湖南省長(zhǎng)沙市xx中學(xué)2022-2023學(xué)年高一上學(xué)期期末數(shù)學(xué)試題（原卷版）

高一上學(xué)期期末數(shù)學(xué)試題時(shí)量：120分鐘滿分：150分一、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.函數(shù)SKIPIF1<0的定義域是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.命題“SKIPIF1<0”的否定是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.用二分法求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上零點(diǎn)的近似值，經(jīng)驗(yàn)證有SKIPIF1<0，取區(qū)間的中點(diǎn)SKIPIF1<0，計(jì)算得SKIPIF1<0，則此時(shí)零點(diǎn)SKIPIF1<0滿足()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.青少年視力是社會(huì)普遍關(guān)注的問(wèn)題，視力情況可借助視力表測(cè)量．通常用五分記錄法和小數(shù)記錄法記錄視力數(shù)據(jù)，五分記錄法的數(shù)據(jù)L和小數(shù)記錄表的數(shù)據(jù)V的滿足SKIPIF1<0．已知某同學(xué)視力的五分記錄法的數(shù)據(jù)為4.9，則其視力的小數(shù)記錄法的數(shù)據(jù)為()(SKIPIF1<0)A.1.5 B.1.2 C.0.8 D.0.65.若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.76.若正實(shí)數(shù)x，y滿足SKIPIF1<0，則x+2y的最小值為()A.7 B.8 C.9 D.107.已知函數(shù)SKIPIF1<0，其中SKIPIF1<0表示不大于x的最大整數(shù)(如SKIPIF1<0，SKIPIF1<0)，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是()A.1 B.2 C.3 D.48.若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0偶函數(shù)，SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0成中心對(duì)稱(chēng)，則下列說(shuō)法正確的個(gè)數(shù)為()①SKIPIF1<0的一個(gè)周期為2②SKIPIF1<0③SKIPIF1<0的一條對(duì)稱(chēng)軸為SKIPIF1<0④SKIPIF1<0A1 B.2 C.3 D.4二、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)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.設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的取值可能是()A.0 B.3 C.SKIPIF1<0 D.210.下列各式中，值為SKIPIF1<0的是()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<011.生活經(jīng)驗(yàn)告訴我們，a克糖水中有b克糖(a>0，b>0，且a>b)，若再添加c克糖(c>0)后，糖水會(huì)更甜，于是得出一個(gè)不等式：SKIPIF1<0.趣稱(chēng)之為“糖水不等式”.根據(jù)生活經(jīng)驗(yàn)和不等式性質(zhì)判斷下列命題一定正確的是()A.若SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的大小關(guān)系隨m的變化而變化B.若SKIPIF1<0，則SKIPIF1<0C.若SKIPIF1<0，則SKIPIF1<0D.若SKIPIF1<0，則一定有SKIPIF1<012.已知函數(shù)SKIPIF1<0是奇函數(shù)，下列選項(xiàng)正確的是()A.SKIPIF1<0B.函數(shù)SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0C.SKIPIF1<0，且SKIPIF1<0，恒有SKIPIF1<0D.若SKIPIF1<0，恒有SKIPIF1<0充分不必要條件為SKIPIF1<0三、填空題(本題共4小題，每小題5分，共20分)13.半徑為SKIPIF1<0，圓心角為SKIPIF1<0的弧長(zhǎng)為_(kāi)__________SKIPIF1<0.14.已知常數(shù)SKIPIF1<0，SKIPIF1<0，假設(shè)無(wú)論SKIPIF1<0為何值，函數(shù)SKIPIF1<0的圖象恒經(jīng)過(guò)一個(gè)定點(diǎn)，則這個(gè)定點(diǎn)的坐標(biāo)是______．15.已知SKIPIF1<0，若“SKIPIF1<0”是“函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上為增函數(shù)”的必要不充分條件，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)__________.16.已知函數(shù)SKIPIF1<0，若至少存在兩個(gè)不相等的實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是________．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.已知集合SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0；(2)若SKIPIF1<0，求實(shí)數(shù)m的取值范圍．18.已知SKIPIF1<0是函數(shù)SKIPIF1<0一個(gè)零點(diǎn).(1)求實(shí)數(shù)SKIPIF1<0的值；(2)求SKIPIF1<0單調(diào)遞減區(qū)間.19.設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0．(1)解關(guān)于x的不等式，SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，求a取值范圍．20.已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0的值；(2)對(duì)任意的SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，恒有SKIPIF1<0，解關(guān)于SKIPIF1<0的不等式SKIPIF1<0.21.在股票市場(chǎng)上，投資者常根據(jù)股價(jià)(每股的價(jià)格)走勢(shì)圖來(lái)操作，股民老張?jiān)谘芯磕持还善睍r(shí)，發(fā)現(xiàn)其在平面直角坐標(biāo)系內(nèi)的走勢(shì)圖有如下特點(diǎn)：每日股價(jià)y(元)與時(shí)間x(天)的關(guān)系在SKIPIF1<0段可近似地用函數(shù)SKIPIF1<0的圖像從最高點(diǎn)A到最低點(diǎn)C的一段來(lái)描述(如圖)，并且從C點(diǎn)到今天的D點(diǎn)在底部橫盤(pán)整理，今天也出現(xiàn)了明顯的底部結(jié)束信號(hào)．老張預(yù)測(cè)這只股票未來(lái)一段時(shí)間的走勢(shì)圖會(huì)如圖中虛線SKIPIF1<0段所示，且SKIPIF1<0段與SKIPIF1<0段關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，點(diǎn)B、D的坐標(biāo)分別是SKIPIF1<0、SKIPIF1<0．(1)請(qǐng)你幫老張確定SKIPIF1<0的值，寫(xiě)出SKIPIF1<0段的函數(shù)表達(dá)式，并指出此時(shí)x的取值范圍；(2)請(qǐng)你幫老張確定虛線SKIPIF1<0段的函數(shù)表達(dá)式，并指出此時(shí)x的取值范圍；(3)如果老張預(yù)測(cè)準(zhǔn)確，且在今天買(mǎi)入該只股票，那么最短買(mǎi)入多少天后，股價(jià)至少是買(mǎi)入價(jià)兩倍？22.已知SKIPIF1<0且SKIPIF1<0，函數(shù)SKIPIF