東華高級中學(xué)  東華松山湖高級中學(xué)2022—2023學(xué)年第二學(xué)期高一2月考數(shù)學(xué)試卷一、單選題：本題共8小題，每小題5分，共40分．在每小題給出的四個選項中，只有一項是符合題目要求的．1. 命題“，”的否定是（）A. ， B. ，C. ， D. ，2. 設(shè)，則“”是“”的（　?。?/span>.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3. 函數(shù)的零點所在的區(qū)間為（）A. B. C. D. 4．若，，向量與向量的夾角為150°，則向量在向量上的投影向量為（    ）A． B． C． D．5. 設(shè)，，則（）A. 且 B. 且C. 且 D. 且6. 要得到函數(shù)的圖象，只需將函數(shù)的圖象進行如下變換得到（）A.向左平移個單位B. 向右平移個單位C. 向右平移個單位D. 向左平移個單位7．已知，是方程的兩根，且，，則的值為（    ）A． B． C．或 D．或8. 若定義上的函數(shù)滿足：對任意有若的最大值和最小值分別為，則的值為（）A. 2022B. 2018 C. 4036D. 4044二、多選題：本題共4小題，每小題5分，共20分．在每小題給出的選項中，有多項符合題目要求．全部選對的得5分，部分選對的得2分，有選錯的得0分．9．在中，為中點，且，則（    ）A．B. C．D．10．已知函數(shù)，則（    ）A．的最大值為B．直線是圖象的一條對稱軸C．在區(qū)間上單調(diào)遞減D．的圖象關(guān)于點對稱11. 若，則下列關(guān)系式中一定成立的是（    ）A.  B. （）C. （是第一象限角） D. 12. 已知函數(shù)，若方程有四個不同的根，且，則下列結(jié)論正確的是（）A.  B. C.  D. 三、填空題：本題共4小題，每小題5分，共20分．13．已知向量，滿足，，，則______．14. 請寫出一個函數(shù)，使它同時滿足下列條件：（1）的最小正周期是4；（2）的最大值為2．____________．15. 若是定義在R上的奇函數(shù)，當(dāng)時，(為常數(shù))，則當(dāng)時，_________.16. 木雕是我國古建筑雕刻中很重要一種藝術(shù)形式，傳統(tǒng)木雕精致細膩?氣韻生動?極富書卷氣．如圖是一扇環(huán)形木雕，可視為扇形OCD截去同心扇形OAB所得部分．已知，，，則該扇環(huán)形木雕的面積為________．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.（本題滿分10分）已知集合(1)    求集合     (2)  若,求實數(shù)的取值范圍. （本題滿分12分）在平面直角坐標(biāo)系中，是坐標(biāo)原點，角的終邊與單位圓的交點坐標(biāo)為，射線繞點按逆時針方向旋轉(zhuǎn)弧度后交單位圓于點，點的縱坐標(biāo)關(guān)于的函數(shù)為.（1）求函數(shù)的解析式，并求的值；（2）若，，求的值. 19.（本題滿分12分） 函數(shù)（1）請用五點作圖法畫出函數(shù)在上的圖象（先列表，再畫圖）（2）設(shè)，，當(dāng)時，試研究函數(shù)的零點的情況.  20.（本題滿分12分）2020年我國面對前所未知，突如其來，來勢洶洶的新冠肺炎疫情，中央出臺了一系列助力復(fù)工復(fù)產(chǎn)好政策．城市快遞行業(yè)運輸能力迅速得到恢復(fù)，市民的網(wǎng)絡(luò)購物也越來越便利．根據(jù)大數(shù)據(jù)統(tǒng)計，某條快遞線路運行時，發(fā)車時間間隔t（單位：分鐘）滿足：，，平均每趟快遞車輛的載件個數(shù)(單位：個)與發(fā)車時間間隔t近似地滿足，其中．（1）若平均每趟快遞車輛的載件個數(shù)不超過1600個，試求發(fā)車時間間隔t的值；（2）若平均每趟快遞車輛每分鐘的凈收益(單位：元)，問當(dāng)發(fā)車時間間隔t為多少時，平均每趟快遞車輛每分鐘的凈收益最大？并求出最大凈收益(結(jié)果取整數(shù))．     21 .（本題滿分12分）已知函數(shù)是定義域上的奇函數(shù)，且滿足（1）          判斷函數(shù)在區(qū)間上的單調(diào)性，并用定義證明（2）          已知，且，若，證明：       22. （本題滿分12分）若函數(shù)對定義域內(nèi)的每一個值，在其定義域內(nèi)都存在唯一的，使成立，則稱函數(shù)具有性質(zhì)．（1）判斷函數(shù)是否具有性質(zhì)，并說明理由；（2）若函數(shù)的定義域為且且具有性質(zhì)，求的值；（3）已知，函數(shù)的定義域為且具有性質(zhì)，若存在實數(shù)，使得對任意的，不等式都成立，求實數(shù)的取值范圍．    東華高級中學(xué)  東華松山湖高級中學(xué)2022—2023學(xué)年第二學(xué)期高一2月考數(shù)學(xué)答案一、選擇題123456789101112DACDBABDBDABCBCBCD二、填空題13.；     14.（答案不唯一）   15. ；    16. 三、解答題17.解：（1），·································································4分 （2）由題意，若，則，···························································5分①時，，解得； ································································· 6分②時，，…………………… 8分 解得；…………………………………………………9分綜上，的取值范圍為.····························································10分18.解：（1）因為，且，所以，····················································2分由此得········································································4分.·············································································5分（2）由知，即··································································7分由于，得，與此同時，所以由平方關(guān)系解得：，·····························································9分············································································ 12分19、（1）·····································································2分按五個關(guān)鍵點列表：0010-1003010描點并將它們用光滑的曲線連接起來如圖1：·············································································7分       （2）因為，所以的零點個數(shù)等價于與圖象交點的個數(shù)，············································8分設(shè)，，則······································································9分當(dāng)，即時，有2個零點；當(dāng)，即時，有1個零點；當(dāng)，即時，有0個零點.      ·························································12分20、解：（1）當(dāng)時，，不滿足題意，舍去．···········································1分當(dāng)時，，即．···································································3分解得（舍）或．·································································4分∵且，∴．····································································5分所以發(fā)車時間間隔為5分鐘．·······················································6分（2）由題意可得．······························································8分當(dāng)時，（元），·································································9分當(dāng)且僅當(dāng)，即時，等號成立，······················································10分當(dāng)時，單調(diào)遞減，時，（元）······················································11分所以發(fā)車時間間隔為6分鐘時，凈收益最大為140（元）．·································12分21.解：（1）由為奇函數(shù)，可得；···················································1分又，得；······································································2分所以. 在上單調(diào)遞增，理由如下：························································3分，且，則······································································4分因為，所以，，，所以，，在上單調(diào)遞增 ···························································6分（2）證法一：由題意，，則有·····················································8分因為，所以，即，······························································10分所以，得證.···································································12 分           證法二：由（1）知，在上單調(diào)遞增，同理可證在上單調(diào)遞減.因為，，所以，，所以···································································8分要證，即證， 即證，即證，···································································9分代入解析式得，即證化簡整理得，即證，····························································10分因為，顯然成立，······························································11分所以原不等式得證，所以.························································ 12 分22、解：（1）對于函數(shù)的定義域內(nèi)任意的，取，則，······································································1分結(jié)合的圖象可知對內(nèi)任意的，是唯一存在的，··········································2分所以函數(shù)具有性質(zhì).（2）因為，且，所以在上是增函數(shù)，················································3分又函數(shù)具有性質(zhì)，所以，即，······················································4分因為，所以且，又，所以，解得，所以．·····························································5分（3）因為，所以，且在定義域上單調(diào)遞增，又因為，在上單調(diào)遞增，所以在上單調(diào)遞增，·····························································6分又因為具有性質(zhì)，從而，即，所以，解得或（舍去），·······························································7分因為存在實數(shù)，使得對任意的，不等式都成立，所以，········································································8分因為在上單調(diào)遞增，所以即對任意的恒成立．·····························································9分所以或，·····································································11分解得或，      綜上可得實數(shù)的取值范圍是………………12分
 

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2023-2024學(xué)年廣東省東莞市東華高級中學(xué)、東華松山湖高級中學(xué)高一上學(xué)期12月月考數(shù)學(xué)試題(含解析）：

這是一份2023-2024學(xué)年廣東省東莞市東華高級中學(xué)、東華松山湖高級中學(xué)高一上學(xué)期12月月考數(shù)學(xué)試題(含解析），共21頁。試卷主要包含了單選題，多選題，填空題，解答題等內(nèi)容，歡迎下載使用。

廣東省東莞市東華高級中學(xué)2023-2024學(xué)年高一上學(xué)期12月月考數(shù)學(xué)試題（Word版附答案）：

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【期中真題】廣東省東莞市東華高級中學(xué)2022-2023學(xué)年高一上學(xué)期期中數(shù)學(xué)試題.zip：

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