题目内容
设y=(a-1)x与y=(
|
试题答案
C
相关题目
已知a>0,函数f(x)=
,x∈({0,+∞}),设0<x1<
,记曲线y=f(x)在点M(x1,f(x1))处的切线为l,
(1)求l的方程;
(2)设l与x轴交点为(x2,0)证明:0<x2≤
.
查看习题详情和答案>>
| 1-ax |
| x |
| 2 |
| a |
(1)求l的方程;
(2)设l与x轴交点为(x2,0)证明:0<x2≤
| 1 |
| a |
已知a>0,函数f(x)=
,x∈(0,+∞).设0<x1<
,记曲线y=f(x)在点M(x1,f(x1))处的切线为l
(1)求l的方程;
(2)设l与x轴交点为(x2,0),求证:①0<x2≤
; ②若0<x1<
,则x1<x2<2x1.
查看习题详情和答案>>
| 1-ax |
| x |
| 2 |
| a |
(1)求l的方程;
(2)设l与x轴交点为(x2,0),求证:①0<x2≤
| 1 |
| a |
| 1 |
| a |
已知a>0,函数f(x)=
,x∈(0,+∞).设0<x1<
,记曲线y=f(x)在点M(x1,f(x1))处的切线为l.
(Ⅰ)求l的方程;
(Ⅱ)设l与x轴交点为(x2,0).证明:
①0<x2≤
;
②若x1<
,则x1<x2<
.
查看习题详情和答案>>
| 1-ax |
| x |
| 2 |
| a |
(Ⅰ)求l的方程;
(Ⅱ)设l与x轴交点为(x2,0).证明:
①0<x2≤
| 1 |
| a |
②若x1<
| 1 |
| a |
| 1 |
| a |