經過原點且與曲線y=相切的方程是 查看更多

 

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經過原點且與曲線y=相切的方程是(    )

A.x+y=0或+y=0                       B.x-y=0或+y=0

C.x+y=0或-y=0                        D.x-y=0或-y=0

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經過原點且與曲線y=相切的方程是(    )

A.  x+y=0或+y=0                B. xy=0或+y=0

C.  x+y=0或y=0               D.  xy=0或y=0

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經過原點且與曲線y=相切的方程是        ▲         。

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經過原點且與曲線y=相切的方程是(  )
A.x+y=0或+y=0B.x-y=0或+y=0
C.x+y=0或-y=0D.x-y=0或-y=0

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經過原點且與曲線y=相切的方程是(  )

A.x+y=0或+y=0 B.x-y=0或+y=0
C.x+y=0或-y=0 D.x-y=0或-y=0

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難點磁場

解:由l過原點,知k=6ec8aac122bd4f6e(x0≠0),點(x0,y0)在曲線C上,y0=x03-3x02+2x0,

6ec8aac122bd4f6e=x02-3x0+2

y′=3x2-6x+2,k=3x02-6x0+2

k=6ec8aac122bd4f6e,∴3x02-6x0+2=x02-3x0+2

2x02-3x0=0,∴x0=0或x0=6ec8aac122bd4f6e

x≠0,知x0=6ec8aac122bd4f6e

y0=(6ec8aac122bd4f6e)3-3(6ec8aac122bd4f6e)2+2?6ec8aac122bd4f6e=-6ec8aac122bd4f6e

k=6ec8aac122bd4f6e=-6ec8aac122bd4f6e

l方程y=-6ec8aac122bd4f6ex 切點(6ec8aac122bd4f6e,-6ec8aac122bd4f6e)

殲滅難點訓練

一、1.解析:y′=esinx[cosxcos(sinx)-cosxsin(sinx)],y′(0)=e0(1-0)=1

答案:B

2.解析:設切點為(x0,y0),則切線的斜率為k=6ec8aac122bd4f6e,另一方面,y′=(6ec8aac122bd4f6e)′=6ec8aac122bd4f6e,故

y′(x0)=k,即6ec8aac122bd4f6ex02+18x0+45=0得x0(1)=-3,y0(2)=-15,對應有y0(1)=3,y0(2)=6ec8aac122bd4f6e,因此得兩個切點A(-3,3)或B(-15,6ec8aac122bd4f6e),從而得y′(A)=6ec8aac122bd4f6e =-1及y′(B)= 6ec8aac122bd4f6e ,由于切線過原點,故得切線:lA:y=-xlB:y=-6ec8aac122bd4f6e.

答案:A

二、3.解析:根據(jù)導數(shù)的定義:f′(x0)=6ec8aac122bd4f6e(這時6ec8aac122bd4f6e)

6ec8aac122bd4f6e

答案:-1

4.解析:設g(x)=(x+1)(x+2)……(x+n),則f(x)=xg(x),于是f′(x)=g(x)+xg′(x),f′(0)=g(0)+0?g′(0)=g(0)=1?2?…n=n!

答案:n!

三、5.解:設lC1相切于點P(x1,x12),與C2相切于Q(x2,-(x2-2)2)

對于C1y′=2x,則與C1相切于點P的切線方程為

yx12=2x1(xx1),即y=2x1xx12                                                                                                                                             

對于C2y′=-2(x-2),與C2相切于點Q的切線方程為y+(x2-2)2=-2(x2-2)(xx2),即y=-2(x2-2)x+x22-4                                                                                             ②

∵兩切線重合,∴2x1=-2(x2-2)且-x12=x22-4,解得x1=0,x2=2或x1=2,x2=0

∴直線l方程為y=0或y=4x-4

6.解:(1)注意到y>0,兩端取對數(shù),得

lny=ln(x2-2x+3)+lne2x=ln(x2-2x+3)+2x

6ec8aac122bd4f6e

 (2)兩端取對數(shù),得

ln|y|=6ec8aac122bd4f6e(ln|x|-ln|1-x|),

兩邊解x求導,得

6ec8aac122bd4f6e

7.解:設經時間t秒梯子上端下滑s米,則s=5-6ec8aac122bd4f6e,當下端移開1.4 m時,t0=6ec8aac122bd4f6e,又s′=-6ec8aac122bd4f6e (25-9t2)6ec8aac122bd4f6e?(-9?2t)=9t6ec8aac122bd4f6e,所以s′(t0)=9×6ec8aac122bd4f6e=0.875(m/s)

8.解:(1)當x=1時,Sn=12+22+32+…+n2=6ec8aac122bd4f6en(n+1)(2n+1),當x≠1時,1+2x+3x2+…+nxn-1?=6ec8aac122bd4f6e,兩邊同乘以x,得

x+2x2+3x2+…+nxn=6ec8aac122bd4f6e兩邊對x求導,得

Sn=12+22x2+32x2+…+n2xn-1?

=6ec8aac122bd4f6e

 


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