新高考數(shù)學(xué)一輪復(fù)習(xí)9.1 切線方程（精練）（基礎(chǔ)版）（解析版）

9.1切線方程（精練）（基礎(chǔ)版）題組一題組一導(dǎo)數(shù)的幾何意義1．（2021高三上·煙臺期中）曲線SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0（）A．-1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】B【解析】SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，由萬能公式得：SKIPIF1<0，所以SKIPIF1<0。故答案為：B2．（2022青海）已知曲線SKIPIF1<0在點SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0的值為（）A．1 B．-1 C．SKIPIF1<0 D．-SKIPIF1<0【答案】A【解析】SKIPIF1<0．當(dāng)SKIPIF1<0時，SKIPIF1<0，因為切線與直線SKIPIF1<0垂直，直線斜率為SKIPIF1<0，所以切線斜率為2，即SKIPIF1<0，得：SKIPIF1<0故答案為：A3（2021·廣西模擬）函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0處的切線斜率為（）A．-8 B．-7 C．-6 D．-5【答案】A【解析】因為SKIPIF1<0，所以所求切線的斜率為SKIPIF1<0.故答案為：A4．（2021·東陽模擬）已知點P在曲線SKIPIF1<0上，SKIPIF1<0為曲線在點P處的切線的傾斜角，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義可知：SKIPIF1<0,所以SKIPIF1<0，故答案為：D.

5．（2022·白山模擬）函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0處的切線的斜率為（）．A．-8 B．-7 C．6 D．-5【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0處的切線的斜率為-8。故答案為：A6．（2022·鄭州模擬）函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處切線的傾斜角為．【答案】SKIPIF1<0【解析】由SKIPIF1<0求導(dǎo)得：SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處切線斜率為-1，傾斜角為SKIPIF1<0。故答案為：SKIPIF1<0。7．（2021·蚌埠模擬）已知曲線SKIPIF1<0在SKIPIF1<0處切線的斜率為SKIPIF1<0，則SKIPIF1<0．【答案】0【解析】對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由已知條件可得SKIPIF1<0，解得SKIPIF1<0.故答案為：0.8．（2021·聊城模擬）曲線SKIPIF1<0在SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0．【答案】SKIPIF1<0【解析】由題得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0題組二題組二在型求切線1．（2021·全國甲卷）曲線SKIPIF1<0在點（-1，-3）處的切線方程為?！敬鸢浮?x-y+2=0【解析】由題意得SKIPIF1<0，所以在點（-1，-3）處的切線斜率k=5，故切線方程為y+3=5(x+1)，即5x-y+2=0故答案為：5x-y+2=0

2．（2022德州）函數(shù)SKIPIF1<0在點SKIPIF1<0處的切線方程為．【答案】SKIPIF1<0【解析】∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴切線方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0．3．（2022·西安模擬）已知傾斜角為SKIPIF1<0的直線SKIPIF1<0與曲線SKIPIF1<0相切，則直線SKIPIF1<0的方程是.【答案】x-y-2+ln2=0【解析】因為直線SKIPIF1<0的傾斜角為SKIPIF1<0，所以直線SKIPIF1<0的斜率為1，將曲線SKIPIF1<0求導(dǎo)，得SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以切點坐標(biāo)為SKIPIF1<0，所以直線SKIPIF1<0：SKIPIF1<0，即x-y-2+ln2=0，故答案為：x-y-2+ln2=0.4．（2022·河南省?？h第一中學(xué)）曲線SKIPIF1<0在SKIPIF1<0處的切線方程為【答案】4x+y+8＝0【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．又當(dāng)SKIPIF1<0時，SKIPIF1<0，故切點坐標(biāo)為SKIPIF1<0，所以切線方程為SKIPIF1<0．5．（2022·河南）已知SKIPIF1<0，則曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為【答案】SKIPIF1<0【解析】∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴y=f(x)在SKIPIF1<0處的切線方程為：SKIPIF1<0，即SKIPIF1<0．6．（2022·安徽·蚌埠二中）已知定義域為SKIPIF1<0的函數(shù)SKIPIF1<0存在導(dǎo)函數(shù)SKIPIF1<0，且滿足SKIPIF1<0，則曲線SKIPIF1<0在點SKIPIF1<0處的切線方程是【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域為SKIPIF1<0，由SKIPIF1<0可知，SKIPIF1<0是偶函數(shù)，由SKIPIF1<0可知，SKIPIF1<0周期為4，因為SKIPIF1<0，故SKIPIF1<0關(guān)于SKIPIF1<0軸對稱，又因為SKIPIF1<0，所以SKIPIF1<0也是SKIPIF1<0的對稱軸，因為SKIPIF1<0在SKIPIF1<0上存在導(dǎo)函數(shù)SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0的極值點，即SKIPIF1<0，曲線SKIPIF1<0在點SKIPIF1<0處的切線斜率為0，故切線方程可能為SKIPIF1<0.題組三題組三過型求切線1．（2022·廣東茂名）已知直線l為函數(shù)SKIPIF1<0的切線，且經(jīng)過原點，則直線l的方程為__________.【答案】SKIPIF1<0【解析】設(shè)切點坐標(biāo)為SKIPIF1<0，所以直線l的斜率為SKIPIF1<0，所以直線l的方程為SKIPIF1<0又直線l過點SKIPIF1<0，所以SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，直線l的斜率SKIPIF1<0，所以直線l的方程為SKIPIF1<0，故答案為：SKIPIF1<0.2．（2022·四川成都）已知函數(shù)f（x）=x3－3x，則過點（1，－2）的切線方程為__________．【答案】SKIPIF1<0和SKIPIF1<0【解析】由函數(shù)SKIPIF1<0，則SKIPIF1<0，當(dāng)點SKIPIF1<0為切點時，則SKIPIF1<0，即切線的斜率SKIPIF1<0，所以切線的方程為SKIPIF1<0，當(dāng)點SKIPIF1<0不是切點時，設(shè)切點SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），所以SKIPIF1<0所以切線的方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0和SKIPIF1<0．3．（2022·四川成都）過點SKIPIF1<0的直線l與曲線SKIPIF1<0相切，則直線l的斜率為___________.【答案】3或SKIPIF1<0【解析】因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0為切點時，SKIPIF1<0，當(dāng)SKIPIF1<0不為切點時，設(shè)切點為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以切線方程為：SKIPIF1<0，過點SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），所以切點為SKIPIF1<0，所以SKIPIF1<0，綜上所述：直線l的斜率為3或SKIPIF1<0，故答案為：3或SKIPIF1<04．（2022·廣東·南海中學(xué)）函數(shù)SKIPIF1<0過原點的切線方程是_______.【答案】SKIPIF1<0.【解析】設(shè)切點為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故切點為SKIPIF1<0的切線方程為SKIPIF1<0，又因此切線過原點，所以SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0過原點的切線方程是SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.題組四題組四根據(jù)切線方程求參1．（2021高三上·普寧月考）若曲線SKIPIF1<0的一條切線SKIPIF1<0與直線SKIPIF1<0垂直，則直線SKIPIF1<0的方程為．【答案】SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0，則由題意得切線l的斜率為SKIPIF1<0

設(shè)切點為(x0，y0)，又y'=ex，則k=ex0=e則x0=1，y0=e則切點為(1，e)則切線l為y-e=e(x-1)，即y=ex故答案為：y=ex

2．（2022·天河模擬）已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處的切線方程為.【答案】0；SKIPIF1<0【解析】由SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線方程為：SKIPIF1<0。故答案為：0；SKIPIF1<0。3．（2022·泰安模擬）已知直線SKIPIF1<0是曲線SKIPIF1<0的一條切線，則SKIPIF1<0.【答案】4【解析】設(shè)SKIPIF1<0，切點為SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故切點為SKIPIF1<0，又切點在切線SKIPIF1<0上，故SKIPIF1<0.故答案為：44．（2022·大連模擬）已知函數(shù)f（x）=axlnx﹣bx（a，b∈R）在點（e，f（e））處的切線方程為y=3x﹣e，則a+b=.【答案】0【解析】∵在點SKIPIF1<0處的切線方程為SKIPIF1<0，SKIPIF1<0，代入SKIPIF1<0得SKIPIF1<0①.又SKIPIF1<0②.聯(lián)立①②解得：SKIPIF1<0.SKIPIF1<0.故答案為：0.5（2022·東莞模擬）已知SKIPIF1<0在SKIPIF1<0的切線方程為SKIPIF1<0，則SKIPIF1<0．【答案】2【解析】【解答】由題意得SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．故答案為：26（2022·遼寧）已知曲線SKIPIF1<0在點SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則實數(shù)a的值為______．【答案】SKIPIF1<0【解析】由題得SKIPIF1<0，所以SKIPIF1<0，所以曲線SKIPIF1<0在點SKIPIF1<0處的切線斜率為3，又曲線在點SKIPIF1<0處的切線與直線SKIPIF1<0垂直，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0.7．（2022·湖南）已知P是曲線SKIPIF1<0上的一動點，曲線C在P點處的切線的傾斜角為SKIPIF1<0，若SKIPIF1<0，則實數(shù)a的取值范圍是【答案】SKIPIF1<0【解析】因為SKIPIF1<0，所以SKIPIF1<0，因為曲線在M處的切線的傾斜角SKIPIF1<0，所以SKIPIF1<0對于任意的SKIPIF1<0恒成立，即SKIPIF1<0對任意SKIPIF1<0恒成立，即SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時，等號成立，故SKIPIF1<0，所以a的取值范圍是SKIPIF1<0．8．（2022·湖北·武漢二中模擬預(yù)測）已知函數(shù)SKIPIF1<0，直線SKIPIF1<0是曲線SKIPIF1<0的一條切線，則SKIPIF1<0的取值范圍是【答案】SKIPIF1<0【解析】設(shè)切點為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0曲線SKIPIF1<0在切點SKIPIF1<0處的切線方程為SKIPIF1<0，整理得SKIPIF1<0，所以SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0．當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增．故SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0．故選：C.8．（2022·河南洛陽）若過點SKIPIF1<0可作出曲線SKIPIF1<0的三條切線，則實數(shù)SKIPIF1<0的取值范圍是【答案】SKIPIF1<0【解析】由已知，曲線SKIPIF1<0，即令SKIPIF1<0，則SKIPIF1<0