Forgot password
 Register account
View 2183|Reply 6

[函数] 分参又失效了

[Copy link]

107

Threads

224

Posts

1

Reputation

Show all posts

facebooker posted 2019-12-9 00:18 |Read mode
Last edited by facebooker 2019-12-9 01:501)对$\forall x>0,a(e^{ax}+1)\geqslant 2(x+\dfrac{1}{x})\ln x$恒成立,求实数$a$的取值范围___
2)函数$f(x)=ae^x-x-\dfrac{x^2}{a}$,对$\forall x\geqslant -1,f(x)\geqslant 1$恒成立,求实数$a$的取值范围___

281

Threads

550

Posts

2

Reputation

Show all posts

力工 posted 2019-12-11 09:49
回复 1# facebooker
答案请贴贴

107

Threads

224

Posts

1

Reputation

Show all posts

original poster facebooker posted 2019-12-12 16:39
Last edited by facebooker 2019-12-12 18:33回复 2# 力工
第一题带人x=e  第二题带人x=-1

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2019-12-12 17:03
回复 3# facebooker

第二题应该不难吧,代 `x=-1` 得 `a\ge\sqrt e`,而 `f` 关于 `a` 递增,故只需证当 `a=\sqrt e` 时成立即可,这你能破吧

107

Threads

224

Posts

1

Reputation

Show all posts

original poster facebooker posted 2019-12-12 17:32
回复 4# kuing
老大 就是在这一步不会证啊 我猜出这个答案了 但是不知道咋证明

673

Threads

110K

Posts

218

Reputation

Show all posts

kuing posted 2019-12-12 17:47
回复 5# facebooker

作代换 `x=-1+t`, `t\ge0`,变成证明 `e^t\ge(1-t)^2+\sqrt e\cdot t`,这可以证明更强的 `e^t\geqslant (1-t)^2+2t=1+t^2`,这样总会了吧?

107

Threads

224

Posts

1

Reputation

Show all posts

original poster facebooker posted 2019-12-12 18:33
回复 6# kuing

谢谢大佬

Quick Reply

Advanced Mode
B Color Image Link Quote Code Smilies
You have to log in before you can reply Login | Register account

$\LaTeX$ formula tutorial

Mobile version

2025-7-15 14:29 GMT+8

Powered by Discuz!

Processed in 0.011214 seconds, 24 queries