廣東省廣州市xx中學(xué)2022-2023學(xué)年高一上學(xué)期期末數(shù)學(xué)試題（含解析）

高一上學(xué)期期末數(shù)學(xué)試題第I卷一?單選題：本大題共8小題，每小題3分，滿分24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.已知集合SKIPIF1<0，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】求出集合SKIPIF1<0，由交集的定義即可得出答案.【詳解】SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0.故選：A.2.“SKIPIF1<0”是“SKIPIF1<0”的()A.充分不必要條件 B.必要不充分條件C.充分必要條件 D.既不充分也不必要條件【答案】B【解析】【分析】解三角函數(shù)的方程，由小范圍能推出大范圍，大范圍不能推出小范圍可得結(jié)果.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0且SKIPIF1<0，∴“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件.故選：B.3.命題“SKIPIF1<0，SKIPIF1<0，”的否定是()A.SKIPIF1<0，SKIPIF1<0 B.SKIPIF1<0，SKIPIF1<0C.SKIPIF1<0，SKIPIF1<0 D.SKIPIF1<0，SKIPIF1<0【答案】B【解析】【分析】根據(jù)全稱量詞命題的否定的知識(shí)求得正確答案.【詳解】原命題的全稱量詞命題，其否定是存在量詞命題,注意到要否定結(jié)論而不是否定條件，所以B選項(xiàng)符合.故選：B4.不等式SKIPIF1<0的解集為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】將原不等式轉(zhuǎn)化為一元二次不等式求解.【詳解】SKIPIF1<0，即SKIPIF1<0，等價(jià)于SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；故選：D.5.已知二次函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)是單調(diào)函數(shù)，則實(shí)數(shù)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】結(jié)合圖像討論對(duì)稱軸位置可得.【詳解】由題知，當(dāng)SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0時(shí)，滿足題意.故選：A6.磚雕是我國(guó)古建筑雕刻中的重要藝術(shù)形式，傳統(tǒng)磚雕精致細(xì)膩、氣韻生動(dòng)、極富書卷氣．如圖所示，一扇環(huán)形磚雕，可視為將扇形SKIPIF1<0截去同心扇形SKIPIF1<0所得圖形，已知SKIPIF1<0，則該扇環(huán)形磚雕的面積為()SKIPIF1<0．A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】根據(jù)扇形的面積公式公式即可求解.【詳解】由SKIPIF1<0以及扇形的面積公式可得:SKIPIF1<0,故選：D7.已知角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為()ASKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】先求得SKIPIF1<0，然后利用誘導(dǎo)公式求得正確答案.【詳解】由于角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故選：D8.已知函數(shù)SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】【分析】分析函數(shù)SKIPIF1<0的單調(diào)性，且SKIPIF1<0.根據(jù)奇偶性可得SKIPIF1<0即為SKIPIF1<0，根據(jù)單調(diào)性即可求解.【詳解】SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，因?yàn)楹瘮?shù)SKIPIF1<0是SKIPIF1<0上的奇函數(shù)，所以SKIPIF1<0在SKIPIF1<0上也單調(diào)遞減.SKIPIF1<0，可轉(zhuǎn)化為SKIPIF1<0,可得SKIPIF1<0.令SKIPIF1<0，可得SKIPIF1<0,故SKIPIF1<0.故由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0.故選:D.二?多選題：本大題共4小題，每小題3分，滿分12分.在每小題給出的四個(gè)選項(xiàng)中，有多項(xiàng)符合要求，全部選對(duì)得3分，選對(duì)但不全的得1分，有選錯(cuò)的得0分.9.下列命題中正確的是()A.SKIPIF1<0時(shí)，SKIPIF1<0的最小值是2B.存在實(shí)數(shù)SKIPIF1<0，使得不等式SKIPIF1<0成立C.若SKIPIF1<0，則SKIPIF1<0D.若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)基本不等式的取等條件可判斷A；取SKIPIF1<0可判斷B；作差可判斷C；利用基本不等式可判斷D.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0時(shí)，SKIPIF1<0取不到最小值2，故A錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0,故B正確；SKIPIF1<0,故SKIPIF1<0，故C正確；SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0,解得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故D正確.故選:BCD.10.下列結(jié)論正確的是()A.函數(shù)SKIPIF1<0且SKIPIF1<0的圖像必過定點(diǎn)SKIPIF1<0B.若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0C.已知函數(shù)SKIPIF1<0，則方程SKIPIF1<0的實(shí)數(shù)解為SKIPIF1<0D.對(duì)任意SKIPIF1<0，都有SKIPIF1<0【答案】AC【解析】【分析】令SKIPIF1<0可判斷A；當(dāng)SKIPIF1<0時(shí)可判斷B；令SKIPIF1<0可得SKIPIF1<0，從而可判斷C；當(dāng)SKIPIF1<0時(shí)可判斷D.【詳解】對(duì)于A，令SKIPIF1<0,可得SKIPIF1<0,故函數(shù)SKIPIF1<0的圖象必過定點(diǎn)SKIPIF1<0，故A正確；對(duì)于B，若SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減，由SKIPIF1<0可得SKIPIF1<0，故B錯(cuò)誤；對(duì)于C，令SKIPIF1<0，可得SKIPIF1<0,解得SKIPIF1<0，故C正確；對(duì)于D，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0,故D錯(cuò)誤.故選:AC.11.下列等式成立的是()A.SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0【答案】ABC【解析】【分析】根據(jù)誘導(dǎo)公式可判斷A；根據(jù)兩角差的余弦公式可判斷B；SKIPIF1<0根據(jù)兩角差的正切公式可判斷C；根據(jù)兩角和的正弦公式可判斷D.【詳解】SKIPIF1<0，故A正確；SKIPIF1<0，故B正確；SKIPIF1<0，故C正確；SKIPIF1<0SKIPIF1<0，故D錯(cuò)誤.故選:ABC.12.已知函數(shù)SKIPIF1<0，則方程SKIPIF1<0的實(shí)根個(gè)數(shù)可能為()A.8 B.7 C.6 D.5【答案】ABC【解析】【分析】以SKIPIF1<0的特殊情形為突破口，解出SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，將SKIPIF1<0看作整體，利用換元的思想進(jìn)一步討論即可.【詳解】由基本不等式可得SKIPIF1<0或SKIPIF1<0，作出函數(shù)SKIPIF1<0圖像，如下：①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；⑤當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；⑥當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；⑦當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故方程SKIPIF1<0的實(shí)數(shù)根個(gè)數(shù)為SKIPIF1<0；故選：ABC【點(diǎn)睛】本題考查了求零點(diǎn)的個(gè)數(shù)，考查了數(shù)形結(jié)合的思想以及分類討論的思想，屬于難題.第II卷三?填空題：本大題共4小題，每小題3分，滿分12分.13.函數(shù)SKIPIF1<0的定義域?yàn)開___________.(用區(qū)間表示)【答案】SKIPIF1<0【解析】【分析】根據(jù)分母不為0，偶次根式的被開方非負(fù)列式可求出結(jié)果.【詳解】由函數(shù)SKIPIF1<0有意義，得SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0.所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<014.已知SKIPIF1<0，則SKIPIF1<0__________.【答案】SKIPIF1<0##0.8【解析】【分析】將條件由輔助角公式化簡(jiǎn)，將條件由二倍角公式化簡(jiǎn)，再代入即可得出答案.【詳解】SKIPIF1<0，由輔助角公式可得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0.15.如果光線每通過一塊玻璃其強(qiáng)度要減少10%，那么至少需要將____________塊這樣的玻璃重疊起來，才能使通過它們的光線強(qiáng)度低于原來的0.5倍．(參考數(shù)據(jù)：SKIPIF1<0．)【答案】SKIPIF1<0【解析】【分析】構(gòu)造不等式SKIPIF1<0，利用對(duì)數(shù)運(yùn)算法則解不等式可求得結(jié)果.【詳解】假設(shè)需要SKIPIF1<0塊這樣的玻璃，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0至少需要7塊這樣的玻璃重疊起來，才能使通過它們的光線強(qiáng)度低于原來的SKIPIF1<0.故答案為：SKIPIF1<0.16.若SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【解析】【分析】將不等式等價(jià)轉(zhuǎn)化為SKIPIF1<0，根據(jù)函數(shù)的單調(diào)性與最值接不等式即可求解.【詳解】根據(jù)不等式SKIPIF1<0恒成立可知SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，先解SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，根據(jù)雙勾函數(shù)的性質(zhì)可得SKIPIF1<0在SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最小值為4，所以SKIPIF1<0，再解SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，根據(jù)雙勾函數(shù)的性質(zhì)可得SKIPIF1<0在SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值為SKIPIF1<0，所以SKIPIF1<0，考慮定義域，所以SKIPIF1<0，故答案:SKIPIF1<0.四?解答題：本大題共6小題，滿分52分.解答應(yīng)寫出文字說明?證明過程或演算過程.17.(1)求值：SKIPIF1<0；(2)設(shè)SKIPIF1<0，且SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)18；(2)500.【解析】【分析】(1)根據(jù)指對(duì)數(shù)的運(yùn)算性質(zhì)即可求解；(2)根據(jù)指對(duì)互化可得SKIPIF1<0，代入SKIPIF1<0，根據(jù)換底公式即可求解.【詳解】(1)SKIPIF1<0.(2)由SKIPIF1<0，可得SKIPIF1<0,所以SKIPIF1<0，解得SKIPIF1<0.18.已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)遞增區(qū)間；(2)若SKIPIF1<0在區(qū)間SKIPIF1<0上的值域?yàn)镾KIPIF1<0，求SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0.【解析】【分析】(1)令SKIPIF1<0即可求得單調(diào)遞增區(qū)間；(2)由SKIPIF1<0，得SKIPIF1<0，畫出SKIPIF1<0在SKIPIF1<0的圖象，可得SKIPIF1<0，從而可求解.【小問1詳解】令SKIPIF1<0，解得SKIPIF1<0.故SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0.【小問2詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.畫出SKIPIF1<0在SKIPIF1<0的圖象如圖所示：所以SKIPIF1<0，解得SKIPIF1<0.故SKIPIF1<0的取值范圍為SKIPIF1<0.19.在密閉培養(yǎng)環(huán)境中，某類細(xì)菌的繁殖在初期會(huì)較快，隨著單位體積內(nèi)細(xì)菌數(shù)量的增加，繁殖速度又會(huì)減慢.在一次實(shí)驗(yàn)中，檢測(cè)到這類細(xì)菌在培養(yǎng)皿中的數(shù)量SKIPIF1<0(單位：百萬個(gè))與培養(yǎng)時(shí)間SKIPIF1<0(單位：小時(shí))的關(guān)系為：SKIPIF1<0234568SKIPIF1<0SKIPIF1<0SKIPIF1<04SKIPIF1<0SKIPIF1<0SKIPIF1<0根據(jù)表格中的數(shù)據(jù)畫出散點(diǎn)圖如下：為了描述從第2小時(shí)開始細(xì)菌數(shù)量隨時(shí)間變化的關(guān)系，現(xiàn)有以下三種模型供選擇：①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0.(1)選出你認(rèn)為最符合實(shí)際的函數(shù)模型，并說明理由；(2)利用SKIPIF1<0和SKIPIF1<0這兩組數(shù)據(jù)求出你選擇的函數(shù)模型的解析式，并預(yù)測(cè)從第2小時(shí)開始，至少再經(jīng)過多少個(gè)小時(shí)，細(xì)菌數(shù)量達(dá)到6百萬個(gè).【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.【解析】【分析】(1)根據(jù)函數(shù)的增長(zhǎng)速度可求解；(2)將所選的兩點(diǎn)坐標(biāo)代入函數(shù)解析式，求出參數(shù)值，可得出函數(shù)模型的解析式，再由SKIPIF1<0即可求解.【小問1詳解】隨著自變量的增加，函數(shù)值的增長(zhǎng)速度變小，而SKIPIF1<0在對(duì)稱軸右方，隨著自變量的增加，函數(shù)值的增長(zhǎng)速度變大，SKIPIF1<0隨著自變量的增加，函數(shù)值的增長(zhǎng)速度變大，故選擇函數(shù)SKIPIF1<0.【小問2詳解】由題意可得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，解得SKIPIF1<0.故至少再經(jīng)過SKIPIF1<0小時(shí)，細(xì)菌數(shù)列達(dá)到6百萬個(gè).20.已知兩個(gè)變量SKIPIF1<0且SKIPIF1<0滿足關(guān)系式SKIPIF1<0，且SKIPIF1<0是SKIPIF1<0的函數(shù).(1)寫出該函數(shù)的表達(dá)式SKIPIF1<0，值域和單調(diào)區(qū)間(不必證明)；(2)在坐標(biāo)系中畫出該函數(shù)的圖象(直接作圖，不必寫過程及理由).【答案】(1)見解析；(2)見解析【解析】【分析】(1)由SKIPIF1<0兩邊取以SKIPIF1<0為底的對(duì)數(shù)可求SKIPIF1<0的解析式，再根據(jù)對(duì)數(shù)函數(shù)的性質(zhì)即可求單調(diào)區(qū)間與值域；(2)根據(jù)解析式與單調(diào)性即可畫出圖象.【小問1詳解】由SKIPIF1<0，可得SKIPIF1<0,即SKIPIF1<0SKIPIF1<0且SKIPIF1<0,故SKIPIF1<0SKIPIF1<0且SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，故SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，故SKIPIF1<0單調(diào)遞減.故SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，無單調(diào)遞減區(qū)間.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0.故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0.小問2詳解】函數(shù)SKIPIF1<0SKIPIF1<0且SKIPIF1<0的圖象如圖所示;21.已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的最小正周期；(2)令SKIPIF1<0，求SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0【解析】【分析】(1)利用同角三角函數(shù)的基本關(guān)系、二倍角公式及輔助角公式可得SKIPIF1<0，從而可求函數(shù)SKIPIF1<0的最小正周期；(2)利用正弦函數(shù)的圖象與性質(zhì)可得SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，根據(jù)二次函數(shù)的性質(zhì)即可求最小值.【小問1詳解】SKIPIF1<0，所以函數(shù)SKIPIF1<0的最小正周期為SKIPIF1<0.【小問2詳解】由SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0,則SKIPIF1<0，令SKIPIF1<0，其對(duì)稱軸為SKIPIF1<0，①當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0；②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0；③當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0.綜上所述，SKIPIF1<0故SKIPIF1<022.給定常數(shù)SKIPIF1<0，定義在SKIPIF1<0上的函數(shù)SKIPIF1<0.(1)若SKIPIF1<0在SKIPIF1<0上的最大值為2，求SKIPIF1<0的值；(2)設(shè)SKIPIF1<0為正整數(shù).如果函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)恰有2022個(gè)零點(diǎn)，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0或SKIPIF1<0.【解析】【分析】(1)根據(jù)誘導(dǎo)公式及二倍角公式可得SKIPIF1<0，設(shè)SKIPIF1<0，分類討論，根據(jù)二次函數(shù)的性質(zhì)即可求解；(2)由題意可得SKIPIF1<0有兩個(gè)不等的實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0.分SKIPIF1