Forgot password?
 Create new account
View 1840|Reply 1

[函数] 一道和三角有关的导数题

[Copy link]

133

Threads

259

Posts

2333

Credits

Credits
2333

Show all posts

郝酒 Posted at 2014-5-3 17:35:57 |Read mode
已知$f(x)=\sin x$,
(1)若$f(x)+1\ge ax+\cos x$在$[0,\pi ]$上恒成立,求实数$a$的取值范围;
(2)证明:$f\left( \frac{\pi }{2n+1} \right)+f\left( \frac{2\pi }{2n+1} \right)+\cdots +f\left( \frac{(n+1)\pi }{2n+1} \right)\ge \frac{3\sqrt{2}(n+1)}{4(2n+1)}$.

两问都不好做啊。

133

Threads

259

Posts

2333

Credits

Credits
2333

Show all posts

 Author| 郝酒 Posted at 2014-5-3 22:03:23
第一问,分离后不好做,对$a$进行讨论.
利用$F(x)=\sin x+1-ax-\cos x$,$F(\pi)=a\pi-2$可得$a\leq \frac{2}{\pi}$
第二问
由$\sqrt{2}\sin (x-\frac{\pi }{4})+1\ge ax$得当$n\ge 2$时$\frac{\pi }{2n+1}+\frac{\pi }{4}<\pi $
\[\sin \left( \frac{\pi }{2n+1} \right)=\sin \left( \frac{\pi }{2n+1}+\frac{\pi }{4}-\frac{\pi }{4} \right)\ge \frac{\sqrt{\text{2}}}{\text{2}}\left( \frac{\pi }{2}\left( \frac{\pi }{2n+1}+\frac{\pi }{4} \right)-1 \right)\]
后面累加即可.

手机版Mobile version|Leisure Math Forum

2025-4-22 04:55 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list