新高考數(shù)學(xué)二輪復(fù)習(xí)專(zhuān)題2-4 構(gòu)造函數(shù)以及切線(xiàn)（原卷版）

專(zhuān)題2-4構(gòu)造函數(shù)以及切線(xiàn)歸類(lèi)目錄TOC\o"1-1"\h\u題型01切線(xiàn)求參 1題型02求“過(guò)點(diǎn)”型切線(xiàn)方程 2題型03“過(guò)點(diǎn)”切線(xiàn)求參 3題型04“過(guò)點(diǎn)”切線(xiàn)條數(shù)的判斷 3題型05由切線(xiàn)條數(shù)求參 4題型06公切線(xiàn) 4題型07特殊構(gòu)造：冪積型構(gòu)造 5題型08特殊構(gòu)造：冪商型構(gòu)造 6題型09特殊構(gòu)造：ex的積型構(gòu)造 6題型10特殊構(gòu)造：ex的商型構(gòu)造 7題型11特殊構(gòu)造：對(duì)數(shù)型構(gòu)造 8題型12特殊構(gòu)造：正弦型構(gòu)造 9題型13特殊構(gòu)造：余弦型構(gòu)造 10題型14復(fù)合型構(gòu)造 11高考練場(chǎng) 12題型01切線(xiàn)求參【解題攻略】求曲線(xiàn)y＝f(x)在點(diǎn)P(x0，f(x0))處的切線(xiàn)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(xiàn)y＝f(x)在點(diǎn)P(x0，f(x0))處切線(xiàn)的斜率．(2)切線(xiàn)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(xiàn)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(xiàn)與直線(xiàn)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【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(xiàn)SKIPIF1<0在點(diǎn)（0，1）處的切線(xiàn)與曲線(xiàn)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【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(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)與直線(xiàn)SKIPIF1<0平行，則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(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程為SKIPIF1<0，則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(guò)定點(diǎn)SKIPIF1<0，若曲線(xiàn)SKIPIF1<0在SKIPIF1<0處的切線(xiàn)經(jīng)過(guò)點(diǎn)SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的值為_(kāi)_____．題型02求“過(guò)點(diǎn)”型切線(xiàn)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(xiàn)方程：y-y0＝k(x－x0)5、過(guò)（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)?？迹┮阎€(xiàn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)的切線(xiàn)，則切線(xiàn)方程．【典例1-2】（2023下·上海浦東新·高三上海市實(shí)驗(yàn)學(xué)校?？奸_(kāi)學(xué)考試）已知曲線(xiàn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)的切線(xiàn)，則切線(xiàn)的方程為.【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(guò)原點(diǎn)的切線(xiàn)方程是_______.【變式1-2】（2023春·河北邢臺(tái)·高三統(tǒng)考）過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)SKIPIF1<0的切線(xiàn)，則該切線(xiàn)的斜率為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【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(xiàn)上檢測(cè)數(shù)學(xué)試題））過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)SKIPIF1<0的切線(xiàn)，則切線(xiàn)方程是__________．.題型03“過(guò)點(diǎn)”切線(xiàn)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€(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0處的切線(xiàn)與曲線(xiàn)SKIPIF1<0相切，則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)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作與SKIPIF1<0軸平行的直線(xiàn)交函數(shù)SKIPIF1<0的圖象于點(diǎn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作SKIPIF1<0的切線(xiàn)交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，則SKIPIF1<0面積的最小值.【變式1-1】（2023·河北保定·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0且平行于SKIPIF1<0軸的直線(xiàn)與曲線(xiàn)SKIPIF1<0的交點(diǎn)為SKIPIF1<0，曲線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0的切線(xiàn)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【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(xiàn)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作該曲線(xiàn)的兩條切線(xiàn)，切點(diǎn)分別為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【變式1-3】.直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的切線(xiàn)，則SKIPIF1<0______.題型04“過(guò)點(diǎn)”切線(xiàn)條數(shù)的判斷【解題攻略】“過(guò)點(diǎn)型”切線(xiàn)條數(shù)判斷：有幾個(gè)切點(diǎn)橫坐標(biāo)，就有幾條切線(xiàn)。切線(xiàn)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?！镜淅?-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(guò)點(diǎn)SKIPIF1<0向曲線(xiàn)SKIPIF1<0可作的切線(xiàn)條數(shù)是（

）A．1 B．2 C．3 D．不確定【典例1-2】已知曲線(xiàn)SKIPIF1<0，則過(guò)點(diǎn)SKIPIF1<0可向SKIPIF1<0引切線(xiàn)，其切線(xiàn)條數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【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(guò)點(diǎn)SKIPIF1<0可作曲線(xiàn)SKIPIF1<0切線(xiàn)的條數(shù)為A．0 B．1 C．2 D．3【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(guò)點(diǎn)（0，0）可作曲線(xiàn)SKIPIF1<0的切線(xiàn)的條數(shù)為(

)A．3 B．0 C．1 D．2【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(guò)點(diǎn)SKIPIF1<0且與曲線(xiàn)SKIPIF1<0相切的直線(xiàn)的條數(shù)有（

）條.A．0 B．1 C．2 D．3題型05由切線(xiàn)條數(shù)求參【典例1-1】若過(guò)點(diǎn)SKIPIF1<0可作出曲線(xiàn)SKIPIF1<0的三條切線(xiàn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________【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(xiàn)SKIPIF1<0有兩條過(guò)坐標(biāo)原點(diǎn)的切線(xiàn)，則a的取值范圍為_(kāi)_________．【變式1-1】過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)SKIPIF1<0的切線(xiàn)，若切線(xiàn)有且只有兩條，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【變式1-2】若曲線(xiàn)SKIPIF1<0有兩條過(guò)坐標(biāo)原點(diǎn)的切線(xiàn)，則a的取值范圍是________________．【變式1-3】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)SKIPIF1<0的切線(xiàn)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題型06公切線(xiàn)【解題攻略】交點(diǎn)處公切線(xiàn)，可以直接參照直線(xiàn)在點(diǎn)處的切線(xiàn)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(xiàn)，只需分別寫(xiě)出兩條切線(xiàn)：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(xiàn)SKIPIF1<0是函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的公切線(xiàn)，若SKIPIF1<0是直線(xiàn)SKIPIF1<0與函數(shù)SKIPIF1<0相切的切點(diǎn)，則SKIPIF1<0____________．【典例1-2】（2023春·高三課時(shí)練習(xí)）已知直線(xiàn)SKIPIF1<0：SKIPIF1<0既是曲線(xiàn)SKIPIF1<0的切線(xiàn)，又是曲線(xiàn)SKIPIF1<0的切線(xiàn)，則SKIPIF1<0（

）A．0 B．SKIPIF1<0 C．0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【變式1-1】（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的切線(xiàn)，也是SKIPIF1<0的切線(xiàn)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022·黑龍江哈爾濱·哈爾濱三中?？寄M預(yù)測(cè)）曲線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0的切線(xiàn)也是曲線(xiàn)SKIPIF1<0的切線(xiàn)，則SKIPIF1<0；若此公切線(xiàn)恒在函數(shù)SKIPIF1<0的圖象上方，則a的取值范圍是.【變式1-3】若曲線(xiàn)SKIPIF1<0與曲線(xiàn)SKIPIF1<0存在2條公共切線(xiàn)，則a的值是_________．.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(mǎn)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【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【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【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．SKIPIF1<0【變式1-3】已知奇函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，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題型08特殊構(gòu)造：冪商型構(gòu)造【解題攻略】?jī)绾瘮?shù)商形式構(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】（江西省宜春市奉新縣第一中學(xué)2019-2020學(xué)年高三第一次月考數(shù)學(xué)試題）已知函數(shù)f（x）是定義在R上的奇函數(shù)，f′（x）為f（x）的導(dǎo)函數(shù)，且滿(mǎn)足當(dāng)x＜0時(shí)，有xf′（x）﹣f（x）＜0，則不等式f（x）﹣xf（1）＞0的解集為（）A．（﹣1，0）∪（1，+∞） B．（﹣∞，0）∪（0，1）C．（﹣∞，﹣1）∪（1，+∞） D．（﹣1，0）∪（0，1）【典例1-2】（2020屆高三1月）》函數(shù)SKIPIF1<0在定義域SKIPIF1<0內(nèi)恒滿(mǎn)足SKIPIF1<0，其中SKIPIF1<0為SKIPIF1<0導(dǎo)函數(shù)，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（四川省宜賓市第四中學(xué)校2019-2020學(xué)年高三考試數(shù)學(xué)試題）設(shè)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，且有SKIPIF1<0，則不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（湖北省仙桃市漢江中學(xué)2018-2019學(xué)年高三試題）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0，則不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（甘肅省張掖市第二中學(xué)2019-2020學(xué)年高三4月線(xiàn)上測(cè)試數(shù)學(xué)（理）試卷）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，其中SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0若SKIPIF1<0，則實(shí)數(shù)m的取值范圍為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型09特殊構(gòu)造：ex的積型構(gòu)造【解題攻略】ex函數(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】（江西省上饒中學(xué)2019-2020學(xué)年高三上學(xué)期第二次月考數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的可導(dǎo)函數(shù)，SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023春·黑龍江哈爾濱·高三哈師大附中?？茧A段練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2023春·重慶沙坪壩·高三重慶八中校考）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0是其導(dǎo)函數(shù)，若SKIPIF1<0，SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023春·河南洛陽(yáng)·高三統(tǒng)考）設(shè)SKIPIF1<0是定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，且SKIPIF1<0．若SKIPIF1<0（e為自然對(duì)數(shù)的底數(shù)），則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，滿(mǎn)足SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)．則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型10特殊構(gòu)造：ex的商型構(gòu)造【解題攻略】ex函數(shù)商形式構(gòu)造：1.SKIPIF1<0，2.SKIPIF1<0【典例1-1】定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，滿(mǎn)足：SKIPIF1<0，SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】已知在SKIPIF1<0上的可導(dǎo)函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，滿(mǎn)足SKIPIF1<0，且SKIPIF1<0為偶函數(shù)，SKIPIF1<0，則不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為SKIPIF1<0，f(0)=1，且SKIPIF1<0,則SKIPIF1<0的解集是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型11特殊構(gòu)造：對(duì)數(shù)型構(gòu)造【解題攻略】1.SKIPIF1<02.授課時(shí)，可以讓學(xué)生寫(xiě)出y=ln（kx+b）與y=f(x)的加、減、乘、除各種結(jié)果【典例1-1】（2023·江西宜春·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0（其中SKIPIF1<0是SKIPIF1<0的導(dǎo)數(shù)），若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列選項(xiàng)中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)函數(shù)SKIPIF1<0是奇函數(shù)SKIPIF1<0SKIPIF1<0的導(dǎo)函數(shù)，SKIPIF1<0時(shí)，SKIPIF1<0，則使得SKIPIF1<0成立的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2020上·河南·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0是奇函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，且滿(mǎn)足SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023上·河南周口·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，導(dǎo)函數(shù)為SKIPIF1<0，不等式SKIPIF1<0SKIPIF1<0恒成立，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2022·廣東梅州·統(tǒng)考二模）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型12特殊構(gòu)造：正弦型構(gòu)造【解題攻略】三角函數(shù)形式構(gòu)造：1.SKIPIF1<0，2.SKIPIF1<03.對(duì)于正切型，可以通分（或者去分母）構(gòu)造正弦或者余弦積商型【典例1-1】（2023春·四川成都·高三階段練習(xí)）記函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0為奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)恒有SKIPIF1<0成立，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2021·貴州遵義·高三遵義航天高級(jí)中學(xué)階段練習(xí)）已知定義在SKIPIF1<0上的函數(shù)，SKIPIF1<0為其導(dǎo)函數(shù)，且SKIPIF1<0恒成立，則A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2023春·重慶·高三統(tǒng)考）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知可導(dǎo)函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2021下·江西·高三校聯(lián)考）已知SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，當(dāng)SKIPIF1<0時(shí)，都有SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型13特殊構(gòu)造：余弦型構(gòu)造【解題攻略】三角函數(shù)形式構(gòu)造：1.SKIPIF1<0，2.SKIPIF1<03.對(duì)于正切型，可以通分（或者去分母）構(gòu)造正弦或者余弦積商型【典例1-1】（2020下·安徽六安·高二六安一中?？计谥校┰O(shè)奇函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0的圖象是連續(xù)不間斷，SKIPIF1<0，有SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2020下·湖南長(zhǎng)沙·高二湖南師大附中?？计谀┮阎婧瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，其導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0成立，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2020下·廣西桂林·高二校考階段練習(xí)）函數(shù)SKIPIF1<0定義在SKIPIF1<0上，SKIPIF1<0是它的導(dǎo)函數(shù)，且SKIPIF1<0在定義域內(nèi)恒成立，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2020下·內(nèi)蒙古巴彥淖爾·高二?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0是定義在R上的奇函數(shù)，且對(duì)于任意的SKIPIF1<0，都有SKIPIF1<0(其中SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù))，則下列不等式成立的是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2021下·江蘇·高二期中）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，其導(dǎo)函數(shù)是SKIPIF1<0.有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題型14復(fù)合型構(gòu)造【典例1-1】已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，其導(dǎo)函數(shù)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0.若對(duì)SKIPIF1<0，不等式SKIPIF1<0恒成立，則正整數(shù)SKIPIF1<0的最大值為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若對(duì)任意的實(shí)數(shù)SKIPIF1<0，都有SKIPIF1<0恒成立，則使SKIPIF1<0成立的實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】設(shè)函數(shù)SKIPIF1<0時(shí)定義在SKIPIF1<0上的奇函數(shù)，記其導(dǎo)函數(shù)為SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0定義域?yàn)镽，導(dǎo)函數(shù)為SKIPIF1<0，SKIPIF1<0滿(mǎn)足下列條件：①任意SKIPIF1<0，SKIPIF1<0恒成立，②SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則關(guān)于t的不等式：SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】已知函數(shù)SKIPIF1<0，其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù).若SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有兩個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0高考練場(chǎng)1.（湖南省永州市2022屆高三下學(xué)期第三次適應(yīng)性考試數(shù)學(xué)試題已知直線(xiàn)SKIPIF1<0:SKIPIF1<0，函數(shù)SKIPIF1<0，若SKIPIF1<0存在切線(xiàn)與SKIPIF1<0關(guān)于直線(xiàn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0__________．2.過(guò)點(diǎn)SKIPIF1<0作曲線(xiàn)SKIPIF1<0的兩條切線(xiàn)，則這兩條切線(xiàn)的斜率之和為_(kāi)_____.3.（2022·全國(guó)·高三專(zhuān)題練習(xí)）過(guò)曲線(xiàn)SKIPIF1<0上一點(diǎn)SKIPIF1<0且與曲線(xiàn)在SKIPIF1<0點(diǎn)處的切線(xiàn)垂直的直線(xiàn)的方程為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04.（2023春·陜西寶雞·高三統(tǒng)考）若過(guò)點(diǎn)SKIPIF1<0可作曲線(xiàn)SKIPIF1<0的兩條切線(xiàn)，則點(diǎn)SKIPIF1<0可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05.已知直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0與SKIPIF1<0的公切線(xiàn)，則SKIPIF1<0__________.6.（內(nèi)蒙古赤峰市、呼倫貝爾市等2022-2023學(xué)年高三上學(xué)期開(kāi)學(xué)考試數(shù)學(xué)（文）試題）若直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0與SKIPIF1<0的公切線(xiàn)，則SKIPIF1<0______.7