高一上學(xué)期期末數(shù)學(xué)試題時(shí)量：120分鐘；分值：150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)符合題目要求．1.命題SKIPIF1<0：SKIPIF1<0，SKIPIF1<0的否定形式SKIPIF1<0為(

)A.SKIPIF1<0，SKIPIF1<0 B.SKIPIF1<0，SKIPIF1<0C.SKIPIF1<0，SKIPIF1<0 D.SKIPIF1<0，SKIPIF1<0【答案】D【解析】【分析】“任意一個(gè)都符合”的否定為“存在一個(gè)不符合”【詳解】由題意，“任意一個(gè)都符合”的否定為“存在一個(gè)不符合”，故SKIPIF1<0為SKIPIF1<0，SKIPIF1<0.故選：D2.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】A【解析】【分析】解不等式確定集合SKIPIF1<0后再求交集即可．【詳解】由題意SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故選：A．3.設(shè)SKIPIF1<0，則“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】【詳解】SKIPIF1<0SKIPIF1<0，但SKIPIF1<0，不滿足SKIPIF1<0，所以充分不必要條件，選A.【考點(diǎn)】充要條件【名師點(diǎn)睛】本題考查充要條件的判斷，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的充分條件，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的必要條件，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的充要條件；從集合的角度看，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的充分條件，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的必要條件，若SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的充要條件，若SKIPIF1<0是SKIPIF1<0的真子集，則SKIPIF1<0是SKIPIF1<0的充分不必要條件，若SKIPIF1<0是SKIPIF1<0的真子集，則SKIPIF1<0是SKIPIF1<0的必要不充分條件.4.SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】利用誘導(dǎo)公式化簡(jiǎn)可得結(jié)果.【詳解】SKIPIF1<0.故選：A.5.設(shè)SKIPIF1<0，則SKIPIF1<0的大小關(guān)系是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】易得SKIPIF1<0，再由SKIPIF1<0，利用冪函數(shù)的單調(diào)性判斷.【詳解】因?yàn)镾KIPIF1<0，且SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0，即SKIPIF1<0，綜上：SKIPIF1<0故選：A6.已知SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】利用誘導(dǎo)公式可得SKIPIF1<0，再由二倍角余弦公式求SKIPIF1<0.【詳解】由SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0.故選：D7.流行病學(xué)基本參數(shù)：基本再生數(shù)SKIPIF1<0指一個(gè)感染者傳染的平均人數(shù)，世代間隔T指相鄰兩代間傳染所需的平均時(shí)間．在新冠肺炎疫情初始階段，可用模型：SKIPIF1<0(其中SKIPIF1<0是開始確診病例數(shù))描述累計(jì)感染病例SKIPIF1<0隨時(shí)間t(單位：天)的變化規(guī)律，指數(shù)增長(zhǎng)率r與SKIPIF1<0，T滿足SKIPIF1<0，有學(xué)者估計(jì)出SKIPIF1<0．據(jù)此，在新冠肺炎疫情初始階段，當(dāng)SKIPIF1<0時(shí)，t的值為(SKIPIF1<0)()A.1.2 B.1.7 C.2.0 D.2.5【答案】B【解析】【分析】根據(jù)所給模型求得SKIPIF1<0，代入已知模型，再由SKIPIF1<0，得SKIPIF1<0，求解SKIPIF1<0值得答案【詳解】解：把SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，兩邊取對(duì)數(shù)得，SKIPIF1<0，得SKIPIF1<0，故選：B【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：此題考查函數(shù)模型的實(shí)際應(yīng)用，考查計(jì)算能力，解題的關(guān)鍵是準(zhǔn)確理解題意，弄清函數(shù)模型中各個(gè)量的關(guān)系，屬于中檔題8.若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)，且在SKIPIF1<0上存在最值，則SKIPIF1<0的取值范圍是()．A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】利用三角函數(shù)的單調(diào)性與周期性的關(guān)系及周期公式，結(jié)合三角函數(shù)的最值即可求解.【詳解】因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)，所以SKIPIF1<0，即SKIPIF1<0,則SKIPIF1<0，由此可得SKIPIF1<0．因?yàn)楫?dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，函數(shù)取得最值，欲滿足在SKIPIF1<0上存在極最點(diǎn)，因?yàn)橹芷赟KIPIF1<0，故在SKIPIF1<0上有且只有一個(gè)最值，故第一個(gè)最值點(diǎn)SKIPIF1<0，得SKIPIF1<0，又第二個(gè)最值點(diǎn)SKIPIF1<0，要使SKIPIF1<0在SKIPIF1<0上單調(diào)，必須SKIPIF1<0，得SKIPIF1<0．綜上可得，SKIPIF1<0的取值范圍是SKIPIF1<0．故選：B．二、多項(xiàng)選擇題：本題共4小題，每小題5分，共20分．在每小題給出的四個(gè)選項(xiàng)中，有多項(xiàng)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.下列命題為真命題的是()A.不論SKIPIF1<0取何實(shí)數(shù)，命題SKIPIF1<0“SKIPIF1<0”為真命題B.不論SKIPIF1<0取何實(shí)數(shù)，命題SKIPIF1<0：“二次函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱”為真命題C.“四邊形SKIPIF1<0的對(duì)角線垂直且相等”是“四邊形SKIPIF1<0是正方形”的充分不必要條件D.“SKIPIF1<0”是“SKIPIF1<0”的既不充分也不必要條件【答案】ABD【解析】【分析】結(jié)合一元二次函數(shù)和一元二次不等式的性質(zhì)可判斷AB；根據(jù)充分條件、必要條件的概念可判斷CD.【詳解】對(duì)于SKIPIF1<0，關(guān)于SKIPIF1<0的一元二次方程SKIPIF1<0滿足SKIPIF1<0，即有不等實(shí)根SKIPIF1<0，顯然SKIPIF1<0，即SKIPIF1<0，因此不等式SKIPIF1<0的解集為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故A正確.對(duì)于SKIPIF1<0，二次函數(shù)SKIPIF1<0圖象的對(duì)稱軸為直線SKIPIF1<0，即SKIPIF1<0軸，故B正確.對(duì)于SKIPIF1<0，對(duì)角線垂直且相等的四邊形不一定是正方形可能為菱形，反之成立.故SKIPIF1<0錯(cuò)誤.對(duì)于SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即充分性不成立，令SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，故必要性也不成立，即“SKIPIF1<0”是“SKIPIF1<0”的既不充分也不必要條件，故D正確.故選：ABD.10.已知SKIPIF1<0，則下列結(jié)論正確的有()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【解析】【分析】根據(jù)同角三角函數(shù)的平方關(guān)系可求出SKIPIF1<0的值，根據(jù)角的范圍得出角，進(jìn)而求解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，也即SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，故選：SKIPIF1<0.11.對(duì)于函數(shù)SKIPIF1<0，下列說法正確的是()A.最小正周期為SKIPIF1<0 B.其圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱C.對(duì)稱軸方程SKIPIF1<0 D.單調(diào)增區(qū)間SKIPIF1<0【答案】AC【解析】【分析】利用余弦型函數(shù)的周期公式可判斷A選項(xiàng)；利用余弦型函數(shù)的對(duì)稱新可判斷BC選項(xiàng)；利用余弦型函數(shù)的單調(diào)性可判斷D選項(xiàng).【詳解】對(duì)于A選項(xiàng)，函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0，A對(duì)；對(duì)于B選項(xiàng)，SKIPIF1<0，B錯(cuò)；對(duì)于C選項(xiàng)，由SKIPIF1<0，可得SKIPIF1<0，即函數(shù)SKIPIF1<0的對(duì)稱軸方程為SKIPIF1<0，C對(duì)；對(duì)于D選項(xiàng)，由SKIPIF1<0，解得SKIPIF1<0，所以，函數(shù)SKIPIF1<0的單調(diào)增區(qū)間SKIPIF1<0，D錯(cuò).故選：AC.12.已知函數(shù)SKIPIF1<0則以下判斷正確的是()A.若函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0B.函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C.直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)公共點(diǎn)D.函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0有且只有一個(gè)公共點(diǎn)【答案】AC【解析】【分析】作出SKIPIF1<0的圖像如圖所示，B可直接由圖像或二次函數(shù)單調(diào)性判斷；AC零點(diǎn)及交點(diǎn)問題均可以通過SKIPIF1<0與SKIPIF1<0交點(diǎn)個(gè)數(shù)判斷；D通過圖像或者聯(lián)立方程求解即可判斷.【詳解】當(dāng)SKIPIF1<0SKIPIF1<0，故SKIPIF1<0的圖像如圖所示，對(duì)AC，函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)，相當(dāng)于SKIPIF1<0與SKIPIF1<0有3個(gè)交點(diǎn)，故SKIPIF1<0的取值范圍是SKIPIF1<0，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)公共點(diǎn)，AC對(duì)；對(duì)B，函數(shù)SKIPIF1<0在SKIPIF1<0上先增后減，B錯(cuò)；對(duì)D，如圖所示，聯(lián)立SKIPIF1<0可得解得SKIPIF1<0或SKIPIF1<0，由圖右側(cè)一定有一個(gè)交點(diǎn)，故函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0不止一個(gè)公共點(diǎn)，D錯(cuò).故選：AC三、填空題：本題共4小題，每小題5分，共20分．13.函數(shù)SKIPIF1<0的定義域?yàn)開_________．【答案】SKIPIF1<0【解析】【分析】根據(jù)題意，列出不等式，即可得到結(jié)果.【詳解】根據(jù)題意可得，SKIPIF1<0，解得SKIPIF1<0即函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為:SKIPIF1<014.SKIPIF1<0___________.【答案】SKIPIF1<0【解析】【分析】根據(jù)誘導(dǎo)公式化簡(jiǎn)后利用二倍角公式求值.【詳解】SKIPIF1<0,故答案為：SKIPIF1<015.寫出不等式SKIPIF1<0成立的一個(gè)必要不充分條件__________.【答案】SKIPIF1<0(不唯一)【解析】【分析】解不等式得到充要條件，再根據(jù)必要不充分條件的定義即可得答案.【詳解】解：由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0，所以不等式SKIPIF1<0成立的一個(gè)必要不充分條件可以是:SKIPIF1<0.故答案為：SKIPIF1<0(不唯一)16.函數(shù)SKIPIF1<0的最大值為__________，當(dāng)且僅當(dāng)SKIPIF1<0__________時(shí)，等號(hào)成立．【答案】①.SKIPIF1<0##SKIPIF1<0②.SKIPIF1<0【解析】【分析】利用基本不等式即可求解.【詳解】SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立．故答案為：SKIPIF1<0；SKIPIF1<0.四、解答題：本題共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SKIPIF1<0，且SKIPIF1<0．(1)求SKIPIF1<0的值；(2)求SKIPIF1<0的值．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】【分析】(1)根據(jù)同角三角函數(shù)基本關(guān)系求SKIPIF1<0的值，進(jìn)而可得SKIPIF1<0的值；(2)利用誘導(dǎo)公式化簡(jiǎn)，再化弦為切，將SKIPIF1<0的值代入即可求解.【小問1詳解】因SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，【小問2詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0．18.已知函數(shù)SKIPIF1<0.(1)判斷函數(shù)SKIPIF1<0的單調(diào)性，并用單調(diào)性定義證明；(2)若SKIPIF1<0為奇函數(shù)，求滿足SKIPIF1<0的SKIPIF1<0的取值范圍.【答案】(1)增函數(shù)，證明見解析；(2)SKIPIF1<0.【解析】【分析】(1)判斷出函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，然后任取SKIPIF1<0、SKIPIF1<0且SKIPIF1<0，作差SKIPIF1<0，因式分解后判斷SKIPIF1<0的符號(hào)，即可證得結(jié)論成立；(2)由奇函數(shù)的定義可求出實(shí)數(shù)SKIPIF1<0的值，再利用函數(shù)SKIPIF1<0的單調(diào)性可得出關(guān)于SKIPIF1<0的不等式，解之即可.【小問1詳解】證明：函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，理由如下：任取SKIPIF1<0、SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，所以，函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù).【小問2詳解】解：若函數(shù)SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，由SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0.因此，滿足SKIPIF1<0的SKIPIF1<0的取值范圍是SKIPIF1<0.19.已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求SKIPIF1<0最小正周期和最大值；(2)設(shè)SKIPIF1<0，求函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間．【答案】(1)SKIPIF1<0，最大值2；(2)SKIPIF1<0．【解析】【分析】(1)根據(jù)題意，由三角恒等變換公式將函數(shù)SKIPIF1<0化簡(jiǎn)，即可得到結(jié)果；(2)根據(jù)題意，得到函數(shù)SKIPIF1<0的解析式，然后由正弦型函數(shù)的單調(diào)區(qū)間，即可得到結(jié)果.【小問1詳解】∵SKIPIF1<0，所以SKIPIF1<0的最小正周期SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值2；【小問2詳解】由(1)知SKIPIF1<0，又SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0．所以，函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0．20.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)的問題轉(zhuǎn)化成方程有解的問題，進(jìn)而求得參數(shù)SKIPIF1<0的取值范圍.【小問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<0.21.某企業(yè)欲做一個(gè)介紹企業(yè)發(fā)展史的銘牌，銘牌的截面形狀是如圖所示的扇形環(huán)面(由扇形OAD挖去扇形OBC后構(gòu)成的).已知SKIPIF1<0，SKIPIF1<0，線段BA，CD與SKIPIF1<0，SKIPIF1<0的長(zhǎng)度之和為30，圓心角為SKIPIF1<0弧度.(1)求SKIPIF1<0關(guān)于x的函數(shù)表達(dá)式；(2)記銘牌的截面面積為y，試問x取何值時(shí)，y的值最大？并求出最大值.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0，SKIPIF1<0.【解析】【分析】(1)根據(jù)扇形的弧長(zhǎng)公式結(jié)合已知條件可得出關(guān)于SKIPIF1<0、SKIPIF1<0的等式，即可得出SKIPIF1<0關(guān)于SKIPIF1<0的函數(shù)解析式；(2)利用扇形的面積公式結(jié)合二次函數(shù)的基本性質(zhì)可求得SKIPIF1<0的最大值，即可得出結(jié)論.【小問1詳解】解：根據(jù)題意，可算得SKIPIF1<0，SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以，SKIPIF1<0.【小問2詳解】解：根據(jù)題意，可知SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.綜上所述，當(dāng)SKIPIF1<0時(shí)銘牌的面積最大，且最大面積為SKIPIF1<0.22.已知SKIPIF1<0，函數(shù)SKIPIF1<0，其中SKIPIF1<0．(1)設(shè)