摘要:13.已知x>1,函数f(x)=x+的最小值是 .
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已知tanα=
-1函数f(x)=x2tan2α+xsin(2α+
)其中α∈(0,
)
(1)求f(x)的解析式;
(2)若数列{an}满足a1=
, an+1=f(an)(n∈N*)求证:
(i)an+1>an(n∈N*);
(ii)1<
+
+…+
<2(n≥2,n∈N*).
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| 2 |
| π |
| 4 |
| π |
| 2 |
(1)求f(x)的解析式;
(2)若数列{an}满足a1=
| 1 |
| 2 |
(i)an+1>an(n∈N*);
(ii)1<
| 1 |
| 1+a1 |
| 1 |
| 1+a2 |
| 1 |
| 1+an |