新高考數(shù)學(xué)二輪復(fù)習(xí)重點(diǎn)突破訓(xùn)練第15講 用導(dǎo)數(shù)的幾何意義研究曲線的切線（含解析）

第15講用導(dǎo)數(shù)的幾何意義研究曲線的切線真題展示2022新高考一卷第15題若曲線SKIPIF1<0有兩條過(guò)坐標(biāo)原點(diǎn)的切線，則SKIPIF1<0的取值范圍是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【思路分析】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)求出切線的斜率，進(jìn)而得到切線方程，再把原點(diǎn)代入可得SKIPIF1<0，因?yàn)榍芯€存在兩條，所以方程有兩個(gè)不等實(shí)根，由△SKIPIF1<0即可求出SKIPIF1<0的取值范圍．【解析】【解法一】(切線方程)SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0切線的斜率SKIPIF1<0，SKIPIF1<0切線方程為SKIPIF1<0，又SKIPIF1<0切線過(guò)原點(diǎn)，SKIPIF1<0，整理得：SKIPIF1<0，SKIPIF1<0切線存在兩條，SKIPIF1<0方程有兩個(gè)不等實(shí)根，SKIPIF1<0△SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【解法二】法二(切線斜率)：設(shè)切點(diǎn)為(m,(m+a)SKIPIF1<0)，易得y'=(x+a+1)SKIPIF1<0，則切線的斜率k=(m+a+1)SKIPIF1<0=SKIPIF1<0，即m(m+a+1)=m+a,SKIPIF1<0+am?a=0，依題意其有兩個(gè)不等實(shí)根，故△=SKIPIF1<0+4a>0，解得a<?4或a>0.【試題評(píng)價(jià)】本題主要考查了利用導(dǎo)數(shù)研究曲線上某點(diǎn)處的切線方程，屬于中檔題．知識(shí)要點(diǎn)整理用導(dǎo)數(shù)求切線方程的四種類型求曲線的切線方程是導(dǎo)數(shù)的重要應(yīng)用之一，用導(dǎo)數(shù)求切線方程的關(guān)鍵在于求出切點(diǎn)SKIPIF1<0及斜率，其求法為：設(shè)SKIPIF1<0是曲線SKIPIF1<0上的一點(diǎn)，則以SKIPIF1<0的切點(diǎn)的切線方程為：SKIPIF1<0．若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0的切線平行于SKIPIF1<0軸（即導(dǎo)數(shù)不存在）時(shí)，由切線定義知，切線方程為SKIPIF1<0．下面例析四種常見(jiàn)的類型及解法．類型一：已知切點(diǎn)，求曲線的切線方程此類題較為簡(jiǎn)單，只須求出曲線的導(dǎo)數(shù)SKIPIF1<0，并代入點(diǎn)斜式方程即可．例1曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（）Ａ．SKIPIF1<0 Ｂ．SKIPIF1<0Ｃ．SKIPIF1<0 Ｄ．SKIPIF1<0解：由SKIPIF1<0則在點(diǎn)SKIPIF1<0處斜率SKIPIF1<0，故所求的切線方程為SKIPIF1<0，即SKIPIF1<0，因而選Ｂ．類型二：已知斜率，求曲線的切線方程此類題可利用斜率求出切點(diǎn)，再用點(diǎn)斜式方程加以解決．例2與直線SKIPIF1<0的平行的拋物線SKIPIF1<0的切線方程是（）Ａ．SKIPIF1<0 Ｂ．SKIPIF1<0Ｃ．SKIPIF1<0 Ｄ．SKIPIF1<0解：設(shè)SKIPIF1<0為切點(diǎn)，則切點(diǎn)的斜率為SKIPIF1<0．SKIPIF1<0．由此得到切點(diǎn)SKIPIF1<0．故切線方程為SKIPIF1<0，即SKIPIF1<0，故選Ｄ．評(píng)注：此題所給的曲線是拋物線，故也可利用SKIPIF1<0法加以解決，即設(shè)切線方程為SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，又因?yàn)镾KIPIF1<0，得SKIPIF1<0，故選Ｄ．類型三：已知過(guò)曲線上一點(diǎn)，求切線方程過(guò)曲線上一點(diǎn)的切線，該點(diǎn)未必是切點(diǎn)，故應(yīng)先設(shè)切點(diǎn)，再求切點(diǎn)，即用待定切點(diǎn)法．例3求過(guò)曲線SKIPIF1<0上的點(diǎn)SKIPIF1<0的切線方程．解：設(shè)想SKIPIF1<0為切點(diǎn)，則切線的斜率為SKIPIF1<0．SKIPIF1<0切線方程為SKIPIF1<0．SKIPIF1<0．又知切線過(guò)點(diǎn)SKIPIF1<0，把它代入上述方程，得SKIPIF1<0．解得SKIPIF1<0，或SKIPIF1<0．故所求切線方程為SKIPIF1<0，或SKIPIF1<0，即SKIPIF1<0，或SKIPIF1<0．評(píng)注：可以發(fā)現(xiàn)直線SKIPIF1<0并不以SKIPIF1<0為切點(diǎn)，實(shí)際上是經(jīng)過(guò)了點(diǎn)SKIPIF1<0且以SKIPIF1<0為切點(diǎn)的直線．這說(shuō)明過(guò)曲線上一點(diǎn)的切線，該點(diǎn)未必是切點(diǎn)，解決此類問(wèn)題可用待定切點(diǎn)法．類型四：已知過(guò)曲線外一點(diǎn)，求切線方程此類題可先設(shè)切點(diǎn)，再求切點(diǎn)，即用待定切點(diǎn)法來(lái)求解．例4求過(guò)點(diǎn)SKIPIF1<0且與曲線SKIPIF1<0相切的直線方程．解：設(shè)SKIPIF1<0為切點(diǎn)，則切線的斜率為SKIPIF1<0．SKIPIF1<0切線方程為SKIPIF1<0，即SKIPIF1<0．又已知切線過(guò)點(diǎn)SKIPIF1<0，把它代入上述方程，得SKIPIF1<0．解得SKIPIF1<0，即SKIPIF1<0．評(píng)注：點(diǎn)SKIPIF1<0實(shí)際上是曲線外的一點(diǎn)，但在解答過(guò)程中卻無(wú)需判斷它的確切位置，充分反映出待定切點(diǎn)法的高效性．例5已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，求此切線方程．解：曲線方程為SKIPIF1<0，點(diǎn)SKIPIF1<0不在曲線上．設(shè)切點(diǎn)為SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)滿足SKIPIF1<0．因SKIPIF1<0，故切線的方程為SKIPIF1<0．點(diǎn)SKIPIF1<0在切線上，則有SKIPIF1<0．化簡(jiǎn)得SKIPIF1<0，解得SKIPIF1<0．所以，切點(diǎn)為SKIPIF1<0，切線方程為SKIPIF1<0．評(píng)注：此類題的解題思路是，先判斷點(diǎn)A是否在曲線上，若點(diǎn)A在曲線上，化為類型一或類型三；若點(diǎn)A不在曲線上，應(yīng)先設(shè)出切點(diǎn)并求出切點(diǎn)．2、求圓錐曲線的切線在初中數(shù)學(xué)中，曲線的切線沒(méi)有一般的定義。例如，圓的切線定義為與圓只有一個(gè)交點(diǎn)的直線，但把這一定義用到其他曲線上就不行了。如直線SKIPIF1<0與拋物線SKIPIF1<0只有一個(gè)交點(diǎn)，SKIPIF1<0是SKIPIF1<0的切線，但SKIPIF1<0與拋物線SKIPIF1<0也只有一個(gè)交點(diǎn)，但SKIPIF1<0卻不是SKIPIF1<0的切線，由此可見(jiàn)，用“一個(gè)交點(diǎn)”來(lái)定義切線并不能用于所有曲線。而學(xué)了微積分的知識(shí)后，就可以給出曲線切線的一般定義了。切線的定義：設(shè)SKIPIF1<0是曲線SKIPIF1<0上一定點(diǎn)，SKIPIF1<0是該曲線上的一動(dòng)點(diǎn)，從而有割線SKIPIF1<0，令SKIPIF1<0沿著曲線無(wú)限趨近于SKIPIF1<0，則割線SKIPIF1<0的極限位置就是曲線SKIPIF1<0在SKIPIF1<0的切線（如果極限存在的話）。這一定義與初等數(shù)學(xué)中圓的切線定義是一致的（用于討論圓的切線時(shí)），用這一定義也容易證明SKIPIF1<0是SKIPIF1<0的切線，而SKIPIF1<0不是SKIPIF1<0的切線，這一切線定義可用于任何曲線SKIPIF1<0。導(dǎo)數(shù)的幾何意義就是曲線SKIPIF1<0在點(diǎn)SKIPIF1<0的切線斜率。故運(yùn)用上述切線的一般定義和結(jié)論，可以處理與切線有關(guān)的許多問(wèn)題。例6求曲線SKIPIF1<0在SKIPIF1<0時(shí)的切線方程。解：SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0又SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，所求的切線方程為：即SKIPIF1<0反思：由此可見(jiàn)，用微積分法解此類問(wèn)題是多么的簡(jiǎn)單容易，可是在初等數(shù)學(xué)中，曲線SKIPIF1<0的切線定義都難得給出，更別說(shuō)討論與SKIPIF1<0的切線有關(guān)的問(wèn)題了。例7已知函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值，過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線，求此切線方程。解：由例4，曲線方程為SKIPIF1<0，點(diǎn)SKIPIF1<0不在曲線上。設(shè)切點(diǎn)為SKIPIF1<0則點(diǎn)SKIPIF1<0的坐標(biāo)滿足SKIPIF1<0，由于SKIPIF1<0，故切線的方程為SKIPIF1<0.注意到點(diǎn)SKIPIF1<0在切線上,有SKIPIF1<0化簡(jiǎn)得SKIPIF1<0,解得SKIPIF1<0.因此,切點(diǎn)為SKIPIF1<0,切線方程為SKIPIF1<0.要點(diǎn)：1.導(dǎo)數(shù)是如何定義2.如何求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程與法線方程。三年真題1.已知函數(shù)SKIPIF1<0，對(duì)于SKIPIF1<0上的任意SKIPIF1<0，有如下條件：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0．其中能使SKIPIF1<0恒成立的條件序號(hào)是____________．【答案】②【詳解】當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增；又因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0是偶函數(shù)，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減；①：當(dāng)SKIPIF1<0時(shí)，取SKIPIF1<0時(shí)，顯然SKIPIF1<0成立，但是SKIPIF1<0，所以本條件不符合題意，②：當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0時(shí)，由單調(diào)性知自變量距離y軸越遠(yuǎn)，函數(shù)值越大，所以SKIPIF1<0，所以本條件符合題意；③：當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，顯然SKIPIF1<0成立，但是SKIPIF1<0，所以本條件不符合題意，故答案為：②.2．某日中午12時(shí)整，甲船自A處以SKIPIF1<0的速度向正東行駛，乙船自A的正北SKIPIF1<0處以SKIPIF1<0的速度向正南行駛，則當(dāng)日12時(shí)30分時(shí)兩船之距離對(duì)時(shí)間的變化率是___________SKIPIF1<0．【答案】-1.6【詳解】SKIPIF1<0中午12時(shí)整，甲船自SKIPIF1<0處以SKIPIF1<0的速度向正東行駛，乙船自SKIPIF1<0的正北SKIPIF1<0處以SKIPIF1<0的速度向正南行駛，當(dāng)日12時(shí)30分時(shí)，乙船沒(méi)有到達(dá)SKIPIF1<0處，故甲乙兩船之間的距離函數(shù)是SKIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)日12時(shí)30分時(shí)，SKIPIF1<0，此時(shí)兩船之間距離對(duì)時(shí)間的變化率是SKIPIF1<0故答案為：SKIPIF1<0．3．曲線SKIPIF1<0與SKIPIF1<0在交點(diǎn)處切線的夾角是____________．（用弧度數(shù)作答）【答案】SKIPIF1<0【詳解】由SKIPIF1<0消元可得，SKIPIF1<0，解得SKIPIF1<0，所以兩曲線只有一個(gè)交點(diǎn)SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，由直線的夾角公式可得SKIPIF1<0，由SKIPIF1<0知，SKIPIF1<0.故答案為：SKIPIF1<04．已知SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的極小值點(diǎn)和極大值點(diǎn)．若SKIPIF1<0，則a的取值范圍是____________．【答案】SKIPIF1<0【詳解】[方法一]：【最優(yōu)解】轉(zhuǎn)化法，零點(diǎn)的問(wèn)題轉(zhuǎn)為函數(shù)圖象的交點(diǎn)因?yàn)镾KIPIF1<0，所以方程SKIPIF1<0的兩個(gè)根為SKIPIF1<0，即方程SKIPIF1<0的兩個(gè)根為SKIPIF1<0，即函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)不同的交點(diǎn)，因?yàn)镾KIPIF1<0分別是函數(shù)SKIPIF1<0的極小值點(diǎn)和極大值點(diǎn)，所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以當(dāng)時(shí)SKIPIF1<0SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0圖象在SKIPIF1<0上方當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0圖象在SKIPIF1<0下方SKIPIF1<0，圖象顯然不符合題意，所以SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，設(shè)過(guò)原點(diǎn)且與函數(shù)SKIPIF1<0的圖象相切的直線的切點(diǎn)為SKIPIF1<0，則切線的斜率為SKIPIF1<0，故切線方程為SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，則切線的斜率為SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)不同的交點(diǎn)，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，綜上所述，SKIPIF1<0的取值范圍為SKIPIF1<0．[方法二]：【通性通法】構(gòu)造新函數(shù)，二次求導(dǎo)SKIPIF1<0=0的兩個(gè)根為SKIPIF1<0因?yàn)镾KIPIF1<0分別是函數(shù)SKIPIF1<0的極小值點(diǎn)和極大值點(diǎn)，所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減，在SKIPIF1<0上遞增，設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，此時(shí)若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，此時(shí)若有SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0且SKIPIF1<0的極小值點(diǎn)和極大值點(diǎn)，則SKIPIF1<0，不符合題意；若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，此時(shí)若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，令SKIPIF1<0，則SKIPIF1<0，此時(shí)若有SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0且SKIPIF1<0的極小值點(diǎn)和極大值點(diǎn)，且SKIPIF1<0，則需滿足SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0故SKIPIF1<0，所以SKIPIF1<0.5．若曲線SKIPIF1<0有兩條過(guò)坐標(biāo)原點(diǎn)的切線，則a的取值范圍是________________．【答案】SKIPIF1<0【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(guò)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<06．函數(shù)SKIPIF1<0的最大值為_(kāi)_____.【答案】SKIPIF1<0##0.25【詳解】當(dāng)SKIPIF1<0時(shí)，求導(dǎo)得：SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則當(dāng)SKIPIF1<0時(shí)，y取得最大值，即SKIPIF1<0，所以函數(shù)SKIPIF1<0的最大值為SKIPIF1<0.故答案為：SKIPIF1<07．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與SKIPIF1<0軸、直線SKIPIF1<0所圍成的三角形的面積為SKIPIF1<0，則SKIPIF1<0________.【答案】SKIPIF1<0【詳解】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0曲線在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0切線與SKIPIF1<0軸，直線SKIPIF1<0所圍成的三角形的面積為SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0．8．曲線SKIPIF1<0在點(diǎn)(0，1)處的切線方程為_(kāi)_______．【答案】SKIPIF1<0【詳解】解：SKIPIF1<0，SKIPIF1<0切線的斜率為SKIPIF1<0則切線方程為SKIPIF1<0，即SKIPIF1<0故答案為：SKIPIF1<09．已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0的兩條切線互相垂直，且分別交y軸于M，N兩點(diǎn)，則SKIPIF1<0取值范圍是_______．【答案】SKIPIF1<0【分析】結(jié)合導(dǎo)數(shù)的幾何意義可得SKIPIF1<0，結(jié)合直線方程及兩點(diǎn)間距離公式可得SKIPIF1<0，SKIPIF1<0，化簡(jiǎn)即可得解.【詳解】由題意，SKIPIF1<0，則SKIPIF1<0，所以點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，同理SKIPIF1<0,所以SKIPIF1<0.故答案為：SKIPIF1<010．寫(xiě)出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)SKIPIF1<0_______．①SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；③SKIPIF1<0是奇函數(shù)．【答案】SKIPIF1<0（答案不唯一，SKIPIF1<0均滿足）【詳解】取SKIPIF1<0，則SKIPIF1<0，滿足①，SKIPIF1<0，SKIPIF1<0時(shí)有SKIPIF1<0，滿足②，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，又SKIPIF1<0，故SKIPIF1<0是奇函數(shù)，滿足③.故答案為：SKIPIF1<0（答案不唯一，SKIPIF1<0均滿足）11．已知函數(shù)SKIPIF1<0，給出下列四個(gè)結(jié)論：①若SKIPIF1<0，SKIPIF1<0恰有2個(gè)零點(diǎn)；②存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有1個(gè)零點(diǎn)；③存在負(fù)數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)；④存在正數(shù)SKIPIF1<0，使得SKIPIF1<0恰有3個(gè)零點(diǎn)．其中所有正確結(jié)論的序號(hào)是_______．【答案】①②④【詳解】對(duì)于①，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，①正確；對(duì)于②，考查直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，所以，存在SKIPIF1<0，使得SKIPIF1<0只有一個(gè)零點(diǎn)，②正確；對(duì)于③，當(dāng)直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)交點(diǎn)，若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)交點(diǎn)，直線SKIPIF1<0與曲線SKIPIF1<0有一個(gè)交點(diǎn)，所以，SKIPIF1<0，此不等式無(wú)解，因此，不存在SKIPIF1<0，使得函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，③錯(cuò)誤；對(duì)于④，考查直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，④正確.故答案為：①②④.12．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為_(kāi)_________．【答案】SKIPIF1<0【詳解】由題，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故點(diǎn)在曲線上．求導(dǎo)得：SKIPIF1<0，所以SKIPIF1<0．故切線方程為SKIPIF1<0．故答案為：SKIPIF1<0．13．函數(shù)SKIPIF1<0的最小值為_(kāi)_____.【答案】1【詳解】由題設(shè)知：SKIPIF1<0定義域?yàn)镾KIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；又SKIPIF1<0在各分段的界點(diǎn)處連續(xù)，∴綜上有：SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；∴SKIPIF1<0故答案為：1.14．在平面直角坐標(biāo)系xOy中，已知SKIPIF1<0，A，B是圓C：SKIPIF1<0上的兩個(gè)動(dòng)點(diǎn)，滿足SKIPIF1<0，則△PAB面積的最大值是__________．【答案】SKIPIF1<0【詳解】SKIPIF1<0設(shè)圓心SKIPIF1<0到直線SKIPIF1<0距離為SKIPIF1<0，則SKIPIF1<0，所以點(diǎn)P到AB的距離為SKIPIF1<0或SKIPIF1<0，且SKIPIF1<0所以SKIPIF1<0令SKIPIF1<0（負(fù)值舍去）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值，即SKIPIF1<0取最大值為SKIPIF1<0，故答案為：SKIPIF1<015．設(shè)函數(shù)SKIPIF1<0．若SKIPIF1<0，則a=_________．【答案】1【詳解】由函數(shù)的解析式可得：SKIPIF1<0，則：SKIPIF1<0，據(jù)此可得：SKIPIF1<0，整理可得：SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<0.16．曲線SKIPIF1<0的一條切線的斜率為2，則該切線的方程為_(kāi)_____________.【答案】SKIPIF1<0【詳解】設(shè)切線的切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，所以切點(diǎn)坐標(biāo)為SKIPIF1<0，所求的切線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.三年模擬1．已知SKIPIF1<0，則曲線SKIPIF1<0在SKIPIF1<0處的切線方程是___________.【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，斜率為SKIPIF1<0，代入點(diǎn)斜式直線方程SKIPIF1<0中則曲線SKIPIF1<0在SKIPIF1<0處的切線方程是SKIPIF1<0.故答案為：SKIPIF1<0.2．若過(guò)點(diǎn)SKIPIF1<0只可以作曲線SKIPIF1<0的一條切線，則SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【詳解】解：函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則切線斜率為SKIPIF1<0，故切線方程為：SKIPIF1<0，又切線過(guò)點(diǎn)SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0得，SKIPIF1<0或SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，又SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0有且只有一個(gè)根，且SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.3．若直線SKIPIF1<0是曲線SKIPIF1<0和SKIPIF1<0的公切線，則實(shí)數(shù)SKIPIF1<0的值是______.【答案】1【詳解】設(shè)直線SKIPIF1<0與曲線SKIPIF1<0分別相切于點(diǎn)SKIPIF1<0,對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0,則SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0,則SKIPIF1<0,曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0，化簡(jiǎn)可得SKIPIF1<0，故答案為：1.4．函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線方程為_(kāi)_____.【答案】SKIPIF1<0【詳解】∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0.故答案為：SKIPIF1<0.5．設(shè)曲線SKIPIF1<0的斜率為3的切線為SKIPIF1<0，則SKIPIF1<0的方程為_(kāi)_____．【答案】SKIPIF1<0【詳解】設(shè)切線SKIPIF1<0與函數(shù)SKIPIF1<0的切點(diǎn)為SKIPIF1<0又因?yàn)镾KIPIF1<0，所以SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)值為SKIPIF1<0所以SKIPIF1<0，又因?yàn)榍悬c(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0上，即SKIPIF1<0所以切點(diǎn)為SKIPIF1<0，所以切線方程SKIPIF1<0，即SKIPIF1<0故答案為：SKIPIF1<06．已知函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是______.【答案】SKIPIF1<0【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率SKIPIF1<0，所以切線方程為：SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<07．已知定義在R上的函數(shù)SKIPIF1<0滿足：①曲線SKIPIF1<0上任意一點(diǎn)處的切線斜率均不小于1；②曲線SKIPIF1<0在原點(diǎn)處的切線與圓SKIPIF1<0相切，請(qǐng)寫(xiě)出一個(gè)符合題意的函數(shù)SKIPIF1<0______．【答案】SKIPIF1<0（答案不唯一）【詳解】由②可設(shè)過(guò)原點(diǎn)且與圓SKIPIF1<0相切的直線為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），結(jié)合①知曲線SKIPIF1<0在原點(diǎn)處的切線為SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，（答案不唯一，只要符合題意即可）SKIPIF1<0，滿足①．因?yàn)镾KIPIF1<0，SKIPIF1<0所以曲線SKIPIF1<0在原點(diǎn)處的切線為SKIPIF1<0，滿足②．故SKIPIF1<0符合題意．故答案為：SKIPIF1<0（答案不唯一）8．已知曲線SKIPIF1<0在某點(diǎn)處的切線的斜率為SKIPIF1<0，則該切線的方程為_(kāi)_____.【答案】SKIPIF1<0【分析】對(duì)函數(shù)求導(dǎo)后，利用導(dǎo)數(shù)的幾何意義列方程求出切點(diǎn)坐標(biāo)，從而可求出切線方程.【詳解】設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0（SKIPIF1<0），由SKIPIF1<0，得SKIPIF1<0（SKIPIF1<0），因?yàn)榍€SKIPIF1<0在SKIPIF1<0處的切線的斜率為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0（舍去），或SKIPIF1<0，所以SKIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0，故答案為：SKIPIF1<0.9．若函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0_________．【答案】SKIPIF1<0【詳解】SKIPIF1<0，所以SKIPIF1<0，所以切線的斜率為3，又因?yàn)镾KIPIF1<0，所以切點(diǎn)的坐標(biāo)為SKIPIF1<0，所以切線方程為SKIPIF1<0即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為:SKIPIF1<0.10．已知函數(shù)SKIPIF1<0的圖像與直線SKIPIF1<0相切，則SKIPIF1<0____________【答案】1【詳解】解：由SKIPIF1<0得SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0．故答案為：1．11．若曲線SKIPIF1<0的圖象總在曲線SKIPIF1<0的圖象上方，則SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【詳解】∵SKIPIF1<0的圖象與SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，即問(wèn)題轉(zhuǎn)化為曲線SKIPIF1<0總在直線SKIPIF1<0下方，當(dāng)直線SKIPIF1<0與曲線SKIPIF1<0相切時(shí)，設(shè)切點(diǎn)SKIPIF1<0，則切線斜率SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0，解得SKIPIF1<0，要滿足題意，SKIPIF1<0，故答案為：SKIPIF1<012．已知曲線SKIPIF1<0與曲線SKIPIF1<0有相同的切線，則這條切線的斜率為_(kāi)__________.【答案】SKIPIF1<0##0.5【詳解】設(shè)曲線SKIPIF1<0與曲線SKIPIF1<0的切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以切線為SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即這條切線的斜率為SKIPIF1<0.故答案為：SKIPIF1<0.13．已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則實(shí)數(shù)SKIPIF1<0______．【答案】SKIPIF1<0【詳解】直線SKIPIF1<0的斜率為：SKIPIF1<0，故切線的斜率為2，SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<014．已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0作曲線SKIPIF1<0的切線SKIPIF1<0，則SKIPIF1<0的方程為_(kāi)__________.【答案】SKIPIF1<0【分析】根據(jù)導(dǎo)數(shù)的幾何意義設(shè)切點(diǎn)坐標(biāo)SKIPIF1<0，利用導(dǎo)數(shù)求切線斜率，從而可得切線方程表達(dá)式，利用切線過(guò)點(diǎn)SKIPIF1<0，解出SKIPIF1<0，即可求得切線方程.【詳解】解：由題意可設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以切線SKIPIF1<0的斜率SKIPIF1<0，則SKIPIF1<0的方程為SKIPIF1<0，又點(diǎn)SKIPIF1<0在切線上，所以SKIPIF1<0解得SKIPIF1<0，所以切線方程為：SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.15．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【詳解】由題意，函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)即SKIPIF1<0有三個(gè)解，即SKIPIF1<0與SKIPIF1<0的交點(diǎn)個(gè)數(shù)為3.作出SKIPIF1<0與SKIPIF1<0的圖象，易得當(dāng)SKIPIF1<0時(shí)不成立，故SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)SKIPIF1<0與SKIPIF1<0必有一個(gè)交點(diǎn)，則當(dāng)SKIPIF1<0有2個(gè)交點(diǎn).當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0，此時(shí)SKIPIF1<0與SKIPIF1<0或S