新高考數(shù)學(xué)二輪復(fù)習(xí)專題2-4 構(gòu)造函數(shù)以及切線（解析版）

專題2-4構(gòu)造函數(shù)以及切線歸類目錄TOC\o"1-1"\h\u題型01切線求參 1題型02求“過點(diǎn)”型切線方程 3題型03“過點(diǎn)”切線求參 5題型04“過點(diǎn)”切線條數(shù)的判斷 7題型05由切線條數(shù)求參 8題型06公切線 10題型07特殊構(gòu)造：冪積型構(gòu)造 12題型08特殊構(gòu)造：冪商型構(gòu)造 15題型09特殊構(gòu)造：ex的積型構(gòu)造 16題型10特殊構(gòu)造：ex的商型構(gòu)造 18題型11特殊構(gòu)造：對(duì)數(shù)型構(gòu)造 21題型12特殊構(gòu)造：正弦型構(gòu)造 23題型13特殊構(gòu)造：余弦型構(gòu)造 26題型14復(fù)合型構(gòu)造 28高考練場(chǎng) 30題型01切線求參【解題攻略】求曲線y＝f(x)在點(diǎn)P(x0，f(x0))處的切線方程：(1)求出函數(shù)y＝f(x)在點(diǎn)x＝x0處的導(dǎo)數(shù)，即曲線y＝f(x)在點(diǎn)P(x0，f(x0))處切線的斜率．(2)切線方程為：y＝y(tǒng)0＋f′(x0)(x－x0)．1、設(shè)切點(diǎn)（或者給出了切點(diǎn)）：P(x0，y0)2、y0=f(x0)3、y＝f′(x)SKIPIF1<0k=f′(x0)4、切線方程：y-y0＝k(x－x0)【典例1-1】（2023春·重慶·高二校聯(lián)考期中）若函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．2或SKIPIF1<0 C．2 D．1或SKIPIF1<0【答案】B【分析】由兩線垂直可知SKIPIF1<0處切線的斜率為5，利用導(dǎo)數(shù)的幾何意義有SKIPIF1<0，即可求SKIPIF1<0的值.【詳解】由題意知：直線SKIPIF1<0的斜率為SKIPIF1<0，則在SKIPIF1<0處切線的斜率為5，又∵SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故選：B．【典例1-2】(山東省煙臺(tái)市2021-2022學(xué)年高三數(shù)學(xué)試題)已知曲線SKIPIF1<0在點(diǎn)（0，1）處的切線與曲線SKIPIF1<0只有一個(gè)公共點(diǎn)，則實(shí)數(shù)a的值為（

）A．SKIPIF1<0 B．1 C．2 D．SKIPIF1<0【答案】A【分析】先求出SKIPIF1<0導(dǎo)數(shù)，求得切線的斜率，可得切線方程，再由切線與曲線SKIPIF1<0只有一個(gè)公共點(diǎn)，進(jìn)而聯(lián)立得到SKIPIF1<0的值.【詳解】SKIPIF1<0的導(dǎo)數(shù)SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處切線斜率SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處切線方程為SKIPIF1<0，即SKIPIF1<0由于切線與曲線SKIPIF1<0只有一個(gè)公共點(diǎn)，聯(lián)立SKIPIF1<0，得SKIPIF1<0即SKIPIF1<0解得SKIPIF1<0故選:A.【變式1-1】（河南省鄭州市2021-2022學(xué)年高三考試數(shù)學(xué)（理科）試題）若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，則SKIPIF1<0___________.【答案】SKIPIF1<0【分析】令SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義得出SKIPIF1<0的值.【詳解】令SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0又該函數(shù)在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，所以SKIPIF1<0故答案為：SKIPIF1<0【變式1-2】（河南省許昌市2021-2022學(xué)年高三數(shù)學(xué)文科試題）已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義可得SKIPIF1<0，根據(jù)切點(diǎn)坐標(biāo)可得SKIPIF1<0，列方程求解．【詳解】SKIPIF1<0，則SKIPIF1<0∵SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0∴可得SKIPIF1<0，解得SKIPIF1<0則SKIPIF1<0故答案為：SKIPIF1<0．【變式1-3】已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象過定點(diǎn)SKIPIF1<0，若曲線SKIPIF1<0在SKIPIF1<0處的切線經(jīng)過點(diǎn)SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為______．【答案】SKIPIF1<0##0.5【分析】先求出SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）所經(jīng)過的定點(diǎn)SKIPIF1<0的坐標(biāo)，然后根據(jù)導(dǎo)數(shù)的幾何意義求出SKIPIF1<0在SKIPIF1<0處的切線方程，最后把點(diǎn)SKIPIF1<0的坐標(biāo)代入切線方程，即可得SKIPIF1<0值．【詳解】函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象恒過點(diǎn)SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0在SKIPIF1<0處的切線的斜率為SKIPIF1<0，又SKIPIF1<0，所以切線方程為SKIPIF1<0，因?yàn)榍芯€經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0題型02求“過點(diǎn)”型切線方程【解題攻略】1、設(shè)切點(diǎn)（或者給出了切點(diǎn)）：P(x0，y0)2、y0=f(x0)3、y＝f′(x)SKIPIF1<0k=f′(x0)4、切線方程：y-y0＝k(x－x0)5、過（a,b）,代入y-y0＝k(x－x0)，得SKIPIF1<0【典例1-1】（2023下·上海嘉定·高三上海市嘉定區(qū)第一中學(xué)校考）已知曲線SKIPIF1<0，過點(diǎn)SKIPIF1<0作曲線的切線，則切線方程．【答案】SKIPIF1<0【分析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，求出切線方程，代入點(diǎn)SKIPIF1<0求出SKIPIF1<0，從而可得切線方程.【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.因?yàn)榍芯€過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.所以切線方程為SKIPIF1<0.故答案為:SKIPIF1<0.【典例1-2】（2023下·上海浦東新·高三上海市實(shí)驗(yàn)學(xué)校校考開學(xué)考試）已知曲線SKIPIF1<0，過點(diǎn)SKIPIF1<0作曲線的切線，則切線的方程為.【答案】SKIPIF1<0【分析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，根據(jù)切線所過的點(diǎn)得到SKIPIF1<0的方程，解出SKIPIF1<0后可得所求的切線方程.【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0,則切線的斜率SKIPIF1<0，故切線方程為SKIPIF1<0，又因?yàn)辄c(diǎn)SKIPIF1<0在切線上，所以SKIPIF1<0SKIPIF1<0，整理得到SKIPIF1<0，解得SKIPIF1<0，所以切線方程為SKIPIF1<0．故答案為:SKIPIF1<0.【變式1-1】）（云南民族大學(xué)附屬中學(xué)2022屆高三高考押題卷二數(shù)學(xué)（理）試題）函數(shù)SKIPIF1<0過原點(diǎn)的切線方程是_______.【答案】SKIPIF1<0.【分析】設(shè)切點(diǎn)為SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義求出函數(shù)切點(diǎn)為SKIPIF1<0的切線方程，再根據(jù)切線過原點(diǎn)求出SKIPIF1<0，即可得解.【詳解】解：設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故切點(diǎn)為SKIPIF1<0的切線方程為SKIPIF1<0，又因此切線過原點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0過原點(diǎn)的切線方程是SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.【變式1-2】（2023春·河北邢臺(tái)·高三統(tǒng)考）過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，則該切線的斜率為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)切點(diǎn)為SKIPIF1<0，然后表示出切線方程，再將SKIPIF1<0代入可求出SKIPIF1<0，然后將SKIPIF1<0代入導(dǎo)函數(shù)中可求得結(jié)果.【詳解】設(shè)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0所以切線方程為SKIPIF1<0，即SKIPIF1<0，將SKIPIF1<0代入得SKIPIF1<0，解得SKIPIF1<0，所以切線的斜率為SKIPIF1<0.故選：C【變式1-3】（（天津市北京師范大學(xué)天津附屬中學(xué)2022-2023學(xué)年高三線上檢測(cè)數(shù)學(xué)試題））過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，則切線方程是__________．【答案】SKIPIF1<0【分析】求解導(dǎo)函數(shù)，設(shè)切點(diǎn)坐標(biāo)，求解SKIPIF1<0，從而設(shè)出切線方程，代入點(diǎn)SKIPIF1<0計(jì)算，即可求出答案.【詳解】函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，所以切線方程為SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，解得：SKIPIF1<0，所以切線方程為SKIPIF1<0，整理得：SKIPIF1<0．故答案為：SKIPIF1<0.題型03“過點(diǎn)”切線求參【典例1-1】（2023上·遼寧錦州·高三渤海大學(xué)附屬高級(jí)中學(xué)?？计谥校┮阎€SKIPIF1<0過點(diǎn)SKIPIF1<0處的切線與曲線SKIPIF1<0相切，則SKIPIF1<0【答案】8【分析】設(shè)切點(diǎn)SKIPIF1<0，并應(yīng)用導(dǎo)數(shù)幾何意義求可得切線為SKIPIF1<0，將切點(diǎn)代入求得SKIPIF1<0得切線方程，再由切線與曲線SKIPIF1<0相切，討論參數(shù)a，聯(lián)立方程有SKIPIF1<0求參數(shù).【詳解】設(shè)過點(diǎn)SKIPIF1<0處的切線在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，而SKIPIF1<0，故切線斜率為SKIPIF1<0，所以切線方程為SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故切線方程為SKIPIF1<0，又切線與曲線SKIPIF1<0相切，聯(lián)立方程，得SKIPIF1<0有且僅有一個(gè)解，當(dāng)SKIPIF1<0時(shí)上述方程無解；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0.綜上，SKIPIF1<0.故答案為：SKIPIF1<0【典例1-2】（2023下·吉林長(zhǎng)春·高二長(zhǎng)春市實(shí)驗(yàn)中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，過點(diǎn)SKIPIF1<0作與SKIPIF1<0軸平行的直線交函數(shù)SKIPIF1<0的圖象于點(diǎn)SKIPIF1<0，過點(diǎn)SKIPIF1<0作SKIPIF1<0的切線交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，則SKIPIF1<0面積的最小值.【答案】SKIPIF1<0【分析】求出SKIPIF1<0的導(dǎo)數(shù)，令x＝a，求得P的坐標(biāo)，可得切線的斜率，運(yùn)用點(diǎn)斜式方程可得切線的方程，令y＝0，可得B的坐標(biāo)，再由三角形的面積公式可得△ABP面積S，求出導(dǎo)數(shù)，利用導(dǎo)數(shù)求最值，即可得到所求值．【詳解】函SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，由題意可令SKIPIF1<0，解得SKIPIF1<0，可得SKIPIF1<0，即有切線的斜率為SKIPIF1<0，切線的方程為SKIPIF1<0,令SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，在直角三角形PAB中，SKIPIF1<0，SKIPIF1<0，則△ABP面積為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，即有SKIPIF1<0處S取得極小值，且為最小值SKIPIF1<0．故答案為：SKIPIF1<0．【變式1-1】（2023·河北保定·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，過點(diǎn)SKIPIF1<0且平行于SKIPIF1<0軸的直線與曲線SKIPIF1<0的交點(diǎn)為SKIPIF1<0，曲線SKIPIF1<0過點(diǎn)SKIPIF1<0的切線交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，則SKIPIF1<0面積的最小值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由已知求得SKIPIF1<0點(diǎn)坐標(biāo)，利用導(dǎo)數(shù)求出過SKIPIF1<0點(diǎn)的切線方程，再求出SKIPIF1<0點(diǎn)坐標(biāo)，寫出三角形SKIPIF1<0的面積，再由導(dǎo)數(shù)求最值得答案．【詳解】SKIPIF1<0，把SKIPIF1<0代入，可得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0,SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0曲線SKIPIF1<0過點(diǎn)SKIPIF1<0的切線方程為SKIPIF1<0，取SKIPIF1<0，得SKIPIF1<0．SKIPIF1<0SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0．則SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0（舍），SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．故選：D．【變式1-2】（2023上·貴州貴陽(yáng)·高三貴陽(yáng)一中?？茧A段練習(xí)）已知曲線SKIPIF1<0，過點(diǎn)SKIPIF1<0作該曲線的兩條切線，切點(diǎn)分別為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】D【分析】求得切線方程為SKIPIF1<0，根據(jù)題意，轉(zhuǎn)化為關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不同的解SKIPIF1<0，結(jié)合二次函數(shù)的性質(zhì)，即可求解.【詳解】由函數(shù)SKIPIF1<0，可得SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，所以切線方程為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)檫^點(diǎn)SKIPIF1<0作該曲線的兩條切線，所以關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不同的解SKIPIF1<0，即關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不同的解SKIPIF1<0，所以SKIPIF1<0.故選：D.【變式1-3】.直線SKIPIF1<0是曲線SKIPIF1<0的切線，則SKIPIF1<0______.【答案】SKIPIF1<0【分析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，利用導(dǎo)數(shù)寫出切線的方程，與直線方程SKIPIF1<0對(duì)比，可出關(guān)于SKIPIF1<0、SKIPIF1<0的方程，解之即可.【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，其中SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以，切線斜率為SKIPIF1<0，所以，曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.題型04“過點(diǎn)”切線條數(shù)的判斷【解題攻略】”過點(diǎn)型“切線條數(shù)判斷：有幾個(gè)切點(diǎn)橫坐標(biāo)，就有幾條切線。切線條數(shù)判斷，轉(zhuǎn)化為關(guān)于切點(diǎn)橫坐標(biāo)的新的函數(shù)零點(diǎn)個(gè)數(shù)判斷。【典例1-1】.（湖南省邵陽(yáng)市武岡市2022-2023學(xué)年高三上學(xué)期數(shù)學(xué)試題）已知SKIPIF1<0是奇函數(shù)，則過點(diǎn)SKIPIF1<0向曲線SKIPIF1<0可作的切線條數(shù)是（

）A．1 B．2 C．3 D．不確定【答案】C【分析】根據(jù)給定條件，求出a，再求出函數(shù)SKIPIF1<0的導(dǎo)數(shù)，設(shè)出切點(diǎn)坐標(biāo)，借助導(dǎo)數(shù)的幾何意義列出方程求解作答.【詳解】因函數(shù)SKIPIF1<0是奇函數(shù)，則由SKIPIF1<0得SKIPIF1<0恒成立，則SKIPIF1<0，即有SKIPIF1<0，SKIPIF1<0，設(shè)過點(diǎn)SKIPIF1<0向曲線SKIPIF1<0所作切線與曲線SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0，而點(diǎn)SKIPIF1<0不在曲線SKIPIF1<0上，則SKIPIF1<0，整理得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即符合條件的切點(diǎn)有3個(gè)，所以過點(diǎn)SKIPIF1<0向曲線SKIPIF1<0可作的切線條數(shù)是3.故選：C【典例1-2】已知曲線SKIPIF1<0，則過點(diǎn)SKIPIF1<0可向SKIPIF1<0引切線，其切線條數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)切點(diǎn)為SKIPIF1<0，利用導(dǎo)數(shù)求出曲線SKIPIF1<0在切點(diǎn)SKIPIF1<0處的切線方程，再將點(diǎn)SKIPIF1<0的坐標(biāo)代入切線方程，可得出關(guān)于SKIPIF1<0的方程，解出該方程，得出該方程根的個(gè)數(shù)，即為所求.【詳解】設(shè)在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0的坐標(biāo)代入切線方程得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.因此，過點(diǎn)SKIPIF1<0可向SKIPIF1<0引切線，有三條.故選：C.【變式1-1】（湖南省長(zhǎng)沙市長(zhǎng)郡中學(xué)2021屆高三第一次暑假作業(yè)檢測(cè)數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0，過點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0切線的條數(shù)為A．0 B．1 C．2 D．3【答案】C【分析】設(shè)出切點(diǎn)坐標(biāo)，根據(jù)導(dǎo)數(shù)的幾何意義及切線所過點(diǎn)求出切點(diǎn)個(gè)數(shù)，從而可得答案.【詳解】設(shè)切點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，整理得SKIPIF1<0；令SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為單調(diào)遞增函數(shù)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為單調(diào)遞減函數(shù)；所以SKIPIF1<0；又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0有兩個(gè)不同的根，即切線的條數(shù)為2，故選：C.【變式1-2】（2021-2022學(xué)年廣東省東莞市高三數(shù)學(xué)A卷）已知函數(shù)SKIPIF1<0，則過點(diǎn)（0，0）可作曲線SKIPIF1<0的切線的條數(shù)為(

)A．3 B．0 C．1 D．2【答案】D【分析】分析可得SKIPIF1<0不是切點(diǎn)，設(shè)切點(diǎn)SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義，求得切線的斜率k，根據(jù)點(diǎn)P和點(diǎn)SKIPIF1<0坐標(biāo)，可求得切線斜率k，聯(lián)立即可得答案.【詳解】∵點(diǎn)SKIPIF1<0不在函數(shù)SKIPIF1<0的圖象上，∴點(diǎn)SKIPIF1<0不是切點(diǎn)，設(shè)切點(diǎn)為SKIPIF1<0（SKIPIF1<0），由SKIPIF1<0，可得SKIPIF1<0，則切線的斜率SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故切線有2條.故選:D．【變式1-3】（北京市北京理工大學(xué)附屬中學(xué)通州校區(qū)2019-2020學(xué)年高三年級(jí)考試數(shù)學(xué)試題）已知過點(diǎn)SKIPIF1<0且與曲線SKIPIF1<0相切的直線的條數(shù)有（

）條.A．0 B．1 C．2 D．3【答案】C【分析】設(shè)出切點(diǎn)的坐標(biāo)，然后根據(jù)導(dǎo)數(shù)的幾何意義求出曲線SKIPIF1<0的切線，根據(jù)切線過點(diǎn)SKIPIF1<0，結(jié)合關(guān)于切點(diǎn)橫坐標(biāo)的方程解的個(gè)數(shù)進(jìn)行求解即可.【詳解】設(shè)曲線SKIPIF1<0的切點(diǎn)的坐標(biāo)為SKIPIF1<0，由SKIPIF1<0，因此該曲線切線的斜率為SKIPIF1<0，所以該曲線切線的方程為：SKIPIF1<0，該切線過點(diǎn)SKIPIF1<0，所以有SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，因此過點(diǎn)SKIPIF1<0且與曲線SKIPIF1<0相切的直線的條數(shù)有2條.故選：C題型05由切線條數(shù)求參【典例1-1】若過點(diǎn)SKIPIF1<0可作出曲線SKIPIF1<0的三條切線，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________【答案】SKIPIF1<0【分析】根據(jù)函數(shù)切線的求解方法，設(shè)切點(diǎn)求切線方程，代入點(diǎn)SKIPIF1<0，根據(jù)方程與函數(shù)的關(guān)系，將問題轉(zhuǎn)化為兩個(gè)函數(shù)求交點(diǎn)問題，利用導(dǎo)數(shù)，作圖，可得答案.【詳解】由已知，曲線SKIPIF1<0，即令SKIPIF1<0，則SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，切線方程的斜率為SKIPIF1<0，所以切線方程為：SKIPIF1<0，將點(diǎn)SKIPIF1<0代入方程得：SKIPIF1<0，整理得SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，過點(diǎn)SKIPIF1<0可作出曲線SKIPIF1<0的三條切線，可知兩個(gè)函數(shù)圖像SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，又因?yàn)镾KIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，則當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0的極大值為SKIPIF1<0，函數(shù)SKIPIF1<0的極小值為SKIPIF1<0，如圖所示，當(dāng)SKIPIF1<0時(shí)，兩個(gè)函數(shù)圖像有三個(gè)不同的交點(diǎn)．故答案為：SKIPIF1<0．【典例1-2】（福建省福州華僑中學(xué)2023屆高三上學(xué)期第二次考試數(shù)學(xué)試題）若曲線SKIPIF1<0有兩條過坐標(biāo)原點(diǎn)的切線，則a的取值范圍為__________．【答案】SKIPIF1<0或SKIPIF1<0.【分析】設(shè)切點(diǎn)坐標(biāo)，根據(jù)導(dǎo)數(shù)的幾何意義得到SKIPIF1<0，再根據(jù)曲線SKIPIF1<0有兩條過坐標(biāo)原點(diǎn)的切線得到方程SKIPIF1<0有兩個(gè)解，讓SKIPIF1<0，解不等式即可.【詳解】由SKIPIF1<0得SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0，因?yàn)榍€SKIPIF1<0有兩條過坐標(biāo)原點(diǎn)的切線，所以方程SKIPIF1<0有兩個(gè)解，故SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0.【變式1-1】過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，若切線有且只有兩條，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【分析】利用導(dǎo)數(shù)幾何意義，求得切線方程，根據(jù)該方程過點(diǎn)SKIPIF1<0，且方程有兩個(gè)根，再構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)，即得.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，設(shè)切點(diǎn)為(SKIPIF1<0)，SKIPIF1<0，所以切線方程為SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0這個(gè)關(guān)于SKIPIF1<0的方程有兩個(gè)解，令SKIPIF1<0(SKIPIF1<0)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有最大值，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.【變式1-2】若曲線SKIPIF1<0有兩條過坐標(biāo)原點(diǎn)的切線，則a的取值范圍是________________．【答案】SKIPIF1<0【分析】設(shè)出切點(diǎn)橫坐標(biāo)SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義求得切線方程，根據(jù)切線經(jīng)過原點(diǎn)得到關(guān)于SKIPIF1<0的方程，根據(jù)此方程應(yīng)有兩個(gè)不同的實(shí)數(shù)根，求得SKIPIF1<0的取值范圍.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0,則SKIPIF1<0,切線斜率SKIPIF1<0,切線方程為：SKIPIF1<0,∵切線過原點(diǎn)，∴SKIPIF1<0,整理得：SKIPIF1<0,∵切線有兩條，∴SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,∴SKIPIF1<0的取值范圍是SKIPIF1<0,故答案為：SKIPIF1<0【變式1-3】（2023·全國(guó)·高三專題練習(xí)）已知過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線有且僅有兩條，則實(shí)數(shù)a的取值可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】設(shè)切線切點(diǎn)為SKIPIF1<0，后由切線幾何意義可得切線方程，代入SKIPIF1<0，可得SKIPIF1<0，則過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線有且僅有兩條，等價(jià)于關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不同實(shí)根，即可得答案.【詳解】設(shè)切線切點(diǎn)為SKIPIF1<0，因SKIPIF1<0，則切線方程為：SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，因SKIPIF1<0，則SKIPIF1<0.因過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線有且僅有兩條，則SKIPIF1<0有且僅有兩個(gè)不等實(shí)根，則SKIPIF1<0或SKIPIF1<0.則SKIPIF1<0符合題意.故選：D題型06公切線【解題攻略】交點(diǎn)處公切線，可以直接參照直線在點(diǎn)處的切線求法設(shè)交點(diǎn)（切點(diǎn)）對(duì)函數(shù)SKIPIF1<0，如果要求它們的圖象的公切線，只需分別寫出兩條切線：SKIPIF1<0)

和SKIPIF1<0再令

SKIPIF1<0

，消去一個(gè)變量后，再討論得到的方程的根的個(gè)數(shù)即可。但在這里需要注意

x1

和

x2

的范圍，例如，若f（x）=lnx，則要求

x1>0

【典例1-1】已知直線SKIPIF1<0是函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的公切線，若SKIPIF1<0是直線SKIPIF1<0與函數(shù)SKIPIF1<0相切的切點(diǎn)，則SKIPIF1<0____________．【答案】SKIPIF1<0【分析】求出導(dǎo)函數(shù)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得切線方程SKIPIF1<0，設(shè)SKIPIF1<0圖象上的切點(diǎn)為SKIPIF1<0，由導(dǎo)數(shù)幾何意義得切線方程，兩直線重合求得SKIPIF1<0，從而得SKIPIF1<0值．【詳解】SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以切線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，設(shè)直線SKIPIF1<0與SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0，SKIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0．【典例1-2】（2023春·高三課時(shí)練習(xí)）已知直線SKIPIF1<0：SKIPIF1<0既是曲線SKIPIF1<0的切線，又是曲線SKIPIF1<0的切線，則SKIPIF1<0（

）A．0 B．SKIPIF1<0 C．0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【分析】本題主要求切線方程，設(shè)兩個(gè)曲線方程的切點(diǎn)，由兩條切線均為SKIPIF1<0，通過等量關(guān)系可得到SKIPIF1<0的取值.【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0，設(shè)切點(diǎn)分別為SKIPIF1<0，則曲線SKIPIF1<0的切線方程為：SKIPIF1<0，化簡(jiǎn)得，SKIPIF1<0，曲線SKIPIF1<0的切線方程為：SKIPIF1<0，化簡(jiǎn)得，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0e或SKIPIF1<0.當(dāng)SKIPIF1<0e，切線方程為SKIPIF1<0，故SKIPIF1<0SKIPIF1<0SKIPIF1<0.當(dāng)SKIPIF1<0，切線方程為SKIPIF1<0，故SKIPIF1<0，則SKIPIF1<0.故SKIPIF1<0的取值為SKIPIF1<0或SKIPIF1<0.故選：D【變式1-1】（2023·全國(guó)·高三專題練習(xí)）若直線SKIPIF1<0是曲線SKIPIF1<0的切線，也是SKIPIF1<0的切線，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】設(shè)直線SKIPIF1<0與SKIPIF1<0和SKIPIF1<0的切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，分別求出切點(diǎn)處的直線方程，由已知切線方程，可得方程組，解方程可得切點(diǎn)的橫坐標(biāo)，即可得到SKIPIF1<0的值.【詳解】設(shè)直線SKIPIF1<0與SKIPIF1<0和SKIPIF1<0的切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，則切線方程分別為，SKIPIF1<0，SKIPIF1<0，化簡(jiǎn)得，SKIPIF1<0SKIPIF1<0依題意上述兩直線與SKIPIF1<0是同一條直線，所以，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．故選：C．【變式1-2】（2022·黑龍江哈爾濱·哈爾濱三中?？寄M預(yù)測(cè)）曲線SKIPIF1<0過點(diǎn)SKIPIF1<0的切線也是曲線SKIPIF1<0的切線，則SKIPIF1<0；若此公切線恒在函數(shù)SKIPIF1<0的圖象上方，則a的取值范圍是.【答案】SKIPIF1<0SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義可求出SKIPIF1<0；將此公切線恒在函數(shù)SKIPIF1<0的圖象上方，轉(zhuǎn)化為SKIPIF1<0恒成立，再構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)求出最小值即可得解.【詳解】由SKIPIF1<0得SKIPIF1<0，設(shè)曲線SKIPIF1<0過點(diǎn)SKIPIF1<0的切線的切點(diǎn)為SKIPIF1<0，則切線的斜率為SKIPIF1<0，切線方程為SKIPIF1<0，由于該切線過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，設(shè)該切線與曲線SKIPIF1<0切于SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以該切線的斜率為SKIPIF1<0，所以切線方程為SKIPIF1<0，將SKIPIF1<0代入得SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0.由以上可知該公切線方程為SKIPIF1<0，即SKIPIF1<0，若此公切線恒在函數(shù)SKIPIF1<0的圖象上方，則SKIPIF1<0，即SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，因?yàn)镾KIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0.所以SKIPIF1<0.【變式1-3】若曲線SKIPIF1<0與曲線SKIPIF1<0存在2條公共切線，則a的值是_________．【答案】SKIPIF1<0【分析】設(shè)公切線在SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，在SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義求出對(duì)應(yīng)的切線方程，有SKIPIF1<0，整理得SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)研究SKIPIF1<0的單調(diào)性，結(jié)合圖像即可得出結(jié)果.【詳解】設(shè)公切線在SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，在SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則曲線在切點(diǎn)的切線方程的斜率分別為SKIPIF1<0，SKIPIF1<0，對(duì)應(yīng)的切線方程分別為SKIPIF1<0、SKIPIF1<0，即SKIPIF1<0、SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，有SKIPIF1<0，則SKIPIF1<0，整理，得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，令SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，因?yàn)閮蓷l曲線有2條公共切線，所以函數(shù)SKIPIF1<0與SKIPIF1<0圖像有兩個(gè)交點(diǎn)，又SKIPIF1<0，且SKIPIF1<0，如圖，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0..題型07特殊構(gòu)造：冪積型構(gòu)造【解題攻略】?jī)绾瘮?shù)積形式構(gòu)造：1.對(duì)于SKIPIF1<0構(gòu)造SKIPIF1<02.對(duì)于SKIPIF1<0構(gòu)造SKIPIF1<0【典例1-1】設(shè)定義在SKIPIF1<0的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且滿足SKIPIF1<0，則關(guān)于x的不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】構(gòu)造函數(shù)SKIPIF1<0,再根據(jù)題意分析SKIPIF1<0的單調(diào)性,再化簡(jiǎn)SKIPIF1<0可得SKIPIF1<0,再利用函數(shù)的單調(diào)性與定義域求解即可.解：令SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,故選：A.【典例1-2】已知定義域?yàn)镾KIPIF1<0的奇函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的大小關(guān)系正確的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用條件構(gòu)造函數(shù)SKIPIF1<0，然后利用導(dǎo)數(shù)研究函數(shù)SKIPIF1<0的單調(diào)性，利用函數(shù)的單調(diào)性比較大小．【詳解】解：根據(jù)題意，設(shè)SKIPIF1<0，若SKIPIF1<0為奇函數(shù)，則SKIPIF1<0，則函數(shù)SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，又由當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，SKIPIF1<0，SKIPIF1<0（2），SKIPIF1<0，且SKIPIF1<0，則有SKIPIF1<0；故選SKIPIF1<0．【變式1-1】已知定義在R上的偶函數(shù)SKIPIF1<0，其導(dǎo)函數(shù)為SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，恒有SKIPIF1<0，若SKIPIF1<0，則不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0也為偶函數(shù)，利用導(dǎo)數(shù)可以判斷SKIPIF1<0在SKIPIF1<0為減函數(shù)，則不等式SKIPIF1<0可轉(zhuǎn)化為SKIPIF1<0，解不等式即可得到答案．【詳解】解：SKIPIF1<0SKIPIF1<0是定義在R上的偶函數(shù)，SKIPIF1<0SKIPIF1<0．SKIPIF1<0SKIPIF1<0時(shí)，恒有SKIPIF1<0，SKIPIF1<0SKIPIF1<0又SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0為減函數(shù)．SKIPIF1<0SKIPIF1<0為偶函數(shù)，SKIPIF1<0SKIPIF1<0也為偶函數(shù)SKIPIF1<0SKIPIF1<0在SKIPIF1<0為增函數(shù)．又SKIPIF1<0SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，得SKIPIF1<0．故選A．【變式1-2】.已知奇函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的可導(dǎo)函數(shù)，其導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，則不等式SKIPIF1<0的解集為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．S