天津市七區(qū)2022-2023學(xué)年高一上學(xué)期期末數(shù)學(xué)試題（含答案詳解）

2022~2023學(xué)年度第一學(xué)期期末練習(xí)高一數(shù)學(xué)第Ⅰ卷（選擇題共40分）一、選擇題：本大題共10小題，每小題4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.已知全集SKIPIF1<0，集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】根據(jù)題意求出SKIPIF1<0，即可求解.【詳解】全集SKIPIF1<0，集合SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0.故選：B2.函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】判斷求解端點(diǎn)的函數(shù)值，利用零點(diǎn)判定定理求解即可．【詳解】解：函數(shù)SKIPIF1<0，因?yàn)镾KIPIF1<0是增函數(shù)，SKIPIF1<0是增函數(shù)，所以函數(shù)SKIPIF1<0是增函數(shù)．SKIPIF1<0．SKIPIF1<0．SKIPIF1<0.SKIPIF1<0．．函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間是：（1，2）．故選：C．【點(diǎn)睛】一是嚴(yán)格把握零點(diǎn)存在性定理的條件；二是連續(xù)函數(shù)在一個(gè)區(qū)間的端點(diǎn)處函數(shù)值異號(hào)是這個(gè)函數(shù)在這個(gè)區(qū)間上存在零點(diǎn)的充分條件，而不是必要條件；三是函數(shù)f(x)在[a，b]上單調(diào)且f(a)f(b)＜0，則f(x)在[a，b]上只有一個(gè)零點(diǎn).3.“SKIPIF1<0”是“SKIPIF1<0”的A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件【答案】A【解析】【分析】由充分條件和必要條件的概念，即可判斷出結(jié)果.【詳解】解:因?yàn)镾KIPIF1<0能推出SKIPIF1<0，而SKIPIF1<0不能推出SKIPIF1<0，所以“SKIPIF1<0”是“SKIPIF1<0”的充分不必要條件,故選:A.【點(diǎn)睛】本題主要考查充分條件、必要條件與充要條件的判斷，屬于基礎(chǔ)題型.4.一個(gè)扇形的面積和弧長的數(shù)值都是2，則這個(gè)扇形中心角的弧度數(shù)為（）A.4 B.3 C.2 D.1【答案】D【解析】【分析】根據(jù)扇形面積和弧長公式計(jì)算即可得出結(jié)果.【詳解】設(shè)扇形中心角的弧度數(shù)為SKIPIF1<0，半徑為SKIPIF1<0，由題意可知，扇形面積SKIPIF1<0，弧長SKIPIF1<0，解得SKIPIF1<0，即扇形中心角的弧度數(shù)為1.故選：D5.已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】根據(jù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的單調(diào)性分別限定a，b，c的取值范圍即可比較出大小.【詳解】由指數(shù)函數(shù)SKIPIF1<0在R上單調(diào)遞增可知，SKIPIF1<0，即SKIPIF1<0；由對(duì)數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增可知，SKIPIF1<0，即SKIPIF1<0;由對(duì)數(shù)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減可知，SKIPIF1<0，即SKIPIF1<0所以，可得SKIPIF1<0故選：C6.把函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長度，得到的圖象所對(duì)應(yīng)的函數(shù)解析式可以是（）A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由題意利用函數(shù)SKIPIF1<0的圖象平移變換規(guī)律，得出結(jié)論．【詳解】將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長度后，所得圖象對(duì)應(yīng)的函數(shù)解析式是SKIPIF1<0，故選：A7.函數(shù)SKIPIF1<0的部分圖象大致為（）A. B.C. D.【答案】A【解析】【分析】首先判斷函數(shù)的奇偶性，再利用特殊值利用排除法判斷即可.【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0定義域?yàn)镾KIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，函數(shù)圖形關(guān)于原點(diǎn)對(duì)稱，故排除C、D，又SKIPIF1<0，故排除B；故選：A8.下列計(jì)算正確的是（）A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】對(duì)選項(xiàng)逐一判斷，A選項(xiàng)注意平方再開方后要取絕對(duì)值，去絕對(duì)值要注意正負(fù)；B選項(xiàng)同底數(shù)冪相乘，底數(shù)不變指數(shù)相加；C選項(xiàng)SKIPIF1<0；D選項(xiàng)由對(duì)數(shù)的運(yùn)算性質(zhì)即可判斷正誤.【詳解】SKIPIF1<0故A錯(cuò)誤；SKIPIF1<0，故B錯(cuò)誤；SKIPIF1<0故C錯(cuò)誤；SKIPIF1<0故D正確.故選：D.9.經(jīng)研究表明，大部分注射藥物的血藥濃度SKIPIF1<0（單位：SKIPIF1<0）隨時(shí)間t（單位：h）的變化規(guī)律可近似表示為SKIPIF1<0，其中SKIPIF1<0表示第一次靜脈注射后人體內(nèi)的初始血藥濃度，k表示該藥物在人體內(nèi)的消除速率常數(shù).已知某麻醉藥的消除速率常數(shù)SKIPIF1<0（單位：SKIPIF1<0），某患者第一次靜脈注射該麻醉藥后即進(jìn)入麻醉狀態(tài)，測(cè)得其血藥濃度為SKIPIF1<0，當(dāng)患者清醒時(shí)測(cè)得其血藥濃度為SKIPIF1<0，則該患者的麻醉時(shí)間約為（SKIPIF1<0）（）A.3.2SKIPIF1<0 B.3.5SKIPIF1<0 C.2.2SKIPIF1<0 D.0.8SKIPIF1<0【答案】A【解析】【分析】依據(jù)題意列出關(guān)于SKIPIF1<0的方程即可求得該患者的麻醉時(shí)間．【詳解】解：由題意得，SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故選：A.10.已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0有四個(gè)不同零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則下列結(jié)論中正確的是（）ASKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】作出函數(shù)SKIPIF1<0圖象，根據(jù)函數(shù)圖象得出4個(gè)零點(diǎn)的關(guān)系及范圍，進(jìn)而得出結(jié)論.【詳解】函數(shù)SKIPIF1<0的四個(gè)不同的零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，就是函數(shù)SKIPIF1<0與SKIPIF1<0兩個(gè)圖象四個(gè)交點(diǎn)的橫坐標(biāo)，作出函數(shù)SKIPIF1<0的圖象，對(duì)于A，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，結(jié)合圖象可知SKIPIF1<0，故A錯(cuò)誤；結(jié)合圖象可知SKIPIF1<0，解得SKIPIF1<0，故B正確；又SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故C錯(cuò)誤；根據(jù)二次函數(shù)的性質(zhì)和圖象得出SKIPIF1<0，所以SKIPIF1<0，故D錯(cuò)誤；故選：B第Ⅱ卷（共80分）二、填空題：本大題共5小題，每小題4分，共20分．11.已知冪函數(shù)y=f（x）的圖象經(jīng)過點(diǎn)（4，2），那么這個(gè)冪函數(shù)的解析式為___________．【答案】SKIPIF1<0【解析】【分析】設(shè)冪函數(shù)SKIPIF1<0，由冪函數(shù)圖象經(jīng)過點(diǎn)SKIPIF1<0，知SKIPIF1<0，由此能求出這個(gè)冪函數(shù)的解析式．【詳解】設(shè)冪函數(shù)SKIPIF1<0，∵冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，∴這個(gè)冪函數(shù)的解析式為SKIPIF1<0．故答案為：SKIPIF1<0．12SKIPIF1<0______．【答案】SKIPIF1<0【解析】【分析】根據(jù)誘導(dǎo)公式求解即可.【詳解】SKIPIF1<0.故答案為：SKIPIF1<0.13.已知SKIPIF1<0，則SKIPIF1<0____________.【答案】SKIPIF1<0【解析】【分析】SKIPIF1<0分子分母同除以SKIPIF1<0，將SKIPIF1<0代入即可得結(jié)果.【詳解】SKIPIF1<0SKIPIF1<0.【點(diǎn)睛】本題主要考查，同角三角函數(shù)之間的關(guān)系的應(yīng)用，屬于中檔題.同角三角函數(shù)之間的關(guān)系包含平方關(guān)系與商的關(guān)系，平方關(guān)系是正弦與余弦值之間的轉(zhuǎn)換，商的關(guān)系是正余弦與正切之間的轉(zhuǎn)換.14.若SKIPIF1<0，則SKIPIF1<0的最小值為______．【答案】SKIPIF1<0【解析】【分析】由于SKIPIF1<0，可將原式整理為SKIPIF1<0，然后利用基本不等式求解即可.【詳解】SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)，取得最小值.故答案為：SKIPIF1<0.15.有下列命題：①函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0；②不等式SKIPIF1<0的解集為SKIPIF1<0，則實(shí)數(shù)k的取值范圍為SKIPIF1<0；③函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．則當(dāng)x＜0時(shí)，SKIPIF1<0．其中正確命題的序號(hào)為______（把正確的答案都填上）．【答案】①③【解析】【分析】對(duì)①②③逐一判斷，①函數(shù)的定義域要滿足分母不為0，對(duì)數(shù)函數(shù)的真數(shù)大于0，②對(duì)不等式的二次項(xiàng)系數(shù)分類討論，分別求的滿足條件的集合，即可求得實(shí)數(shù)k的取值范圍，③有函數(shù)的奇偶性可知SKIPIF1<0，又知當(dāng)當(dāng)SKIPIF1<0時(shí)函數(shù)的解析式，即可求得當(dāng)SKIPIF1<0時(shí)函數(shù)的解析式.【詳解】對(duì)于①函數(shù)SKIPIF1<0定義域滿足SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0，①正確；對(duì)于②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0滿足題意，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0開口向上，SKIPIF1<0解集為SKIPIF1<0不成立，當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0解集為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜上若不等式SKIPIF1<0的解集為SKIPIF1<0，則數(shù)k的取值范圍為SKIPIF1<0，故②錯(cuò)誤；對(duì)于③SKIPIF1<0函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，SKIPIF1<0又SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)x＜0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0即當(dāng)x＜0時(shí)，SKIPIF1<0故③正確.故答案為：①③.三、解答題：本大題共5小題，共60分．解答應(yīng)寫出文字說明、證明過程或演算步驟．16.已知SKIPIF1<0，SKIPIF1<0是第三象限的角．（1）求SKIPIF1<0；（2）求SKIPIF1<0的值．【答案】（1）SKIPIF1<0（2）SKIPIF1<0【解析】【分析】（1）先求出SKIPIF1<0，后由SKIPIF1<0可得答案；（2）由（1）可得SKIPIF1<0，后由兩角差的正弦公式可得答案.【小問1詳解】∵SKIPIF1<0，且SKIPIF1<0是第三象限的角，∴SKIPIF1<0，∴SKIPIF1<0；【小問2詳解】SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.17.已知函數(shù)SKIPIF1<0（1）求SKIPIF1<0，SKIPIF1<0的值；（2）若SKIPIF1<0，求實(shí)數(shù)a的值；（3）直接寫出SKIPIF1<0的單調(diào)區(qū)間．【答案】（1）SKIPIF1<0；SKIPIF1<0（2）SKIPIF1<0（3）單調(diào)遞增區(qū)間SKIPIF1<0，單調(diào)遞減區(qū)間SKIPIF1<0，SKIPIF1<0【解析】【分析】（1）根據(jù)分段函數(shù)定義直接代入計(jì)算即可；（2）分類討論實(shí)數(shù)a的取值范圍，解方程即可得出符合題意的a的值；（3）畫出函數(shù)圖象即可直接寫出單調(diào)區(qū)間．【小問1詳解】根據(jù)分段函數(shù)解析式可得SKIPIF1<0，易知SKIPIF1<0；所以SKIPIF1<0即SKIPIF1<0.【小問2詳解】①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0（舍）.②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0（舍）.綜上可得SKIPIF1<0.即實(shí)數(shù)a的值為SKIPIF1<0【小問3詳解】畫出函數(shù)圖象如下所示：所以，單調(diào)遞增區(qū)間SKIPIF1<0，單調(diào)遞減區(qū)間SKIPIF1<0，SKIPIF1<018.已知指數(shù)函數(shù)SKIPIF1<0（a＞0，且SKIPIF1<0）的圖象過點(diǎn)SKIPIF1<0．（1）求a的值；（2）若SKIPIF1<0，SKIPIF1<0，求m＋n的值；（3）求不等式SKIPIF1<0的解集．【答案】（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0【解析】【分析】（1）由于函數(shù)過點(diǎn)SKIPIF1<0，將點(diǎn)代入函數(shù)解析式即可求得a的值.（2）將SKIPIF1<0，SKIPIF1<0分別代入函數(shù)中，分別求得SKIPIF1<0，再用對(duì)數(shù)的運(yùn)算性質(zhì)求得SKIPIF1<0的值。（3）將SKIPIF1<0中的SKIPIF1<0代換成SKIPIF1<0，再由函數(shù)的單調(diào)性即可求得不等式SKIPIF1<0的解集.【小問1詳解】函數(shù)SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)的圖象過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.又SKIPIF1<0，故a的值為SKIPIF1<0.【小問2詳解】由（1）知，SKIPIF1<0因?yàn)镾KIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，SKIPIF1<0SKIPIF1<0【小問3詳解】不等式SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)椋琒KIPIF1<0在SKIPIF1<0上單調(diào)遞減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，所以不等式的解集為SKIPIF1<0.19.已知函數(shù)SKIPIF1<0，SKIPIF1<0．（1）求SKIPIF1<0的最小正周期；（2）求SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值；（3）若SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值．【答案】（1）SKIPIF1<0（2）最大值為SKIPIF1<0，最小值為－SKIPIF1<0.（3）SKIPIF1<0【解析】【分析】(1)根據(jù)兩角和差，二倍角公式及輔助角公式化簡函數(shù)解析式，求周期即可；(2)根據(jù)自變量范圍求，結(jié)合單調(diào)性求最值；(3)由已知條件結(jié)合兩角和差公式求值.【小問1詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0的最小正周期為SKIPIF1<0.【小問2詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0的最大值為SKIPIF1<0，最小值為－SKIPIF1<0.【小問3詳解】因?yàn)?，SKIPIF1<0，所以，SKIPIF1<0，又因?yàn)镾KIPIF1<0所以，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0SKIPIF1<0SKIPIF1<0.20.已知函數(shù)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，且SKIPIF1<0．（1）求SKIPIF1<0的解析式；（2）用函數(shù)單調(diào)性定義證明SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)