新高考數(shù)學(xué)二輪復(fù)習(xí)專題培優(yōu)練習(xí)專題09 導(dǎo)數(shù)的幾何意義及應(yīng)用（原卷版）

專題09導(dǎo)數(shù)的幾何意義及應(yīng)用一、單選題1．函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．函數(shù)SKIPIF1<0與直線SKIPIF1<0相切，則實數(shù)a的值為（

）A．1 B．2 C．e D．SKIPIF1<03．已知SKIPIF1<0是奇函數(shù)，則SKIPIF1<0在SKIPIF1<0處的切線方程是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．若直線SKIPIF1<0與曲線SKIPIF1<0相切，直線SKIPIF1<0與曲線SKIPIF1<0相切，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<05．若存在SKIPIF1<0，SKIPIF1<0，使得直線SKIPIF1<0與SKIPIF1<0，SKIPIF1<0的圖象均相切，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．直線l經(jīng)過點SKIPIF1<0，且與直線SKIPIF1<0平行，如果直線l與曲線SKIPIF1<0相切，那么b等于（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<07．已知曲線SKIPIF1<0與直線SKIPIF1<0相切，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．過點SKIPIF1<0作曲線SKIPIF1<0的兩條切線，切點分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．39．若函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象存在公切線，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<010．已知過點SKIPIF1<0不可能作曲線SKIPIF1<0的切線．對于滿足上述條件的任意的b，函數(shù)SKIPIF1<0恒有兩個不同的極值點，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題11．已知直線SKIPIF1<0與曲線SKIPIF1<0相切，則下列直線中可能與SKIPIF1<0垂直的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．若過點SKIPIF1<0可作3條直線與函數(shù)SKIPIF1<0的圖象相切，則實數(shù)SKIPIF1<0可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<013．已知SKIPIF1<0，若過點SKIPIF1<0恰能作兩條直線與曲線SKIPIF1<0相切，其中SKIPIF1<0，則m與n可能滿足的關(guān)系式為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<014．已知函數(shù)SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0分別與曲線SKIPIF1<0和曲線SKIPIF1<0相切于點SKIPIF1<0，SKIPIF1<0，且直線SKIPIF1<0也與曲線SKIPIF1<0，SKIPIF1<0都相切，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題15．已知直線SKIPIF1<0與曲線SKIPIF1<0相切，則實數(shù)SKIPIF1<0.16．若曲線SKIPIF1<0在點SKIPIF1<0處的切線與曲線SKIPIF1<0相切于點SKIPIF1<0，則SKIPIF1<0.17．在平面直角坐標(biāo)系SKIPIF1<0中，若過點SKIPIF1<0且同時與曲線SKIPIF1<0，曲線SKIPIF1<0都相切的直線有兩條，則點SKIPIF1<0的坐標(biāo)為．18．已知函數(shù)SKIPIF1<0，若過點SKIPIF1<0可作曲線SKIPIF1<0的三條切線，則SKIPIF1<0的取值范圍是．19．若過SKIPIF1<0軸上任意點SKIPIF1<0可作曲線SKIPIF1<0兩條切線，則SKIPIF1<0的取值范圍．20．已知函數(shù)SKIPIF1<0有兩條與直線SKIPIF1<0平行的切線，且切點坐標(biāo)分別為SKIPIF1<0，則SKIPIF1<0的取值范圍是．四、解答題21．已知函數(shù)SKIPIF1<0．(1)求證：曲線SKIPIF1<0僅有一條過原點的切線；(2)若SKIPIF1<0時，關(guān)于SKIPIF1<0的方程SKIPIF