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高一上學(xué)期期末數(shù)學(xué)試題一、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,只有一個(gè)符合題目要求1.函數(shù)SKIPIF1<0的定義域是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】C【解析】【分析】由解析式有意義列不等式求SKIPIF1<0的取值范圍即可.【詳解】因?yàn)镾KIPIF1<0有意義,所以SKIPIF1<0,解不等式可得SKIPIF1<0,所以函數(shù)SKIPIF1<0的定義域是SKIPIF1<0,故選:C.2.已知點(diǎn)SKIPIF1<0在第三象限,則角SKIPIF1<0的終邊位置在()A.第一象限 B.第二象限 C.第三象限 D.第四象限【答案】B【解析】【分析】由SKIPIF1<0所在的象限有SKIPIF1<0,即可判斷SKIPIF1<0所在的象限.【詳解】因?yàn)辄c(diǎn)SKIPIF1<0在第三象限,所以SKIPIF1<0,由SKIPIF1<0,可得角SKIPIF1<0的終邊在第二、四象限,由SKIPIF1<0,可得角SKIPIF1<0的終邊在第二、三象限或SKIPIF1<0軸非正半軸上,所以角SKIPIF1<0終邊位置在第二象限,故選:B.3.設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的大小關(guān)系為(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由指數(shù)函數(shù),對(duì)數(shù)函數(shù)單調(diào)性分析SKIPIF1<0和SKIPIF1<0與1和0的關(guān)系,由正切函數(shù)性質(zhì)分析SKIPIF1<0與1和0的關(guān)系,即可得出答案.【詳解】SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,即SKIPIF1<0,由正切函數(shù)性質(zhì)可知SKIPIF1<0,即SKIPIF1<0,故SKIPIF1<0,故選:A.4.函數(shù)SKIPIF1<0的零點(diǎn)所在的區(qū)間為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】【分析】判斷函數(shù)的單調(diào)性,計(jì)算區(qū)間端點(diǎn)處函數(shù)值,由局零點(diǎn)存在定理即可判斷答案.【詳解】函數(shù)SKIPIF1<0,SKIPIF1<0是單調(diào)遞增函數(shù),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0故函數(shù)的零點(diǎn)所在的區(qū)間為SKIPIF1<0,故選:B5.奇函數(shù)SKIPIF1<0滿足SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0=()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】由SKIPIF1<0,可得到函數(shù)SKIPIF1<0的周期是4,利用函數(shù)的周期性和奇偶性,將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0,代入函數(shù)解析式求解即可.【詳解】解:已知奇函數(shù)SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0是以4為周期奇函數(shù),又當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,故選:A.6.函數(shù)SKIPIF1<0的部分圖像大致是()A. B.C. D.【答案】C【解析】【分析】根據(jù)函數(shù)基本性質(zhì)及函數(shù)圖像特征分別判斷即可.【詳解】因?yàn)镾KIPIF1<0,SKIPIF1<0.所以SKIPIF1<0為奇函數(shù),故SKIPIF1<0選項(xiàng)錯(cuò);SKIPIF1<0SKIPIF1<0,故SKIPIF1<0選項(xiàng)錯(cuò);故選:SKIPIF1<0.7.已知函數(shù)SKIPIF1<0(SKIPIF1<0),若SKIPIF1<0在SKIPIF1<0上有兩個(gè)零點(diǎn),則SKIPIF1<0取值范圍是()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】【分析】求出SKIPIF1<0的范圍,數(shù)形結(jié)合得到關(guān)于SKIPIF1<0的范圍,求出SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0,解得:SKIPIF1<0.故選:A8.已知函數(shù)f(x)=|log2(x?1)|,1<x≤3x2?8x+16,x>3,若方程SKIPIF1<0有4個(gè)不同的零點(diǎn)SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0().A.10 B.8 C.6 D.4【答案】B【解析】【分析】作出f(x)圖像,由圖可知方程SKIPIF1<0的4個(gè)不同的零點(diǎn)為函數(shù)y=f(x)與函數(shù)y=m圖像的四個(gè)交點(diǎn)的橫坐標(biāo),由圖可知,SKIPIF1<0且SKIPIF1<0SKIPIF1<0.【詳解】作函數(shù)SKIPIF1<0SKIPIF1<0的圖像如圖,SKIPIF1<0有四個(gè)不同的實(shí)根SKIPIF1<0且SKIPIF1<0,可得SKIPIF1<0SKIPIF1<0,且SKIPIF1<0,即為SKIPIF1<0,即有SKIPIF1<0,即為SKIPIF1<0,可得SKIPIF1<0.故選:B.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)的的0分.9.下列命題為真命題的是(

)A.若SKIPIF1<0,則SKIPIF1<0 B.若SKIPIF1<0,則SKIPIF1<0C.若SKIPIF1<0,則SKIPIF1<0 D.若SKIPIF1<0,則SKIPIF1<0【答案】BD【解析】【分析】利用不等式的運(yùn)算法則與性質(zhì)即可求解.【詳解】對(duì)于A:當(dāng)SKIPIF1<0,SKIPIF1<0,故A錯(cuò)誤;對(duì)于B:SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故B正確;對(duì)于C:當(dāng)SKIPIF1<0,SKIPIF1<0時(shí),則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,故C錯(cuò)誤;對(duì)于D:SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故D正確;故選:BD.10.下列說法正確的是(

)A.命題SKIPIF1<0的否定為:SKIPIF1<0.B.SKIPIF1<0與SKIPIF1<0為同一函數(shù)C.若冪函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0,則SKIPIF1<0D.函數(shù)SKIPIF1<0和SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱【答案】AD【解析】【分析】根據(jù)全稱量詞的否定是存在量詞,可知A正確;根據(jù)兩個(gè)函數(shù)的定義域不同,可知B不正確;利用待定系數(shù)法求出SKIPIF1<0的解析式,再根據(jù)解析式求出SKIPIF1<0,可知C不正確;根據(jù)函數(shù)SKIPIF1<0與SKIPIF1<0互為反函數(shù),可知D正確.【詳解】對(duì)于A,命題SKIPIF1<0的否定為:SKIPIF1<0,故A正確;對(duì)于B,SKIPIF1<0與SKIPIF1<0的定義域不同,所以不為同一函數(shù),故B不正確;對(duì)于C,設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故C不正確;對(duì)于D,函數(shù)SKIPIF1<0與SKIPIF1<0互為反函數(shù),它們的圖象關(guān)于直線SKIPIF1<0對(duì)稱,故D正確.故選:AD11.已知函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱,則()A.函數(shù)SKIPIF1<0為奇函數(shù)B.函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C.若SKIPIF1<0,則SKIPIF1<0的最小值為SKIPIF1<0D.函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到函數(shù)SKIPIF1<0的圖象【答案】AC【解析】【分析】利用SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱,即可求出SKIPIF1<0的值,從而得出SKIPIF1<0的解析式,再利用三角函數(shù)的性質(zhì)逐一判斷四個(gè)選項(xiàng)即可.【詳解】因?yàn)镾KIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱,所以SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,對(duì)于A:SKIPIF1<0,所以SKIPIF1<0為奇函數(shù)成立,故選項(xiàng)A正確;對(duì)于B:SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0在SKIPIF1<0上不是單調(diào)函數(shù);故選項(xiàng)B不正確;對(duì)于C:因?yàn)镾KIPIF1<0,SKIPIF1<0,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0的最小值為半個(gè)周期,即SKIPIF1<0,故選項(xiàng)C正確;對(duì)于D:函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0,故選項(xiàng)D不正確;故選:AC【點(diǎn)睛】本題主要考查了利用三角函數(shù)對(duì)稱軸求函數(shù)解析式,考查了三角函數(shù)平移變換、三角函數(shù)的周期、單調(diào)性、最值,屬于中檔題12.已知函數(shù)SKIPIF1<0,則下列說法正確的是(

)A.函數(shù)SKIPIF1<0有3個(gè)零點(diǎn)B.關(guān)于x的方程SKIPIF1<0有SKIPIF1<0個(gè)不同的解C.對(duì)于實(shí)數(shù)SKIPIF1<0,不等式SKIPIF1<0恒成立D.當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0的圖象與x軸圍成的圖形的面積為SKIPIF1<0【答案】ACD【解析】【分析】根據(jù)題意求出函數(shù)的解析式,再畫出函數(shù)的圖象,然后結(jié)合圖象逐個(gè)分析判斷即可.【詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,SKIPIF1<0,依次類推,可得函數(shù)的解析式,作出函數(shù)的大致圖象如圖所示,對(duì)于A,由SKIPIF1<0,得SKIPIF1<0,令SKIPIF1<0,由圖象可知SKIPIF1<0與SKIPIF1<0的圖象只有3個(gè)交點(diǎn),所以函數(shù)SKIPIF1<0有3個(gè)零點(diǎn),所以A正確,對(duì)于B,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,由圖象可知SKIPIF1<0與SKIPIF1<0的圖象只有3個(gè)交點(diǎn),所以關(guān)于x的方程SKIPIF1<0有3個(gè)不同的解,而當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以B錯(cuò)誤,對(duì)于C,對(duì)于實(shí)數(shù)SKIPIF1<0,不等式SKIPIF1<0恒成立,即SKIPIF1<0恒成立,由圖可知函數(shù)SKIPIF1<0的圖象的每一個(gè)上頂點(diǎn)都在曲線SKIPIF1<0上,所以SKIPIF1<0恒成立,所以C正確,對(duì)于D,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,此時(shí)函數(shù)SKIPIF1<0的圖象與x軸圍成的圖形的面積為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,此時(shí)函數(shù)SKIPIF1<0的圖象與x軸圍成的圖形的面積為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,此時(shí)函數(shù)SKIPIF1<0的圖象與x軸圍成的圖形的面積為SKIPIF1<0,……,當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0的圖象與x軸圍成的圖形的面積為SKIPIF1<0,所以D正確,故選:ACD三、填空題:本題共4小題,每小題5分,共20分.13.SKIPIF1<0_____________.【答案】SKIPIF1<0【解析】【分析】由于SKIPIF1<0,進(jìn)而結(jié)合誘導(dǎo)公式求解即可.【詳解】由誘導(dǎo)公式可得SKIPIF1<0.故答案為:SKIPIF1<0.14.已知函數(shù)SKIPIF1<0的圖象如圖所示.則函數(shù)SKIPIF1<0的解析式為_________.【答案】SKIPIF1<0【解析】【分析】根據(jù)最值可求SKIPIF1<0,根據(jù)周期可求SKIPIF1<0,代入特殊值可求SKIPIF1<0.【詳解】由圖可知,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,解得SKIPIF1<0.故答案為:SKIPIF1<0.15.以等邊三角形每個(gè)頂點(diǎn)為圓心,以邊長(zhǎng)為半徑,在另兩個(gè)頂點(diǎn)間作一段弧,三段弧圍成曲邊三角形就是勒洛三角形.勒洛三角形是由德國(guó)機(jī)械工程專家、機(jī)構(gòu)運(yùn)動(dòng)學(xué)家勒洛首先發(fā)現(xiàn),所以以他的名字命名.一些地方的市政檢修井蓋、方孔轉(zhuǎn)機(jī)等都有應(yīng)用勒洛三角形.如圖,已知某勒洛三角形的一段弧SKIPIF1<0的長(zhǎng)度為SKIPIF1<0,則該勒洛三角形的面積為___________.【答案】SKIPIF1<0【解析】【分析】計(jì)算出等邊SKIPIF1<0的邊長(zhǎng),計(jì)算出由弧SKIPIF1<0與SKIPIF1<0所圍成的弓形的面積,進(jìn)而可求得勒洛三角形的面積.【詳解】設(shè)等邊三角形SKIPIF1<0的邊長(zhǎng)為SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,所以,由弧SKIPIF1<0與SKIPIF1<0所圍成的弓形的面積為SKIPIF1<0,所以該勒洛三角形的面積SKIPIF1<0.故答案:SKIPIF1<0.16.函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù),且關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立,則實(shí)數(shù)SKIPIF1<0的取值范圍為________.【答案】SKIPIF1<0【解析】【分析】先利用函數(shù)的奇偶性求解實(shí)數(shù)SKIPIF1<0;再利用定義證明函數(shù)的單調(diào)性,利用奇偶性和單調(diào)性將不等式恒成立問題轉(zhuǎn)化為分離參數(shù)問題,利用基本不等式以及雙勾函數(shù)的單調(diào)性求解即可.【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,由函數(shù)SKIPIF1<0為SKIPIF1<0上的奇函數(shù),可得SKIPIF1<0,即SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0;所以SKIPIF1<0,任取SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,則函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù);又函數(shù)SKIPIF1<0為SKIPIF1<0上的奇函數(shù),所以不等式SKIPIF1<0恒成立,轉(zhuǎn)化為SKIPIF1<0,即SKIPIF1<0對(duì)SKIPIF1<0恒成立,所以SKIPIF1<0對(duì)SKIPIF1<0恒成立,即SKIPIF1<0,令SKIPIF1<0,因SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),由雙勾函數(shù)的單調(diào)性知:SKIPIF1<0,函數(shù)單調(diào)遞減,SKIPIF1<0,函數(shù)單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,故實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為:SKIPIF1<0.【點(diǎn)睛】關(guān)鍵點(diǎn)睛:本題考查函數(shù)奇偶性的定義,以及利用奇偶性,單調(diào)性解不等式恒成立問題,利用奇偶性和單調(diào)性將不等式恒成立問題轉(zhuǎn)化為分離參數(shù)問題是解決本題的關(guān)鍵.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.已知角SKIPIF1<0的始邊與SKIPIF1<0軸的非負(fù)半軸重合,終邊與單位圓的交點(diǎn)SKIPIF1<0的坐標(biāo)為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)由三角函數(shù)的定義與三角函數(shù)的象限符號(hào)即可求解;(2)由同角三角函數(shù)的關(guān)系即可求解.【小問1詳解】∵角SKIPIF1<0的終邊與單位圓的交點(diǎn)為SKIPIF1<0SKIPIF1<0∴SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<0.【小問2詳解】原式SKIPIF1<0又∵SKIPIF1<0∴原式SKIPIF1<018.某居民小區(qū)欲在一塊空地上建一面積為SKIPIF1<0的矩形停車場(chǎng),停車場(chǎng)的四周留有人行通道,設(shè)計(jì)要求停車場(chǎng)外側(cè)南北的人行通道寬3m,東西的人行通道寬4m,如圖所示(圖中單位:m),問如何設(shè)計(jì)停車場(chǎng)的邊長(zhǎng),才能使人行通道占地面積最???最小面積是多少?【答案】設(shè)計(jì)矩形停車場(chǎng)南北側(cè)邊長(zhǎng)為30SKIPIF1<0,則其東西側(cè)邊長(zhǎng)為40SKIPIF1<0,人行通道占地面積最小528SKIPIF1<0.【解析】【分析】設(shè)矩形停車場(chǎng)南北側(cè)邊長(zhǎng)為SKIPIF1<0,則其東西側(cè)邊長(zhǎng)為SKIPIF1<0m,人行通道占地面積為SKIPIF1<0,再由基本不等式可得答案.【詳解】設(shè)矩形停車場(chǎng)南北側(cè)邊長(zhǎng)為SKIPIF1<0,則其東西側(cè)邊長(zhǎng)為SKIPIF1<0m,人行通道占地面積為SKIPIF1<0,由均值不等式,得SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)SKIPIF1<0.所以,設(shè)計(jì)矩形停車場(chǎng)南北側(cè)邊長(zhǎng)為30m,則其東西側(cè)邊長(zhǎng)為40m,人行通道占地面積最小528m2.19.設(shè)函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的最小正周期和單調(diào)遞減區(qū)間;(2)求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值和最小值.【答案】(1)SKIPIF1<0,SKIPIF1<0(2)最大值為1;最小值為SKIPIF1<0【解析】【分析】(1)代入正弦函數(shù)的周期公式與單調(diào)遞減區(qū)間即可求解;(2)根據(jù)正弦函數(shù)的單調(diào)區(qū)間與定義域即可求出最大值和最小值.【小問1詳解】由題知,SKIPIF1<0所以函數(shù)SKIPIF1<0的最小正周期SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.【小問2詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0有最大值,最大值為1;當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0有最小值,最小值為SKIPIF1<0.20.中國(guó)地大物博,大興安嶺的雪花還在飛舞,長(zhǎng)江兩岸的柳枝已經(jīng)發(fā)芽,海南島上盛開著鮮花.燕子每年秋天都要從北方飛向南方過冬,專家發(fā)現(xiàn),某種兩歲燕子在飛行時(shí)的耗氧量與飛行速度SKIPIF1<0米SKIPIF1<0秒SKIPIF1<0之間滿足關(guān)系:SKIPIF1<0,其中SKIPIF1<0表示燕子耗氧量的單位數(shù).(1)當(dāng)該燕子的耗氧量為SKIPIF1<0個(gè)單位時(shí),它的飛行速度大約是多少?(2)若某只兩歲燕子飛行時(shí)的耗氧量變?yōu)樵瓉淼腟KIPIF1<0倍,則它的飛行速度大約增加多少?SKIPIF1<0參考數(shù)據(jù):SKIPIF1<0,SKIPIF1<0【答案】(1)SKIPIF1<0(米/秒)(2)SKIPIF1<0(米/秒)【解析】【分析】(1)由耗氧量和飛行速度的關(guān)系可將SKIPIF1<0表示為對(duì)數(shù),然后求出SKIPIF1<0即可.(2)記燕子原來的耗氧量為SKIPIF1<0,飛行速度為SKIPIF1<0,現(xiàn)在的耗氧量為SKIPIF1<0,飛行速度為SKIPIF1<0,則可得SKIPIF1<0,然后化為對(duì)數(shù)運(yùn)算即可.【小問1詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,即它的飛行速度大約是SKIPIF1<0米SKIPIF1<0秒SKIPIF1<0.【小問2詳解】記燕子原來的耗氧量為SKIPIF1<0,飛行速度為SKIPIF1<0,現(xiàn)在的耗氧量為SKIPIF1<0,飛行速度為SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以它的飛行速度大約增加SKIPIF1<0米SKIPIF1<0秒SKIPIF1<0.21.已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0,且函數(shù)SKIPIF1<0有零點(diǎn),求實(shí)數(shù)SKIPIF1<0的取值范圍;(2)當(dāng)SKIPIF1<0時(shí),解關(guān)于SKIPIF1<0的不等式SKIPIF1<0;(3)若正數(shù)SKIPIF1<0滿足SKIPIF1<0,且對(duì)于任意的SKIPIF1<0恒成立,求實(shí)數(shù)SKIPIF1<0的值.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0時(shí)SKIPIF1<0;SKIPIF1<0時(shí)SKIPIF1<0;SKIPIF1<0時(shí)SKIPIF1<0;(3)SKIPIF1<0;【解析】【分析】(1)由SKIPIF1<0可得結(jié)果;(2)SKIPIF1<0時(shí),SKIPIF1<0SKIPIF1<0,分三種情況討論,分別利用一元二次不等式的解法求解即可;(3)SKIPIF1<0時(shí)SKIPIF1<0恒成立,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,解不等式即可的結(jié)果.【詳解】(1)SKIPIF1<0時(shí),SKIPIF1<0,由函數(shù)SKIPIF1<0有零點(diǎn),可得SKIPIF1<0,即SKIPIF1<0或SKIPIF1<0;(2)SKIPIF1<0時(shí),SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0,當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0,當(dāng)SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0;(3)二次函數(shù)SKIPIF1<0開口響上,對(duì)稱軸SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0在SKIPIF1<0單調(diào)遞增,SKIPIF1<0時(shí)SKIPIF1<0恒成立,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,由SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0,此時(shí)SKIPIF1<0,則SKIPIF1<0.【點(diǎn)睛】本題主要考查函數(shù)的零點(diǎn)、一元二次不等式的解法、二次函數(shù)的性質(zhì)以及分類討論思想的應(yīng)用,屬于中檔題.分類討論思想解決高中數(shù)學(xué)問題的一種重要思想方法,是中學(xué)數(shù)學(xué)四種重要的數(shù)學(xué)思想之一,尤其在解決含參數(shù)問題發(fā)揮著奇特功效,大大提高了解題能力與速度.運(yùn)用這種方法的關(guān)鍵是將題設(shè)條件研究透,這樣才能快速找準(zhǔn)突破點(diǎn).充分利用分類討論思想方法能夠使問題條理清晰,進(jìn)而順利解答,希望同學(xué)們能夠熟練掌握并應(yīng)用與解題當(dāng)中.22.設(shè)函數(shù)SKIPIF1<0(SKIPIF1<0為實(shí)數(shù)).(1)當(dāng)SKIPIF1<0時(shí),求方程SKIPIF1<0的實(shí)數(shù)解;(2)當(dāng)SKIPIF1<0時(shí),(ⅰ)存在SKIPIF1<0使不等式SKIPIF1<0成立,求SKIPIF1<0的范圍;(ⅱ)設(shè)函數(shù)SKIPIF1<0若對(duì)任意的SKIPIF1<0總存在SKIPIF1<0使SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0或SKIPIF1<0(2)(ⅰ)SKIPIF1<0;(ⅱ)SKIPIF1<0【解析】【分析】(1)將SKIPIF1<0代入SKIPIF1<0中,直接求

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