版權(quán)說(shuō)明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請(qǐng)進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡(jiǎn)介
答案第=page11頁(yè),共=sectionpages22頁(yè)專題16一元二次不等式和基本不等式問(wèn)題【練基礎(chǔ)】一、單選題1.(2023·山西忻州·統(tǒng)考模擬預(yù)測(cè))已知SKIPIF1<0,則SKIPIF1<0的最小值是(
)A.6 B.8 C.10 D.12【答案】D【分析】利用基本不等式性質(zhì)求解即可.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),等號(hào)成立.所以SKIPIF1<0的最小值為SKIPIF1<0.故選:D2.(2023·山東濰坊·統(tǒng)考一模)“SKIPIF1<0”是“SKIPIF1<0,SKIPIF1<0成立”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【分析】由不等式SKIPIF1<0恒成立,可求得SKIPIF1<0,即可得出答案.【詳解】因?yàn)镾KIPIF1<0,SKIPIF1<0成立,則SKIPIF1<0,即SKIPIF1<0.所以,“SKIPIF1<0”是“SKIPIF1<0,SKIPIF1<0成立”的充分不必要條件.故選:A.3.(2023·河南·長(zhǎng)葛市第一高級(jí)中學(xué)統(tǒng)考模擬預(yù)測(cè))已知命題“SKIPIF1<0,SKIPIF1<0”為真命題,則實(shí)數(shù)SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【分析】由題知SKIPIF1<0時(shí),SKIPIF1<0,再根據(jù)二次函數(shù)求最值即可得答案.【詳解】解:因?yàn)槊}“SKIPIF1<0,SKIPIF1<0”為真命題,所以,命題“SKIPIF1<0,SKIPIF1<0”為真命題,所以,SKIPIF1<0時(shí),SKIPIF1<0,因?yàn)?,SKIPIF1<0,所以,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取得等號(hào).所以,SKIPIF1<0時(shí),SKIPIF1<0,即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0故選:C4.(2023·四川南充·四川省南充高級(jí)中學(xué)??寄M預(yù)測(cè))已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0,且SKIPIF1<0,若不等式SKIPIF1<0恒成立,則實(shí)數(shù)SKIPIF1<0的最大值為(
)A.9 B.12 C.16 D.25【答案】D【分析】由SKIPIF1<0得到SKIPIF1<0,從而利用基本不等式“1”的妙用求出SKIPIF1<0的最小值,從而得到SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),等號(hào)成立.因不等式SKIPIF1<0恒成立,只需SKIPIF1<0,因此SKIPIF1<0,故實(shí)數(shù)SKIPIF1<0的最大值為25.故選:D5.(2023·遼寧·校聯(lián)考模擬預(yù)測(cè))古印度數(shù)學(xué)家婆什伽羅在《麗拉沃蒂》一書中提出如下問(wèn)題:某人給一個(gè)人布施,初日施2子安貝(古印度貨幣單位),以后逐日倍增,問(wèn)一月共施幾何?在這個(gè)問(wèn)題中,以一個(gè)月31天計(jì)算,記此人第n日布施了SKIPIF1<0子安貝(其中SKIPIF1<0,SKIPIF1<0),數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0.若關(guān)于n的不等式SKIPIF1<0恒成立,則實(shí)數(shù)t的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】SKIPIF1<0是等比數(shù)列,求出SKIPIF1<0與SKIPIF1<0,代入不等式SKIPIF1<0中,結(jié)合基本不等式實(shí)數(shù)t的取值范圍.【詳解】由題意可知,數(shù)列SKIPIF1<0是以2為首項(xiàng),2為公比的等比數(shù)列,故SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0.由SKIPIF1<0,得SKIPIF1<0,整理得SKIPIF1<0對(duì)任意SKIPIF1<0,且SKIPIF1<0恒成立,又SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即n=2時(shí)等號(hào)成立,所以SKIPIF1<0,即實(shí)數(shù)t的取值范圍是SKIPIF1<0故選:B.6.(2023秋·廣西河池·高三統(tǒng)考期末)如圖,在△ABC中,M為線段BC的中點(diǎn),G為線段AM上一點(diǎn)且SKIPIF1<0,過(guò)點(diǎn)G的直線分別交直線AB、AC于P、Q兩點(diǎn),SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.1 C.SKIPIF1<0 D.4【答案】B【分析】由SKIPIF1<0可得SKIPIF1<0,根據(jù)三點(diǎn)共線向量性質(zhì)可得SKIPIF1<0,再結(jié)合均值不等式即可求出結(jié)果.【詳解】由于M為線段BC的中點(diǎn),則SKIPIF1<0又SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0所以SKIPIF1<0,則SKIPIF1<0因?yàn)镾KIPIF1<0三點(diǎn)共線,則SKIPIF1<0,化得SKIPIF1<0由SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),等號(hào)成立,SKIPIF1<0的最小值為1故選:B7.(2022秋·黑龍江牡丹江·高三牡丹江一中??计谀┮阎猄KIPIF1<0,SKIPIF1<0,SKIPIF1<0是SKIPIF1<0與SKIPIF1<0的等比中項(xiàng),則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】根據(jù)等比中項(xiàng)定義可求得SKIPIF1<0,將所求式子化為SKIPIF1<0,利用基本不等式可求得最小值.【詳解】由等比中項(xiàng)定義知:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0時(shí)取等號(hào)),即SKIPIF1<0的最小值為SKIPIF1<0.故選:B.8.(2023·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0,若存在SKIPIF1<0使得SKIPIF1<0,則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】SKIPIF1<0,易得SKIPIF1<0與SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱,由SKIPIF1<0大小關(guān)系易判斷SKIPIF1<0,再將SKIPIF1<0全部代換為含a的式子得SKIPIF1<0,令SKIPIF1<0,利用換元法和對(duì)勾函數(shù)性質(zhì)進(jìn)而得解.【詳解】∵SKIPIF1<0,∴SKIPIF1<0與SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱,作出SKIPIF1<0的大致圖象如圖所示,易知SKIPIF1<0,由SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,得SKIPIF1<0,∴SKIPIF1<0.設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0.SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0取到等號(hào),故當(dāng)SKIPIF1<0時(shí),令SKIPIF1<0,SKIPIF1<0單減,SKIPIF1<0,故SKIPIF1<0.故選:A9.(2022·廣西·校聯(lián)考模擬預(yù)測(cè))雙曲線SKIPIF1<0的左右頂點(diǎn)分別為SKIPIF1<0,曲線SKIPIF1<0上的一點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)為SKIPIF1<0,若直線SKIPIF1<0的斜率為SKIPIF1<0,直線SKIPIF1<0的斜率為SKIPIF1<0,則當(dāng)SKIPIF1<0取到最小值時(shí),雙曲線離心率為()A.SKIPIF1<0 B.2 C.3 D.6【答案】B【分析】由題意SKIPIF1<0利用均值定理可得SKIPIF1<0,再利用雙曲線的幾何性質(zhì)求解即可.【詳解】設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,將曲線方程SKIPIF1<0代入得SKIPIF1<0,又由均值定理得SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立,所以離心率SKIPIF1<0,故選:B二、多選題10.(2023·安徽宿州·統(tǒng)考一模)已知SKIPIF1<0,且SKIPIF1<0,則下列不等關(guān)系成立的是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】ABC【分析】利用基本不等式易知選項(xiàng)AB正確;利用對(duì)數(shù)運(yùn)算法則和重要不等式可知C正確;將不等式SKIPIF1<0化簡(jiǎn)整理可得SKIPIF1<0,構(gòu)造函數(shù)SKIPIF1<0利用函數(shù)單調(diào)性即可證明D錯(cuò)誤.【詳解】由基本不等式可知,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立,即A正確;易知SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立,即B正確;由重要不等式和對(duì)數(shù)運(yùn)算法則可得:SKIPIF1<0,當(dāng)且僅當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立,即C正確;由SKIPIF1<0可得SKIPIF1<0,所以SKIPIF1<0,若SKIPIF1<0,即證明SKIPIF1<0,即SKIPIF1<0即需證明SKIPIF1<0,令函數(shù)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0時(shí),解不等式SKIPIF1<0可得SKIPIF1<0即可,即SKIPIF1<0時(shí)不等式SKIPIF1<0成立;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,解不等式SKIPIF1<0可得SKIPIF1<0,即SKIPIF1<0時(shí)不等式SKIPIF1<0才成立;綜上可知,當(dāng)SKIPIF1<0時(shí),不等式SKIPIF1<0才成立,所以D錯(cuò)誤.故選:ABC11.(2023·全國(guó)·模擬預(yù)測(cè))已知m,n為正實(shí)數(shù),且滿足SKIPIF1<0,則下列不等式正確的是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】ABD【分析】依題意得SKIPIF1<0,直接用基本不等式即可判斷A;根據(jù)SKIPIF1<0,再結(jié)合基本不等式即可判斷B;根據(jù)SKIPIF1<0,再結(jié)合基本不等式即可判斷C;結(jié)合C即可判斷D.【詳解】由SKIPIF1<0,得SKIPIF1<0.對(duì)于A,因?yàn)閙>0,n>0,所以SKIPIF1<0,當(dāng)且僅當(dāng)n=2時(shí)等號(hào)成立,故A正確;對(duì)于B,SKIPIF1<0,又SKIPIF1<0,當(dāng)且僅當(dāng)n=4時(shí)等號(hào)成立,故SKIPIF1<0,故B正確;對(duì)于С,SKIPIF1<0,當(dāng)且僅當(dāng)n=2時(shí)等號(hào)成立,故C錯(cuò)誤;對(duì)于D,SKIPIF1<0,由選項(xiàng)C可知SKIPIF1<0,故SKIPIF1<0,當(dāng)且僅當(dāng)n=2時(shí)等號(hào)成立,故D正確.故選:ABD.12.(2022秋·山東·高三山東聊城一中校聯(lián)考階段練習(xí))已知命題SKIPIF1<0:關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為R,那么命題SKIPIF1<0的一個(gè)必要不充分條件是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】CD【分析】求出命題p成立時(shí)a的取值范圍,再根據(jù)必要不充分條件的定義判斷即可.【詳解】命題p:關(guān)于x的不等式SKIPIF1<0的解集為R,則SKIPIF1<0,解得SKIPIF1<0又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故選:CD.13.(2023·全國(guó)·高三專題練習(xí))若SKIPIF1<0,使得SKIPIF1<0成立是假命題,則實(shí)數(shù)SKIPIF1<0可能取值是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.3 D.SKIPIF1<0【答案】AB【解析】首先由條件可知命題的否定是真命題,參變分離后,轉(zhuǎn)化為最值問(wèn)題求SKIPIF1<0的取值范圍.【詳解】由條件可知SKIPIF1<0,SKIPIF1<0是真命題,即SKIPIF1<0,即SKIPIF1<0,設(shè)SKIPIF1<0等號(hào)成立的條件是SKIPIF1<0,所以SKIPIF1<0的最小值是SKIPIF1<0,即SKIPIF1<0,滿足條件的有AB.故選:AB【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:本題的關(guān)鍵首先是寫出特稱命題的否定,第二個(gè)關(guān)鍵是參變分離,轉(zhuǎn)化為函數(shù)的最值求參數(shù)的取值范圍.14.(2022秋·江蘇淮安·高三??奸_學(xué)考試)已知曲線SKIPIF1<0上存在兩條斜率為3的不同切線,且切點(diǎn)的橫坐標(biāo)都大于零,則實(shí)數(shù)SKIPIF1<0可能的取值(
)A.SKIPIF1<0 B.3 C.SKIPIF1<0 D.SKIPIF1<0【答案】AC【分析】本題先求導(dǎo)函數(shù)并根據(jù)題意建立關(guān)于SKIPIF1<0的方程,再根據(jù)根的分布求SKIPIF1<0的取值范圍,最后判斷得到答案即可.【詳解】解:∵SKIPIF1<0,∴SKIPIF1<0,可令切點(diǎn)的橫坐標(biāo)為SKIPIF1<0,且SKIPIF1<0,可得切線斜率SKIPIF1<0即SKIPIF1<0,由題意,可得關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不等的正根,且可知SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,解得:SKIPIF1<0,所以SKIPIF1<0的取值可能為SKIPIF1<0,SKIPIF1<0.故選:AC.【點(diǎn)睛】本題考查求導(dǎo)函數(shù),導(dǎo)數(shù)的幾何意義,根的分布,是中檔題.15.(2023·遼寧·校聯(lián)考模擬預(yù)測(cè))設(shè)SKIPIF1<0均為正數(shù),且SKIPIF1<0,則(
)A.SKIPIF1<0 B.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0可能成立C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【分析】利用基本不等式相關(guān)公式逐項(xiàng)分析即可求解.【詳解】對(duì)于A:因?yàn)镾KIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立,又SKIPIF1<0,所以SKIPIF1<0,所以A選項(xiàng)正確;對(duì)于B:若SKIPIF1<0,則SKIPIF1<0,因?yàn)镾KIPIF1<0為正數(shù),所以SKIPIF1<0,所以B選項(xiàng)錯(cuò)誤;對(duì)于C:由SKIPIF1<0,且SKIPIF1<0為正數(shù),得SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以C選項(xiàng)正確;對(duì)于D:SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號(hào)成立,所以SKIPIF1<0,所以D選項(xiàng)正確.故選:ACD.16.(2023·福建·統(tǒng)考一模)已知正實(shí)數(shù)x,y滿足SKIPIF1<0,則(
)A.SKIPIF1<0的最小值為SKIPIF1<0 B.SKIPIF1<0的最小值為8C.SKIPIF1<0的最大值為SKIPIF1<0 D.SKIPIF1<0沒(méi)有最大值【答案】AC【分析】將SKIPIF1<0代入SKIPIF1<0,根據(jù)二次函數(shù)的性質(zhì)即可判斷A;根據(jù)SKIPIF1<0及基本不等式可判斷B;SKIPIF1<0,根據(jù)基本不等式可判斷C;SKIPIF1<0,SKIPIF1<0,根據(jù)基本不等式可判斷D.【詳解】因?yàn)閤,y為正實(shí)數(shù),且SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的最小值為SKIPIF1<0,故A正確;SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立,故B錯(cuò)誤;SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立,故SKIPIF1<0,即SKIPIF1<0的最大值為SKIPIF1<0,故C正確;SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立,所以SKIPIF1<0.所以SKIPIF1<0有最大值SKIPIF1<0,故D錯(cuò)誤.故選:AC.17.(2023·山西·統(tǒng)考一模)設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則下列結(jié)論正確的是(
)A.SKIPIF1<0的最大值為SKIPIF1<0 B.SKIPIF1<0的最小值為SKIPIF1<0C.SKIPIF1<0的最小值為9 D.SKIPIF1<0的最小值為SKIPIF1<0【答案】ABC【分析】對(duì)于AD,利用基本不等式判斷即可;對(duì)于B,利用不等式SKIPIF1<0判斷即可,對(duì)于C,利用基本不等式“1”的妙用判斷即可.【詳解】對(duì)于A,因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),故A正確;對(duì)于B,因?yàn)镾KIPIF1<0,故SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),即SKIPIF1<0的最小值SKIPIF1<0,故B正確;對(duì)于C,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0時(shí)取等號(hào),所以SKIPIF1<0的最小值為9,故C正確;對(duì)于D,SKIPIF1<0,故SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),即SKIPIF1<0的最大值SKIPIF1<0,故D錯(cuò)誤.故選:ABC.三、填空題18.(2022·上海松江·統(tǒng)考一模)對(duì)任意SKIPIF1<0,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.【分析】由SKIPIF1<0得SKIPIF1<0的最小值,轉(zhuǎn)化為解關(guān)于a的一元二次不等式.【詳解】由題意知,SKIPIF1<0,又∵SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,解得:SKIPIF1<0,故答案為:SKIPIF1<0.19.(2022·四川遂寧·射洪中學(xué)??寄M預(yù)測(cè))若命題:“SKIPIF1<0,使SKIPIF1<0”是假命題,則實(shí)數(shù)m的取值范圍為____.【答案】SKIPIF1<0或SKIPIF1<0【分析】先得出存在量詞命題的否定,即為恒成立問(wèn)題,結(jié)合二次函數(shù)的圖象與性質(zhì)對(duì)SKIPIF1<0的符號(hào)分類討論即可【詳解】由題意得,“SKIPIF1<0,使SKIPIF1<0”是真命題,當(dāng)SKIPIF1<0時(shí),易得SKIPIF1<0時(shí)命題成立;當(dāng)SKIPIF1<0時(shí),由拋物線開口向下,命題不成立;當(dāng)SKIPIF1<0時(shí),則命題等價(jià)于SKIPIF1<0,即SKIPIF1<0或SKIPIF1<0故答案為:SKIPIF1<0或SKIPIF1<020.(2022秋·四川內(nèi)江·高三??茧A段練習(xí))已知函數(shù)SKIPIF1<0,關(guān)于SKIPIF1<0的方程SKIPIF1<0有三個(gè)不等的實(shí)根,則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【分析】求函數(shù)SKIPIF1<0的導(dǎo)數(shù),判斷其單調(diào)性,作出其大致圖象,數(shù)形結(jié)合,將關(guān)于SKIPIF1<0的方程SKIPIF1<0有三個(gè)不等的實(shí)根轉(zhuǎn)化為SKIPIF1<0有兩個(gè)不等的實(shí)根,結(jié)合二次方程根的分布,求得答案.【詳解】由題意得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0遞增;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0遞減,且SKIPIF1<0;可知函數(shù)SKIPIF1<0的圖象如圖所示,令SKIPIF1<0,則方程SKIPIF1<0有三個(gè)不等的實(shí)根,即為SKIPIF1<0有兩個(gè)不等的實(shí)根,令SKIPIF1<0,則SKIPIF1<0有兩個(gè)不等的實(shí)根,則SKIPIF1<0,所以不妨令SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,故答案為:SKIPIF1<0【點(diǎn)睛】本題考查了利用導(dǎo)數(shù)解決方程的根的個(gè)數(shù)問(wèn)題,考查求參數(shù)的范圍,解答時(shí)要注意利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性,進(jìn)而作出函數(shù)圖象,數(shù)形結(jié)合,將方程根的問(wèn)題轉(zhuǎn)化為二次方程的根的分布問(wèn)題.21.(2023·全國(guó)·模擬預(yù)測(cè))已知向量SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,且a,b均為正數(shù).則ab的最大值為______.【答案】SKIPIF1<0##SKIPIF1<0【分析】根據(jù)SKIPIF1<0求得SKIPIF1<0的關(guān)系式,結(jié)合基本不等式求得SKIPIF1<0的最大值.【詳解】SKIPIF1<0,由于SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.故答案為:SKIPIF1<022.(2023·浙江·永嘉中學(xué)校聯(lián)考模擬預(yù)測(cè))已知實(shí)數(shù)SKIPIF1<0,滿足SKIPIF1<0,則SKIPIF1<0的最小值是______.【答案】9【分析】將已知條件SKIPIF1<0通過(guò)恒等變形,再利用基本不等式即可求解.【詳解】由已知條件得SKIPIF1<0,∵SKIPIF1<0,∴SKIPIF1<0,又∵SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立.故答案為:9.23.(2023·湖北·校聯(lián)考模擬預(yù)測(cè))設(shè)SKIPIF1<0且SKIPIF1<0,若對(duì)SKIPIF1<0都有SKIPIF1<0恒成立,則實(shí)數(shù)a的取值范圍為______.【答案】SKIPIF1<0【分析】由原不等式結(jié)合基本不等式可得SKIPIF1<0,再由SKIPIF1<0可得SKIPIF1<0,則得SKIPIF1<0,然后由SKIPIF1<0結(jié)合指數(shù)的運(yùn)算可得SKIPIF1<0,再通過(guò)構(gòu)造函數(shù)利用導(dǎo)數(shù)證明在SKIPIF1<0,有SKIPIF1<0即可.【詳解】因?yàn)镾KIPIF1<0且SKIPIF1<0,因?yàn)镾KIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),故SKIPIF1<0,所以SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),所以SKIPIF1<0.又SKIPIF1<0,所以SKIPIF1<0,顯然SKIPIF1<0,所以有SKIPIF1<0,即SKIPIF1<0恒成立,又SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0恒成立,即SKIPIF1<0恒成立,與SKIPIF1<0矛盾.下面證明:在SKIPIF1<0,有SKIPIF1<0,令SKIPIF1<0要使SKIPIF1<0,即SKIPIF1<0即SKIPIF1<0由SKIPIF1<0知SKIPIF1<0,得SKIPIF1<0從而需證:SKIPIF1<0即需證明:SKIPIF1<0,記SKIPIF1<0從而只需證:SKIPIF1<0①而SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上遞減,在SKIPIF1<0上遞增,所以SKIPIF1<0,即SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上遞增,又SKIPIF1<0,∴在SKIPIF1<0遞減,SKIPIF1<0,SKIPIF1<0遞增,SKIPIF1<0,而SKIPIF1<0,從而在SKIPIF1<0時(shí)總有SKIPIF1<0∴①式恒成立,不等式SKIPIF1<0得證.綜上所述,SKIPIF1<0.故答案為:SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:此題考查導(dǎo)數(shù)的綜合應(yīng)用,考查不等式恒成立問(wèn)題,解題的關(guān)鍵是根據(jù)已知條件結(jié)合基本不等式確定出SKIPIF1<0的范圍,然后通過(guò)構(gòu)造函數(shù)再證明其正確性即可,考查數(shù)學(xué)轉(zhuǎn)化思想和計(jì)算能力,屬于難題.24.(2023·四川內(nèi)江·統(tǒng)考一模)已知正實(shí)數(shù)a、b滿足SKIPIF1<0,則a、b一定滿足的關(guān)系有______.(填序號(hào))①SKIPIF1<0;②SKIPIF1<0;③SKIPIF1<0;④SKIPIF1<0.【答案】①③.【分析】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,可得SKIPIF1<0,SKIPIF1<0,結(jié)合基本不等式即可求解最值,進(jìn)而判斷可得答案.【詳解】因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,可得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0對(duì)于①,SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立,所以SKIPIF1<0,故①正確.對(duì)于②,SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立,所以SKIPIF1<0,故②錯(cuò)誤.對(duì)于③,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,SKIPIF1<0時(shí)等號(hào)成立,所以SKIPIF1<0,故③正確.對(duì)于④,SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0,SKIPIF1<0時(shí)等號(hào)成立,所以SKIPIF1<0,故④錯(cuò)誤.綜上所述:正確的序號(hào)為①③.故答案為:①③.【提能力】一、單選題25.(2023·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0,若SKIPIF1<0恰有兩個(gè)零點(diǎn),則實(shí)數(shù)SKIPIF1<0的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】函數(shù)SKIPIF1<0,SKIPIF1<0均有有兩個(gè)零點(diǎn),分類討論每部分的零點(diǎn)個(gè)數(shù),結(jié)合零點(diǎn)分布處理.【詳解】∵SKIPIF1<0,則二次函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)若SKIPIF1<0恰有兩個(gè)零點(diǎn),則SKIPIF1<0,得SKIPIF1<0此時(shí)SKIPIF1<0無(wú)零點(diǎn),則SKIPIF1<0,解得SKIPIF1<0則SKIPIF1<0若SKIPIF1<0無(wú)零點(diǎn),則SKIPIF1<0,得SKIPIF1<0此時(shí)SKIPIF1<0有兩個(gè)零點(diǎn),則SKIPIF1<0,得SKIPIF1<0則SKIPIF1<0若SKIPIF1<0有且僅有一個(gè)零點(diǎn),則SKIPIF1<0得SKIPIF1<0,或SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,經(jīng)檢驗(yàn)SKIPIF1<0不合題意則SKIPIF1<0此時(shí)SKIPIF1<0有且僅有一個(gè)零點(diǎn),則SKIPIF1<0,解得SKIPIF1<0且SKIPIF1<0則SKIPIF1<0且SKIPIF1<0綜上所述:SKIPIF1<0故選:B.26.(2022·山西運(yùn)城·統(tǒng)考模擬預(yù)測(cè))已知橢圓SKIPIF1<0的上頂點(diǎn)為A,離心率為e,若在C上存在點(diǎn)P,使得SKIPIF1<0,則SKIPIF1<0的最小值是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【分析】設(shè)出SKIPIF1<0,利用SKIPIF1<0得到SKIPIF1<0在區(qū)間SKIPIF1<0上有解,結(jié)合端點(diǎn)值的符號(hào)得到SKIPIF1<0,求出SKIPIF1<0的最小值.【詳解】易知SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,即方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解,令SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以只需SKIPIF1<0,即SKIPIF1<0解得:SKIPIF1<0.故選:C.27.(2022·全國(guó)·高三專題練習(xí))已知函數(shù)SKIPIF1<0,若SKIPIF1<0,則實(shí)數(shù)SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】由題意,畫出圖形,結(jié)合SKIPIF1<0,分SKIPIF1<0和SKIPIF1<0進(jìn)行討論,解得SKIPIF1<0的范圍,從而即可得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】解:作出函數(shù)SKIPIF1<0的圖象如圖,因?yàn)镾KIPIF1<0,若SKIPIF1<0,由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,且SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0;若SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0;綜上,SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選:A.28.(2022秋·安徽六安·高三六安一中校考階段練習(xí))在正四棱臺(tái)SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0.當(dāng)該正四棱臺(tái)的體積最大時(shí),其外接球的表面積為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【分析】根據(jù)正棱臺(tái)的性質(zhì),表示出棱臺(tái)的高與邊長(zhǎng)之間的關(guān)系,根據(jù)棱臺(tái)的體積公式,將體積函數(shù)式子表示出來(lái),利用不等式求解最值,得到棱臺(tái)的高.因?yàn)橥饨忧虻那蛐囊欢ㄔ诶馀_(tái)上下底面中心的連線及其延長(zhǎng)線上,通過(guò)作圖,數(shù)形結(jié)合,求出外接球的半徑,得到表面積.【詳解】圖1設(shè)底邊長(zhǎng)為a,原四棱錐的高為h,如圖1,SKIPIF1<0分別是上下底面的中心,連結(jié)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,根據(jù)邊長(zhǎng)關(guān)系,知該棱臺(tái)的高為SKIPIF1<0,則SKIPIF1<0,由SKIPIF1<0,且四邊形SKIPIF1<0為直角梯形,SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號(hào)成立,此時(shí)棱臺(tái)的高為1.上底面外接圓半徑SKIPIF1<0,下底面半徑SKIPIF1<0,設(shè)球的半徑為R,顯然球心M在SKIPIF1<0所在的直線上.顯然球心M在SKIPIF1<0所在的直線上.圖2當(dāng)棱臺(tái)兩底面在球心異側(cè)時(shí),即球心M在線段SKIPIF1<0上,如圖2,設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,顯然SKIPIF1<0則,有SKIPIF1<0,即SKIPIF1<0解得SKIPIF1<0,舍去.圖3當(dāng)棱臺(tái)兩底面在球心異側(cè)時(shí),顯然球心M在線段SKIPIF1<0的延長(zhǎng)線上,如圖3,設(shè)SKIPIF1<0,則SKIPIF1<0,顯然SKIPIF1<0即SKIPIF1<0,即SKIPIF1<0解得SKIPIF1<0,SKIPIF1<0,此時(shí),外接球的表面積為SKIPIF1<0.故選:D.29.(2022·河南鶴壁·鶴壁高中??寄M預(yù)測(cè))已知函數(shù)SKIPIF1<0,若不等式SKIPIF1<0對(duì)任意SKIPIF1<0均成立,則SKIPIF1<0的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】利用函數(shù)奇偶性的定義及導(dǎo)數(shù)可得函數(shù)SKIPIF1<0為R上單調(diào)遞增的奇函數(shù),化簡(jiǎn)不等式,然后將SKIPIF1<0分離,利用基本不等式,即可求出答案.【詳解】因?yàn)镾KIPIF1<0的定義域?yàn)镽,SKIPIF1<0,所以函數(shù)SKIPIF1<0是奇函數(shù),由SKIPIF1<0,可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以函數(shù)SKIPIF1<0為R上單調(diào)遞增的奇函數(shù),所以不等式SKIPIF1<0對(duì)任意SKIPIF1<0均成立等價(jià)于SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0對(duì)任意SKIPIF1<0均成立,又SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),所以SKIPIF1<0的取值范圍為SKIPIF1<0.故選:A.30.(2022·河南鶴壁·鶴壁高中??寄M預(yù)測(cè))已知橢圓和雙曲線有共同的焦點(diǎn)SKIPIF1<0,SKIPIF1<0,P是它們的一個(gè)交點(diǎn),且SKIPIF1<0,記橢圓和雙曲線的離心率分別為SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最小值為(
)A.SKIPIF1<0 B.SKIPIF1<0 C.1 D.SKIPIF1<0【答案】B【分析】利用橢圓和雙曲線的定義及SKIPIF1<0可以列出關(guān)于SKIPIF1<0,SKIPIF1<0的方程,再利用均值定理即可得到SKIPIF1<0的最小值【詳解】設(shè)橢圓長(zhǎng)軸長(zhǎng)為SKIPIF1<0,雙曲線實(shí)軸長(zhǎng)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,(SKIPIF1<0),SKIPIF1<0則SKIPIF1<0,解之得SKIPIF1<0又SKIPIF1<0則SKIPIF1<0則SKIPIF1<0,則SKIPIF1<0則SKIPIF1<0,則SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立)則SKIPIF1<0的最小值為SKIPIF1<0故選:B31.(2022秋·福建福州·高三??茧A段練習(xí))已知SKIPIF1<0且SKIPIF1<0,則SKIPIF1<0的取值范圍是(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C【分析】首先求得SKIPIF1<0及SKIPIF1<0的取值范圍,再把SKIPIF1<0轉(zhuǎn)化為關(guān)于SKIPIF1<0的代數(shù)式SKIPIF1<0,利用函數(shù)SKIPIF1<0的單調(diào)性去求SKIPIF1<0的取值范圍即可解決【詳解】由SKIPIF1<0,可得SKIPIF1<0,SKIPIF1<0則SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,SKIPIF1<0又SKIPIF1<0在SKIPIF1<0單調(diào)遞增,在SKIPIF1<0單調(diào)遞減SKIPIF1<0,SKIPIF1<0,SKIPIF1<0則SKIPIF1<0,即SKIPIF1<0故選:C32.(2022·黑龍江哈爾濱·哈九中??寄M預(yù)測(cè))已知函數(shù)SKIPIF1<0,(SKIPIF1<0)的三個(gè)零點(diǎn)分別為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,其中SKIPIF1<0,SKIPIF1<0的取值范圍為(
)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】設(shè)SKIPIF1<0SKIPIF1<0,則SKIPIF1<0有兩個(gè)不等于SKIPIF1<0的零點(diǎn),再根據(jù)SKIPIF1<0,可知SKIPIF1<0的零點(diǎn)互為倒數(shù),則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,利用SKIPIF1<0可求出結(jié)果.【詳解】因?yàn)镾KIPIF1<0SKIPIF1<0恒有零點(diǎn)1,令SKIPIF1<0SKIPIF1<0,則SKIPIF1<0有兩個(gè)不等于SKIPIF1<0的零點(diǎn),因?yàn)镾KIPIF1<0SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0的零點(diǎn)互為倒數(shù),則必然一個(gè)大于0小于1,另一個(gè)大于1,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,且SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,令SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0的取值范圍為SKIPIF1<0.故選:B【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:設(shè)SKIPIF1<0SKIPIF1<0,根據(jù)SKIPIF1<0推出SKIPIF1<0的零點(diǎn)互為倒數(shù)是解題關(guān)鍵.二、多選題33.(2022·浙江·模擬預(yù)測(cè))已知a,b為正數(shù),且SKIPIF1<0,則(
)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】ACD【分析】對(duì)于A選項(xiàng),配成完全平方后驗(yàn)證取等條件即可判斷A選項(xiàng)正誤;對(duì)于B選項(xiàng),根據(jù)均值定理中的“1”的妙用即可判斷B選項(xiàng)正誤;對(duì)于C選項(xiàng),將SKIPIF1<0代入,整理成二次函數(shù),借助二次函數(shù)值域即可判斷C選項(xiàng)的正誤;對(duì)于D選項(xiàng),將SKIPIF1<0代入,整理成分式函數(shù),借助分式函數(shù)值域即可判斷D選項(xiàng)的正誤.【詳解】對(duì)于A選項(xiàng),SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立,當(dāng)SKIPIF1<0時(shí),由于SKIPIF1<0,得SKIPIF1<0,與SKIPIF1<0為正數(shù)矛盾,故SKIPIF1<0,即得SKIPIF1<0,故A選項(xiàng)正確;對(duì)于B選項(xiàng),SKIPIF1<0,SKIPIF1<0.又SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等
溫馨提示
- 1. 本站所有資源如無(wú)特殊說(shuō)明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請(qǐng)下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請(qǐng)聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁(yè)內(nèi)容里面會(huì)有圖紙預(yù)覽,若沒(méi)有圖紙預(yù)覽就沒(méi)有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫(kù)網(wǎng)僅提供信息存儲(chǔ)空間,僅對(duì)用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對(duì)用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對(duì)任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請(qǐng)與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對(duì)自己和他人造成任何形式的傷害或損失。
最新文檔
- 2023雙方汽車租賃協(xié)議書七篇
- 色素性癢疹病因介紹
- 臂叢神經(jīng)損傷病因介紹
- 個(gè)體防護(hù)用品基礎(chǔ)知識(shí)
- 《模具設(shè)計(jì)與制造李集仁》課件-第6章
- (2024)清潔汽油項(xiàng)目可行性研究報(bào)告寫作范本(一)
- 2024-2025年遼寧省錦州市第十二中學(xué)第三次月考英語(yǔ)問(wèn)卷-A4
- 天津市五區(qū)縣重點(diǎn)校聯(lián)考2022-2023學(xué)年高二下學(xué)期期中考試語(yǔ)文試卷
- 電氣施工對(duì)土建工程的 要求與配合- 電氣施工技術(shù)98課件講解
- 2023年監(jiān)護(hù)病房項(xiàng)目籌資方案
- 湘科版小學(xué)二年級(jí)上冊(cè)科學(xué)全冊(cè)教案
- 消化系統(tǒng)常見疾病課件(完美版)
- ISO13485質(zhì)量手冊(cè)+全套程序文件
- 人教版數(shù)學(xué)八年級(jí)上冊(cè)15.2.2.1《分式的加減》說(shuō)課稿1
- 宴會(huì)廳租賃合同
- AQ/T 2080-2023 金屬非金屬地下礦山在用人員定位系統(tǒng)安全檢測(cè)檢驗(yàn)規(guī)范(正式版)
- 事業(yè)編藥學(xué)類考試真題
- 蛋白質(zhì)組學(xué)知識(shí)考試題庫(kù)與答案
- 紅色文化教育教案與反思(3篇模板)
- JTT 1499-2024 公路水運(yùn)工程臨時(shí)用電技術(shù)規(guī)程(正式版)
- 南京市鼓樓區(qū)2022-2023學(xué)年九年級(jí)上學(xué)期期末物理試題
評(píng)論
0/150
提交評(píng)論