùn)算進(jìn)行求解即可.【詳解】設(shè)在最壞的情況下，這臺(tái)計(jì)算機(jī)破譯該密碼所需時(shí)間為SKIPIF1<0秒，則有SKIPIF1<0；兩邊取常用對(duì)數(shù)，得SKIPIF1<0；SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0；所以SKIPIF1<0.故選：D.5.設(shè)SKIPIF1<0，SKIPIF1<0，則下列命題正確的是()．A.若SKIPIF1<0，則SKIPIF1<0 B.若SKIPIF1<0，則SKIPIF1<0C.若SKIPIF1<0，則SKIPIF1<0 D.若SKIPIF1<0，則SKIPIF1<0【答案】D【解析】【分析】列舉特殊數(shù)值，排除選項(xiàng).【詳解】A.SKIPIF1<0時(shí)，SKIPIF1<0，故A不成立；B.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故B不成立；C.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故C不成立；D.若SKIPIF1<0，根據(jù)函數(shù)SKIPIF1<0在SKIPIF1<0的單調(diào)性可知，SKIPIF1<0成立，故D正確.故選：D6.已知函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)，SKIPIF1<0是其圖象上的兩點(diǎn)，那么SKIPIF1<0的解集是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】不等式轉(zhuǎn)化為SKIPIF1<0，根據(jù)函數(shù)的單調(diào)性得到答案.【詳解】SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)，故SKIPIF1<0，解得SKIPIF1<0.故選：A7.已知SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，滿足SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.0 C.2 D.50【答案】C【解析】【分析】利用奇函數(shù)的性質(zhì)及SKIPIF1<0，推出函數(shù)SKIPIF1<0的周期為4，然后得出SKIPIF1<0得出結(jié)果．【詳解】由函數(shù)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0是周期函數(shù)，且周期為4，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0故選：C8.已知函數(shù)SKIPIF1<0，SKIPIF1<0，則圖象如圖的函數(shù)可能是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】結(jié)合函數(shù)圖像的奇偶性和單調(diào)性即可判斷.【詳解】由圖可知，該函數(shù)為奇函數(shù)，SKIPIF1<0和SKIPIF1<0為非奇非偶函數(shù)，故A、B不符；當(dāng)x＞0時(shí)，SKIPIF1<0單調(diào)遞增，與圖像不符，故C不符；SKIPIF1<0為奇函數(shù)，當(dāng)x→＋時(shí)，∵y＝SKIPIF1<0的增長(zhǎng)速度快于y＝lnx的增長(zhǎng)速度，故SKIPIF1<0＞0且單調(diào)遞減，故圖像應(yīng)該在x軸上方且無(wú)限靠近x軸，與圖像相符.故選：D.二?多項(xiàng)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)是符合題目要求.全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)的得0分.)9.下面說(shuō)法中，錯(cuò)誤的是()A.“SKIPIF1<0中至少有一個(gè)小于零”是“SKIPIF1<0”的充要條件；B.“SKIPIF1<0”是“SKIPIF1<0且SKIPIF1<0”的充要條件；C.“SKIPIF1<0”是“SKIPIF1<0或SKIPIF1<0”的充要條件；D.若集合SKIPIF1<0是全集SKIPIF1<0的子集，則命題“SKIPIF1<0”與“SKIPIF1<0”是等價(jià)命題.【答案】AC【解析】【分析】從充分性和必要性的角度，結(jié)合題意，對(duì)選項(xiàng)進(jìn)行逐一判斷即可.【詳解】對(duì)SKIPIF1<0：若SKIPIF1<0，滿足SKIPIF1<0中至少有一個(gè)小于零，但無(wú)法推出SKIPIF1<0，故SKIPIF1<0錯(cuò)誤；對(duì)SKIPIF1<0：若SKIPIF1<0，則只能是SKIPIF1<0；若SKIPIF1<0，則一定有SKIPIF1<0，故“SKIPIF1<0”是“SKIPIF1<0且SKIPIF1<0”的充要條件，則SKIPIF1<0正確；對(duì)SKIPIF1<0：若SKIPIF1<0且SKIPIF1<0，是SKIPIF1<0的充分非必要條件，又因?yàn)槿鬝KIPIF1<0，則SKIPIF1<0或SKIPIF1<0，是命題：若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的逆否命題，故其真假一致，則SKIPIF1<0，是SKIPIF1<0或SKIPIF1<0的充分非必要條件，故SKIPIF1<0錯(cuò)誤；對(duì)SKIPIF1<0：因?yàn)榧蟂KIPIF1<0是全集SKIPIF1<0的子集，故可得SKIPIF1<0，故命題“SKIPIF1<0”與“SKIPIF1<0”是等價(jià)命題，則SKIPIF1<0正確.綜上所述：A、C錯(cuò)誤.故選：AC.【點(diǎn)睛】本題考查充分條件和必要條件的判定，注意細(xì)節(jié)處理即可.10.已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】BC【解析】分析】利用給定條件結(jié)合基本不等式判斷A，C；利用二次函數(shù)性質(zhì)判斷B；取特值判斷D作答.【詳解】因SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則有SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”，A不正確；因SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”，B正確；因SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”，C正確；因SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則取SKIPIF1<0，即有SKIPIF1<0，于是得SKIPIF1<0，D不正確.故選：BC11.已知函數(shù)SKIPIF1<0，下列關(guān)于函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)的說(shuō)法中，正確的是()A.當(dāng)SKIPIF1<0，有1個(gè)零點(diǎn) B.當(dāng)SKIPIF1<0時(shí)，有3個(gè)零點(diǎn)C.當(dāng)SKIPIF1<0，有4個(gè)零點(diǎn) D.當(dāng)SKIPIF1<0時(shí)，有7個(gè)零點(diǎn)【答案】ABD【解析】【分析】令SKIPIF1<0得SKIPIF1<0，利用換元法將函數(shù)分解為SKIPIF1<0和SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象，利用數(shù)形結(jié)合即可得到結(jié)論．【詳解】令SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，則方程SKIPIF1<0等價(jià)為SKIPIF1<0，函數(shù)SKIPIF1<0，開(kāi)口向上，過(guò)點(diǎn)SKIPIF1<0，對(duì)稱(chēng)軸為SKIPIF1<0對(duì)于A，當(dāng)SKIPIF1<0時(shí)，作出函數(shù)SKIPIF1<0的圖象：SKIPIF1<0，此時(shí)方程SKIPIF1<0有一個(gè)根SKIPIF1<0，由SKIPIF1<0可知，此時(shí)x只有一解，即函數(shù)SKIPIF1<0有1個(gè)零點(diǎn)，故A正確；對(duì)于B，當(dāng)SKIPIF1<0時(shí)，作出函數(shù)SKIPIF1<0的圖象：SKIPIF1<0，此時(shí)方程SKIPIF1<0有一個(gè)根SKIPIF1<0，由SKIPIF1<0可知，此時(shí)x有3個(gè)解，即函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，故B正確；對(duì)于C，當(dāng)SKIPIF1<0時(shí)，圖像如A，故只有1個(gè)零點(diǎn)，故C錯(cuò)誤；對(duì)于D，當(dāng)SKIPIF1<0時(shí)，作出函數(shù)SKIPIF1<0的圖象：SKIPIF1<0，此時(shí)方程SKIPIF1<0有3個(gè)根，其中SKIPIF1<0，SKIPIF1<0，SKIPIF1<0由SKIPIF1<0可知，此時(shí)x有3個(gè)解，由SKIPIF1<0，此時(shí)x有3個(gè)解，由SKIPIF1<0，此時(shí)x有1個(gè)解，即函數(shù)SKIPIF1<0有7個(gè)零點(diǎn)，故D正確；故選：ABD．【點(diǎn)睛】方法點(diǎn)睛：本題考查分段函數(shù)的應(yīng)用，考查復(fù)合函數(shù)的零點(diǎn)的判斷，利用換元法和數(shù)形結(jié)合是解決本題的關(guān)鍵，已知函數(shù)有零點(diǎn)(方程有根)求參數(shù)值(取值范圍)常用的方法：(1)直接法：直接求解方程得到方程的根，再通過(guò)解不等式確定參數(shù)范圍；(2)分離參數(shù)法：先將參數(shù)分離，轉(zhuǎn)化成求函數(shù)的值域問(wèn)題加以解決；(3)數(shù)形結(jié)合法：先對(duì)解析式變形，進(jìn)而構(gòu)造兩個(gè)函數(shù)，然后在同一平面直角坐標(biāo)系中畫(huà)出函數(shù)的圖象，利用數(shù)形結(jié)合的方法求解，屬于難題．12.定義“正對(duì)數(shù)”：SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則下列結(jié)論中正確的是.A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】AD【解析】【分析】根據(jù)所給的定義及對(duì)數(shù)的運(yùn)算性質(zhì)對(duì)四個(gè)命題進(jìn)行判斷，由于在不同的定義域中函數(shù)的解析式不一樣，故需要對(duì)SKIPIF1<0進(jìn)行分類(lèi)討論，判斷出每個(gè)命題的真假.【詳解】對(duì)A，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．所以當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，故A正確．對(duì)B，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)滿足SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B錯(cuò)誤；對(duì)C，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0，故C錯(cuò)誤；對(duì)D，由“正對(duì)數(shù)”的定義知，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．綜上所述，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，故D正確．故選AD．【點(diǎn)睛】本題考查新定義及對(duì)數(shù)的運(yùn)算性質(zhì)，理解定義所給的運(yùn)算規(guī)則是解題的關(guān)鍵，考查分類(lèi)討論思想、轉(zhuǎn)化與化歸思想的靈活運(yùn)用，考查運(yùn)算求解能力，注意本題容易因?yàn)槔斫獠磺宥x及忘記分類(lèi)論論的方法使解題無(wú)法入手致錯(cuò).第II卷(共90分)三?填空題(本題共4小題，每小題5分，共20分；)13.計(jì)算SKIPIF1<0____________【答案】5【解析】【分析】由分?jǐn)?shù)指數(shù)冪的運(yùn)算及對(duì)數(shù)的運(yùn)算即可得解.【詳解】解：原式SKIPIF1<0，故答案為：5.【點(diǎn)睛】本題考查了分?jǐn)?shù)指數(shù)冪的運(yùn)算及對(duì)數(shù)的運(yùn)算，屬基礎(chǔ)題.14.設(shè)函數(shù)SKIPIF1<0，則使得SKIPIF1<0成立的SKIPIF1<0的取值范圍是_______________.【答案】SKIPIF1<0【解析】【詳解】試題分析：當(dāng)時(shí)，，∴，∴；當(dāng)時(shí)，，∴，∴，綜上，使得SKIPIF1<0成立的的取值范圍是．故答案為．考點(diǎn)：分段函數(shù)不等式及其解法.【方法點(diǎn)晴】本題考查不等式的解法，在分段函數(shù)中結(jié)合指數(shù)函數(shù)不等式與冪函數(shù)不等式，考查學(xué)生的計(jì)算能力，屬于基礎(chǔ)題．利用分段函數(shù)，結(jié)合SKIPIF1<0分為兩段當(dāng)時(shí)，根據(jù)單調(diào)性，解指數(shù)函數(shù)不等式，取交集；當(dāng)時(shí)，解冪函數(shù)不等式，取交集，綜合取上述兩者的并集，即可求出使得SKIPIF1<0成立的的取值范圍．15.已知函數(shù)SKIPIF1<0定義域?yàn)镾KIPIF1<0，且對(duì)于任意SKIPIF1<0，都有SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)________.【答案】SKIPIF1<0【解析】【分析】根據(jù)不等式的結(jié)構(gòu)構(gòu)新函數(shù)，利用新函數(shù)的單調(diào)性進(jìn)行求解即可.【詳解】任意SKIPIF1<0，不妨設(shè)SKIPIF1<0，由SKIPIF1<0，構(gòu)造新函數(shù)SKIPIF1<0，由SKIPIF1<0，所以函數(shù)SKIPIF1<0增函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0，故答案為：SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)睛：根據(jù)不等式形式構(gòu)造新函數(shù)進(jìn)而判斷新函數(shù)的單調(diào)性是解題的關(guān)鍵.16.對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0_________.【答案】SKIPIF1<0【解析】【分析】由對(duì)數(shù)有意義可得：SKIPIF1<0，將不等式SKIPIF1<0等價(jià)轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上恒成立，構(gòu)造函數(shù)SKIPIF1<0，由函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，再根據(jù)二次函數(shù)的圖象和性質(zhì)即可求出實(shí)數(shù)SKIPIF1<0的值，最后取交集即可求解.【詳解】由題意可知：SKIPIF1<0且SKIPIF1<0成立，則SKIPIF1<0，因?yàn)閷?duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，也即SKIPIF1<0在SKIPIF1<0上恒成立，記SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0恒成立，則SKIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式顯然成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0恒成立，則SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0，因?yàn)閷?duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，則綜上可知：實(shí)數(shù)SKIPIF1<0的值為SKIPIF1<0.故答案為：SKIPIF1<0.四?解答題：(本題共6小題，共70分，解答應(yīng)寫(xiě)出文字說(shuō)明?證明過(guò)程或演算步驟.)17.為了預(yù)防新型冠狀病毒，唐徠回民中學(xué)對(duì)教室進(jìn)行藥熏消毒，室內(nèi)每立方米空氣中的含藥量y(單位：毫克)隨時(shí)間x(單位：h)的變化情況如圖所示，在藥物釋放過(guò)程中，y與x成正比，藥物釋放完畢后，y與x的函數(shù)關(guān)系式為SKIPIF1<0(a為常數(shù))，根據(jù)圖中提供的信息，回答下列問(wèn)題：(1)寫(xiě)出從藥物釋放開(kāi)始，y與x的之間的函數(shù)關(guān)系；(2)據(jù)測(cè)定，當(dāng)空氣中每立方米的含藥量降低至0.25毫克以下時(shí)，學(xué)生方可進(jìn)入教室，那么從藥物釋放開(kāi)始，至少需要經(jīng)過(guò)多少小時(shí)后，學(xué)生才能回到教室．【答案】(1)SKIPIF1<0(2)0.6【解析】【分析】(1)利用函數(shù)圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0，分段討論即可得出結(jié)論；(2)利用指數(shù)函數(shù)的單調(diào)性解不等式SKIPIF1<0．【小問(wèn)1詳解】解：依題意，當(dāng)SKIPIF1<0時(shí)，可設(shè)SKIPIF1<0，且SKIPIF1<0，解得SKIPIF1<0又由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0；【小問(wèn)2詳解】解：令SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，即至少需要經(jīng)過(guò)SKIPIF1<0后，學(xué)生才能回到教室．18.已知函數(shù)SKIPIF1<0，其中SKIPIF1<0為常數(shù)且滿足SKIPIF1<0．(1)求SKIPIF1<0的值；(2)證明函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，并判斷SKIPIF1<0在SKIPIF1<0上的單調(diào)性；(3)若對(duì)任意的SKIPIF1<0，總有SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)詳見(jiàn)解析(3)SKIPIF1<0【解析】分析】(1)根據(jù)條件列方程組求解(2)由單調(diào)性的定義證明(3)不等式恒成立，轉(zhuǎn)化為最值問(wèn)題【詳解】(1)由SKIPIF1<0解得SKIPIF1<0(2)由(1)得SKIPIF1<0任取SKIPIF1<0，則SKIPIF1<0SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0故函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)同理若SKIPIF1<0，則SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增(3)由題意SKIPIF1<0由(2)知SKIPIF1<0故SKIPIF1<0的取值范圍是SKIPIF1<019.已知函數(shù)SKIPIF1<0是偶函數(shù)(1)求實(shí)數(shù)SKIPIF1<0的值；(2)設(shè)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有公共點(diǎn)，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)根據(jù)函數(shù)SKIPIF1<0解析式以及偶函數(shù)的定義可求得實(shí)數(shù)SKIPIF1<0的值；(2)利用函數(shù)與方程的思想，把函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有公共點(diǎn)的問(wèn)題轉(zhuǎn)化成方程有解的問(wèn)題，進(jìn)而求得參數(shù)SKIPIF1<0的取值范圍.【小問(wèn)1詳解】由函數(shù)SKIPIF1<0，得SKIPIF1<0，又因?yàn)镾KIPIF1<0是偶函數(shù)，所以滿足SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0對(duì)于一切SKIPIF1<0恒成立，所以SKIPIF1<0，故SKIPIF1<0；【小問(wèn)2詳解】由SKIPIF1<0得SKIPIF1<0若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有公共點(diǎn)，等價(jià)于方程SKIPIF1<0有解，即SKIPIF1<0，所以SKIPIF1<0，即方程SKIPIF1<0在SKIPIF1<0上有解，由指數(shù)函數(shù)值域可知，SKIPIF1<0，所以SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.20.已知函數(shù)SKIPIF1<0，且SKIPIF1<0.(1)若函數(shù)SKIPIF1<0的圖像與函數(shù)SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，且點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖像上，求實(shí)數(shù)SKIPIF1<0的值；(2)已知SKIPIF1<0，函數(shù)SKIPIF1<0.若SKIPIF1<0的最大值為8，求實(shí)數(shù)SKIPIF1<0的值.【答案】(1)4

(2)2

【解析】【分析】(1)先求出SKIPIF1<0，將P點(diǎn)坐標(biāo)代入即可求出a；(2)將SKIPIF1<0轉(zhuǎn)化為二次函數(shù)，根據(jù)條件即可算出a.【小問(wèn)1詳解】由題意SKIPIF1<0，將SKIPIF1<0代入得：SKIPIF1<0；【小問(wèn)2詳解】SKIPIF1<0，其中SKIPIF1<0，令SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0是關(guān)于t的開(kāi)口向上，對(duì)稱(chēng)軸為SKIPIF1<0的拋物線，SKIPIF1<0，并且SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，又SKIPIF1<0SKIPIF1<0；21.已知函數(shù)SKIPIF1<0為自然對(duì)數(shù)的底數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，判斷函數(shù)SKIPIF1<0零點(diǎn)個(gè)數(shù)，并證明你的結(jié)論；(2)當(dāng)SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍【答案】(1)一個(gè)，證明見(jiàn)解析；(2)SKIPIF1<0.【解析】【分析】(1)根據(jù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的單調(diào)性得到SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，再利用零點(diǎn)存在性定理和SKIPIF1<0，SKIPIF1<0，即可得到零點(diǎn)的個(gè)數(shù)；(2)將SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，轉(zhuǎn)化為SKIPIF1<0，然后根據(jù)單調(diào)性求最小值得到SKIPIF1<0，根據(jù)函數(shù)SKIPIF1<0的單調(diào)性和特殊值SKIPIF1<0解不等式即可.【小問(wèn)1詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上存在零點(diǎn)，且只有一個(gè)零點(diǎn)，所以SKIPIF1<0只有一個(gè)零點(diǎn).【小問(wèn)2詳解】由題意得，當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0恒成立，即SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，令SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0單調(diào)遞增，