新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專題35 高考新題型開(kāi)放性試題綜合問(wèn)題（解析版）

試卷第=page11頁(yè)，共=sectionpages33頁(yè)專題35高考新題型開(kāi)放性試題綜合問(wèn)題1．（2022秋·廣東揭陽(yáng)·高三?？茧A段練習(xí)）寫(xiě)出一個(gè)導(dǎo)函數(shù)恒大于等于2的函數(shù)SKIPIF1<0____________.【答案】SKIPIF1<0（答案不唯一）【分析】保證SKIPIF1<0即可.【詳解】可設(shè)SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）2．（2023·江蘇泰州·統(tǒng)考一模）寫(xiě)出一個(gè)同時(shí)滿足下列條件①②的等比數(shù)列{SKIPIF1<0}的通項(xiàng)公式SKIPIF1<0＝___．①SKIPIF1<0；②SKIPIF1<0【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)題目所給條件以及等比數(shù)列的知識(shí)求得正確答案.【詳解】依題意，SKIPIF1<0是等比數(shù)列，設(shè)其公比為SKIPIF1<0，由于①SKIPIF1<0，所以SKIPIF1<0，由于②SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0符合題意.故答案為：SKIPIF1<0（答案不唯一）3．（2023·重慶沙坪壩·重慶南開(kāi)中學(xué)校考模擬預(yù)測(cè)）已知向量SKIPIF1<0滿足SKIPIF1<0，請(qǐng)寫(xiě)出一個(gè)符合題意的向量SKIPIF1<0的坐標(biāo)______.【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)題意，由數(shù)量積的計(jì)算公式可得SKIPIF1<0，分析SKIPIF1<0、SKIPIF1<0的關(guān)系，利用特殊值法可得答案.【詳解】根據(jù)題意，向量SKIPIF1<0，且SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）4．（2023·安徽宿州·統(tǒng)考一模）若拋物線C：SKIPIF1<0存在以點(diǎn)SKIPIF1<0為中點(diǎn)的弦，請(qǐng)寫(xiě)出一個(gè)滿足條件的拋物線方程為_(kāi)______.【答案】SKIPIF1<0（答案不唯一）【分析】拋物線存在以點(diǎn)SKIPIF1<0為中點(diǎn)的弦，則該點(diǎn)在拋物線開(kāi)口內(nèi)，列式求解即可.【詳解】拋物線存在以點(diǎn)SKIPIF1<0為中點(diǎn)的弦，則該點(diǎn)在拋物線開(kāi)口內(nèi)，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.可取SKIPIF1<0，則滿足條件的拋物線方程為SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）5．（2023·浙江·校聯(lián)考三模）寫(xiě)出一個(gè)滿足下列條件的正弦型函數(shù)，SKIPIF1<0____________．①最小正周期為SKIPIF1<0；

②SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；

③SKIPIF1<0成立．【答案】SKIPIF1<0（答案不唯一）【分析】設(shè)SKIPIF1<0，SKIPIF1<0，根據(jù)SKIPIF1<0，則可設(shè)SKIPIF1<0，根據(jù)最小正周期為SKIPIF1<0，可得SKIPIF1<0，通過(guò)整體換元法則可得到SKIPIF1<0，取SKIPIF1<0即可.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0，不妨設(shè)SKIPIF1<0因?yàn)镾KIPIF1<0最小正周期為SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不妨設(shè)SKIPIF1<0所以滿足條件之一的SKIPIF1<0．故答案為：SKIPIF1<0.6．（2022秋·廣東廣州·高三?？计谥校┖瘮?shù)SKIPIF1<0的最大值為2，則常數(shù)SKIPIF1<0的一個(gè)取值可為_(kāi)_____.【答案】SKIPIF1<0（答案不唯一）【分析】由三角函數(shù)的有界性得到SKIPIF1<0同時(shí)成立，不妨令SKIPIF1<0，求出SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0，要想SKIPIF1<0的最大值為2，需要SKIPIF1<0同時(shí)成立，由SKIPIF1<0得到SKIPIF1<0，SKIPIF1<0，不妨取SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，故答案為：SKIPIF1<0（答案不唯一）7．（2022秋·江蘇·高三校聯(lián)考階段練習(xí)）寫(xiě)出一個(gè)同時(shí)滿足下列性質(zhì)①②的函數(shù)SKIPIF1<0_____________．①SKIPIF1<0；②SKIPIF1<0在定義域上單調(diào)遞增．【答案】SKIPIF1<0（滿足SKIPIF1<0均可）【分析】由基本初等函數(shù)性質(zhì)篩選判斷即可【詳解】SKIPIF1<0，且SKIPIF1<0單調(diào)遞增．故答案為：SKIPIF1<0（滿足SKIPIF1<0均可）8．（2022秋·福建·高三校聯(lián)考階段練習(xí)）已知奇函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，則函數(shù)SKIPIF1<0的解析式可以為SKIPIF1<0=______.（寫(xiě)出一個(gè)符合題意的函數(shù)即可）【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)正弦函數(shù)的周期和單調(diào)性的性質(zhì)，直接寫(xiě)出符合題意的解析式即可.【詳解】因?yàn)镾KIPIF1<0，所以奇函數(shù)SKIPIF1<0的周期為4，所以可得SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，可知此時(shí)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.故答案為：SKIPIF1<09．（2023春·浙江紹興·高三統(tǒng)考開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0滿足：SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，請(qǐng)你寫(xiě)出符合上述條件的一個(gè)函數(shù)SKIPIF1<0__________.【答案】SKIPIF1<0（答案不唯一）【分析】由SKIPIF1<0，可得對(duì)數(shù)函數(shù)具有此性質(zhì)，從而可得函數(shù).【詳解】對(duì)于函數(shù)SKIPIF1<0，SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0滿足條件，故答案為：SKIPIF1<0（答案不唯一）.10．（2023·黑龍江大慶·統(tǒng)考一模）已知直線SKIPIF1<0與圓SKIPIF1<0相離，則整數(shù)SKIPIF1<0的一個(gè)取值可以是______．【答案】SKIPIF1<0或SKIPIF1<0或SKIPIF1<0（注意：只需從SKIPIF1<0中寫(xiě)一個(gè)作答即可）【分析】利用直線與圓的位置關(guān)系列出不等式組，解出整數(shù)SKIPIF1<0的范圍．【詳解】因?yàn)閳ASKIPIF1<0的圓心為SKIPIF1<0，所以圓心到直線SKIPIF1<0的距離SKIPIF1<0，因?yàn)閳ASKIPIF1<0的方程可化簡(jiǎn)為SKIPIF1<0，即半徑為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故整數(shù)SKIPIF1<0的取值可能是SKIPIF1<0．故答案為：SKIPIF1<0或SKIPIF1<0或SKIPIF1<0（注意：只需從SKIPIF1<0中寫(xiě)一個(gè)作答即可）11．（2023春·遼寧本溪·高三?？茧A段練習(xí)）請(qǐng)寫(xiě)出與曲線SKIPIF1<0在SKIPIF1<0處具有相同切線的另一個(gè)函數(shù)：______．【答案】SKIPIF1<0（答案不唯一）【分析】利用導(dǎo)數(shù)的幾何意義可求得在SKIPIF1<0處的切線斜率，由此可得切線方程；若兩曲線在原點(diǎn)處具有相同切線，只需滿足過(guò)點(diǎn)SKIPIF1<0且在SKIPIF1<0處的導(dǎo)數(shù)值SKIPIF1<0即可，由此可得曲線方程.【詳解】SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，又SKIPIF1<0過(guò)原點(diǎn)，SKIPIF1<0在原點(diǎn)SKIPIF1<0處的切線斜率SKIPIF1<0，SKIPIF1<0在原點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0；所求曲線只需滿足過(guò)點(diǎn)SKIPIF1<0且在SKIPIF1<0處的導(dǎo)數(shù)值SKIPIF1<0即可，如SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0過(guò)原點(diǎn)，SKIPIF1<0在原點(diǎn)處的切線斜率SKIPIF1<0，SKIPIF1<0在原點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.故答案為:SKIPIF1<0（答案不唯一）.12．（2023春·廣東揭陽(yáng)·高三?？茧A段練習(xí)）在某數(shù)學(xué)活動(dòng)課上，數(shù)學(xué)教師把一塊三邊長(zhǎng)分別為6，8，10的三角板SKIPIF1<0放在直角坐標(biāo)系中，則SKIPIF1<0外接圓的方程可以為_(kāi)____________．（寫(xiě)出其中一個(gè)符合條件的即可）【答案】SKIPIF1<0（答案不唯一）【分析】由題意邊長(zhǎng)分別為6，8，10的SKIPIF1<0為直角三角形，且外接圓的半徑為SKIPIF1<0，可以將斜邊的中點(diǎn)與坐標(biāo)原點(diǎn)重合時(shí)，即可得解.【詳解】邊長(zhǎng)分別為6，8，10的SKIPIF1<0為直角三角形，且外接圓的半徑為SKIPIF1<0，若將斜邊的中點(diǎn)與坐標(biāo)原點(diǎn)重合時(shí)，則圓心為SKIPIF1<0，所以其外接圓方程可以為SKIPIF1<0，若將直角頂點(diǎn)與坐標(biāo)原點(diǎn)重合，邊長(zhǎng)為SKIPIF1<0的直角邊落在SKIPIF1<0軸的正半軸時(shí)，則圓心為SKIPIF1<0，所以其外接圓方程可以為SKIPIF1<0，若將直角頂點(diǎn)與坐標(biāo)原點(diǎn)重合，邊長(zhǎng)為SKIPIF1<0的直角邊落在SKIPIF1<0軸的負(fù)半軸時(shí)，則圓心為SKIPIF1<0，所以其外接圓方程可以為SKIPIF1<0，若將直角頂點(diǎn)與坐標(biāo)原點(diǎn)重合，邊長(zhǎng)為SKIPIF1<0的直角邊落在SKIPIF1<0軸的正半軸時(shí)，則圓心為SKIPIF1<0，所以其外接圓方程可以為SKIPIF1<0，若將直角頂點(diǎn)與坐標(biāo)原點(diǎn)重合，邊長(zhǎng)為SKIPIF1<0的直角邊落在SKIPIF1<0軸的負(fù)半軸時(shí)，則圓心為SKIPIF1<0，所以其外接圓方程可以為SKIPIF1<0.故答案為：SKIPIF1<0.（答案不唯一）13．（2023·福建莆田·統(tǒng)考二模）直線l經(jīng)過(guò)點(diǎn)SKIPIF1<0，且與曲線SKIPIF1<0相切，寫(xiě)出l的一個(gè)方程_______．【答案】SKIPIF1<0（答案不唯一）【分析】先對(duì)SKIPIF1<0求導(dǎo)，再假設(shè)直線l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，斜率為SKIPIF1<0，從而得到關(guān)于SKIPIF1<0的方程組，解之即可求得直線l的方程.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，不妨設(shè)直線l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，斜率為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，直線l為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，直線l為SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，直線l為SKIPIF1<0，即SKIPIF1<0；綜上：直線l的方程為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）.14．（2023春·重慶沙坪壩·高三重慶八中?？茧A段練習(xí)）已知圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，圓SKIPIF1<0，請(qǐng)寫(xiě)出一條與圓SKIPIF1<0都相切的直線方程：_____________.(寫(xiě)一條即可)【答案】SKIPIF1<0(或SKIPIF1<0或SKIPIF1<0，答案不唯一，寫(xiě)一條即可)【分析】根據(jù)圓與直線對(duì)稱求得SKIPIF1<0，進(jìn)而判斷兩圓外切，從而確定公切線有三條.根據(jù)直線與圓相切的幾何條件建立方程從而可解.【詳解】因?yàn)閳ASKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，故圓心SKIPIF1<0在直線SKIPIF1<0上，得SKIPIF1<0，解得SKIPIF1<0，故圓SKIPIF1<0，圓心SKIPIF1<0半徑SKIPIF1<0而圓SKIPIF1<0的圓心SKIPIF1<0，半徑SKIPIF1<0所以兩圓的圓心距為SKIPIF1<0所以兩圓外切，公切線有三條.顯然公切線的斜率存在，設(shè)方程為SKIPIF1<0，于是有：SKIPIF1<0兩式相除得：SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0，代入SKIPIF1<0可解得SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，代入SKIPIF1<0可解得SKIPIF1<0，所以三條公切線方程分別為：SKIPIF1<0，SKIPIF1<0.故答案為：SKIPIF1<0(或SKIPIF1<0或SKIPIF1<0，答案不唯一，寫(xiě)一條即可)15．（2023春·廣東·高三校聯(lián)考階段練習(xí)）已知SKIPIF1<0、SKIPIF1<0，直線SKIPIF1<0上有且只有一個(gè)點(diǎn)SKIPIF1<0滿足SKIPIF1<0，寫(xiě)出滿足條件的其中一條直線SKIPIF1<0的方程__________．【答案】SKIPIF1<0（答案不唯一，只需滿足直線SKIPIF1<0與圓SKIPIF1<0相切即可）【分析】設(shè)點(diǎn)SKIPIF1<0，由SKIPIF1<0，求出點(diǎn)SKIPIF1<0的軌跡方程，可知點(diǎn)SKIPIF1<0的軌跡為圓，且圓心為SKIPIF1<0，半徑SKIPIF1<0，分析可知直線SKIPIF1<0與圓SKIPIF1<0相切即可.【詳解】設(shè)點(diǎn)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，整理可得SKIPIF1<0，即點(diǎn)SKIPIF1<0的軌跡為圓，且圓心為SKIPIF1<0，半徑SKIPIF1<0，直線SKIPIF1<0上有且只有一個(gè)點(diǎn)SKIPIF1<0滿足SKIPIF1<0，所以，直線SKIPIF1<0與圓SKIPIF1<0相切，所以，直線SKIPIF1<0的方程可為SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一，只需滿足直線SKIPIF1<0與圓SKIPIF1<0相切即可）.16．（2023·吉林白山·統(tǒng)考三模）寫(xiě)出一條與圓SKIPIF1<0和曲線SKIPIF1<0都相切的直線的方程：___________.【答案】SKIPIF1<0（答案不唯一）【分析】設(shè)切線SKIPIF1<0與圓SKIPIF1<0相切于點(diǎn)SKIPIF1<0，得到切線SKIPIF1<0的方程，與SKIPIF1<0聯(lián)立，由判別式為零求解.【詳解】解：設(shè)切線SKIPIF1<0與圓SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0，切線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，將SKIPIF1<0與SKIPIF1<0聯(lián)立，可得SKIPIF1<0，令SKIPIF1<0，聯(lián)立解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0所以切線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）17．（2023春·湖南長(zhǎng)沙·高三雅禮中學(xué)校考階段練習(xí)）已知SKIPIF1<0，寫(xiě)出滿足條件①②的一個(gè)SKIPIF1<0的值__________.①SKIPIF1<0；②SKIPIF1<0.【答案】SKIPIF1<0或11.（答案不唯一）【分析】令SKIPIF1<0，得到SKIPIF1<0，再由SKIPIF1<0求解.【詳解】解：令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，由條件②知SKIPIF1<0.又SKIPIF1<0的值可以為SKIPIF1<0或11.（答案不唯一）故答案為：SKIPIF1<0或11.（答案不唯一）18．（2022秋·遼寧撫順·高三校聯(lián)考階段練習(xí)）寫(xiě)出一個(gè)同時(shí)滿足下列三個(gè)性質(zhì)的函數(shù)：SKIPIF1<0__________.①SKIPIF1<0為奇函數(shù)；②SKIPIF1<0為偶函數(shù)；③SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0.【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)①②可知SKIPIF1<0是周期為4的周期函數(shù)，可根據(jù)三角函數(shù)的周期關(guān)系寫(xiě)出符合題意的函數(shù)形式.【詳解】由②可知SKIPIF1<0，由此可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0是周期為4的奇函數(shù)，SKIPIF1<0是周期為4的偶函數(shù)，因此不妨假設(shè)SKIPIF1<0，則SKIPIF1<0，由③可知SKIPIF1<0或SKIPIF1<0均可.故答案為：SKIPIF1<0（答案不唯一）19．（2022秋·山東濰坊·高三統(tǒng)考階段練習(xí)）已知函數(shù)SKIPIF1<0，試舉出一個(gè)SKIPIF1<0的值，使得SKIPIF1<0成立，則SKIPIF1<0可以為_(kāi)_________．（寫(xiě)出一個(gè)即可）【答案】-1或7【分析】根據(jù)給出的分段函數(shù)自變量的范圍,分情況討論計(jì)算即可.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0,可得當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0當(dāng)SKIPIF1<0且SKIPIF1<0時(shí),SKIPIF1<0與SKIPIF1<0矛盾,不合題意;當(dāng)SKIPIF1<0且SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0當(dāng)SKIPIF1<0時(shí),則SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0.故答案為:-1或7.20．（2022秋·江蘇徐州·高三期末）寫(xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)解析式為SKIPIF1<0__________.①不是常數(shù)函數(shù)；②SKIPIF1<0；③SKIPIF1<0.【答案】SKIPIF1<0（答案不唯一）【分析】首先理解條件中的性質(zhì)，再寫(xiě)出滿足條件的函數(shù).【詳解】因?yàn)镾KIPIF1<0，即SKIPIF1<0，所以函數(shù)是偶函數(shù)，因?yàn)镾KIPIF1<0，所以函數(shù)關(guān)于SKIPIF1<0對(duì)稱，且函數(shù)不是常函數(shù)，所以滿足條件的函數(shù)SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）21．（2023秋·遼寧沈陽(yáng)·高三沈陽(yáng)二中?？计谀?xiě)出與圓SKIPIF1<0和圓SKIPIF1<0都相切的一條直線的方程______.【答案】SKIPIF1<0（答案不唯一，寫(xiě)其它三條均可）【分析】先判斷兩圓的位置關(guān)系，可設(shè)公切線方程為SKIPIF1<0，根據(jù)圓心到直線的距離等于半徑列出方程組，解之即可得出答案.【詳解】解：圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為SKIPIF1<0，圓SKIPIF1<0的圓心為SKIPIF1<0，半徑為SKIPIF1<0，則SKIPIF1<0，所以兩圓外離，由兩圓的圓心都在SKIPIF1<0軸上，則公切線的斜率一定存在，設(shè)公切線方程為SKIPIF1<0，即SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0所以公切線方程為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0.（答案不唯一，寫(xiě)其它三條均可）22．（2022秋·福建·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0的部分圖象如圖所示，則滿足圖象的一個(gè)解析式為_(kāi)_____.【答案】SKIPIF1<0（答案不唯一）【分析】設(shè)函數(shù)解析式為SKIPIF1<0，根據(jù)函數(shù)得最值可求得SKIPIF1<0，再根據(jù)函數(shù)的對(duì)稱性結(jié)合圖象可得函數(shù)的最小正周期，從而可求得SKIPIF1<0，再利用待定系數(shù)法求得SKIPIF1<0即可.【詳解】解：設(shè)函數(shù)解析式為SKIPIF1<0，由圖可知SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，可取SKIPIF1<0，所以滿足圖象的一個(gè)解析式可以為SKIPIF1<0.故答案為：SKIPIF1<0.（答案不唯一）23．（2023秋·廣東清遠(yuǎn)·高三統(tǒng)考期末）已知P為雙曲線C：SKIPIF1<0上異于頂點(diǎn)SKIPIF1<0，SKIPIF1<0的任意一點(diǎn)，直線SKIPIF1<0，SKIPIF1<0的斜率分別為SKIPIF1<0，SKIPIF1<0，寫(xiě)出滿足C的焦距小于8且SKIPIF1<0的C的一個(gè)標(biāo)準(zhǔn)方程：_________.【答案】SKIPIF1<0（答案不唯一）【分析】首先設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，根據(jù)條件轉(zhuǎn)化為關(guān)于SKIPIF1<0的不等式組，再寫(xiě)出滿足條件的一個(gè)標(biāo)準(zhǔn)方程.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以滿足條件的雙曲線的標(biāo)準(zhǔn)方程是SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）24．（2023秋·山東濰坊·高三統(tǒng)考期末）寫(xiě)出一個(gè)同時(shí)滿足下列三個(gè)性質(zhì)的函數(shù)SKIPIF1<0______.①SKIPIF1<0是奇函數(shù)；②SKIPIF1<0在SKIPIF1<0單調(diào)遞增；③SKIPIF1<0有且僅有3個(gè)零點(diǎn).【答案】SKIPIF1<0（答案不唯一）【分析】根據(jù)奇函數(shù)圖像關(guān)于原點(diǎn)對(duì)稱，若函數(shù)有且僅有3個(gè)零點(diǎn)則原點(diǎn)兩側(cè)各有一個(gè)，再保證SKIPIF1<0單調(diào)遞增即可寫(xiě)出解析式.【詳解】由SKIPIF1<0是奇函數(shù)，不妨取SKIPIF1<0，且函數(shù)圖象關(guān)于原點(diǎn)對(duì)稱；又SKIPIF1<0有且僅有3個(gè)零點(diǎn)，所以原點(diǎn)兩側(cè)各有一個(gè)零點(diǎn)，且關(guān)于原點(diǎn)對(duì)稱，若保證SKIPIF1<0在SKIPIF1<0單調(diào)遞增，顯然SKIPIF1<0滿足.故答案為：SKIPIF1<0（答案不唯一）25．（2022秋·江蘇常州·高三?？计谥校┮阎瘮?shù)SKIPIF1<0SKIPIF1<0的最小值為0，且SKIPIF1<0，則SKIPIF1<0圖象的一個(gè)對(duì)稱中心的坐標(biāo)為_(kāi)_______．【答案】SKIPIF1<0(答案不唯一)【分析】由二倍角公式化簡(jiǎn)，根據(jù)三角函數(shù)性質(zhì)結(jié)合條件列式得SKIPIF1<0，再求對(duì)稱中心坐標(biāo).【詳解】由題意得SKIPIF1<0，所以SKIPIF1<0最小值為SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0圖象的對(duì)稱中心為SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0（答案不唯一）26．（2023秋·江蘇南通·高三統(tǒng)考期末）經(jīng)過(guò)坐標(biāo)原點(diǎn)的圓SKIPIF1<0與圓SKIPIF1<0相外切，則圓SKIPIF1<0的標(biāo)準(zhǔn)方程可以是__________SKIPIF1<0寫(xiě)出一個(gè)滿足題意的方程即可SKIPIF1<0【答案】SKIPIF1<0．（答案不唯一）【分析】根據(jù)題意易知圓SKIPIF1<0過(guò)坐標(biāo)原點(diǎn)，圓SKIPIF1<0與圓SKIPIF1<0的切點(diǎn)即為坐標(biāo)原點(diǎn)，則圓SKIPIF1<0的圓心在直線SKIPIF1<0上，且其圓心在第一象限，可設(shè)圓SKIPIF1<0的圓心坐標(biāo)為SKIPIF1<0，則可求得圓SKIPIF1<0的半徑，再根據(jù)圓的標(biāo)準(zhǔn)方程，即可求得結(jié)果．【詳解】設(shè)經(jīng)過(guò)坐標(biāo)原點(diǎn)的圓SKIPIF1<0圓心為SKIPIF1<0，半徑為SKIPIF1<0，則圓SKIPIF1<0方程：SKIPIF1<0，圓SKIPIF1<0經(jīng)過(guò)原點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，圓SKIPIF1<0：SKIPIF1<0可化為SKIPIF1<0，則圓SKIPIF1<0圓心為SKIPIF1<0，半徑SKIPIF1<0，顯然圓SKIPIF1<0經(jīng)過(guò)坐標(biāo)原點(diǎn)，由題意，圓SKIPIF1<0與圓SKIPIF1<0相外切，則圓SKIPIF1<0與圓SKIPIF1<0的切點(diǎn)即為坐標(biāo)原點(diǎn)，則圓SKIPIF1<0的圓心在直線SKIPIF1<0上，且圓心在第一象限，所以SKIPIF1<0，可令SKIPIF1<0，則圓SKIPIF1<0的圓心為SKIPIF1<0，則點(diǎn)SKIPIF1<0到圓SKIPIF1<0圓心SKIPIF1<0的距離SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，則圓SKIPIF1<0方程：SKIPIF1<0，故答案為：SKIPIF1<027．（2023·山西·統(tǒng)考一模）寫(xiě)出一個(gè)同時(shí)滿足下列三個(gè)條件的函數(shù)SKIPIF1<0的解析式______.①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.【答案】SKIPIF1<0（答案不唯一，滿足條件即可）【分析】根據(jù)題意得SKIPIF1<0圖像關(guān)于直線SKIPIF1<0對(duì)稱，點(diǎn)SKIPIF1<0對(duì)稱，進(jìn)而結(jié)合三角函數(shù)性質(zhì)和條件③求解即可.【詳解】解：由①SKIPIF1<0可知，函數(shù)SKIPIF1<0圖像關(guān)于直線SKIPIF1<0對(duì)稱；由②SKIPIF1<0可知函數(shù)SKIPIF1<0圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱；所以，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0的周期為SKIPIF1<0，故考慮余弦型函數(shù)，不妨令SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，滿足性質(zhì)①②，由③SKIPIF1<0在SKIPIF1<0上單調(diào)遞增可得SKIPIF1<0，故不妨取SKIPIF1<0，即SKIPIF1<0，此時(shí)滿足已知三個(gè)條件.故答案為：SKIPIF1<028．（2023春·重慶萬(wàn)州·高三重慶市萬(wàn)州第二高級(jí)中學(xué)?？茧A段練習(xí)）寫(xiě)出一個(gè)同時(shí)滿足下列條件①②的等比數(shù)列SKIPIF1<0的通項(xiàng)公式SKIPIF1<0__________.①SKIPIF1<0；②SKIPIF1<0【答案】SKIPIF1<0（答案不唯一）【分析】可構(gòu)造等比數(shù)列，設(shè)公比為SKIPIF1<0，由條件，可知公比SKIPIF1<0為負(fù)數(shù)且SKIPIF1<0，再取符合的值即可得解.【詳解】可構(gòu)造等比數(shù)列，設(shè)公比為SKIPIF1<0，由SKIPIF1<0，可知公比SKIPIF1<0為負(fù)數(shù)，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0可取SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0.