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[函数] 宁波2021高一期末第12题怎么解释?

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realnumber posted 2022-1-21 10:37 |Read mode
若实数满足$3^a+4a=4^b+3b$,则下列关系式中可能成立的是(ABD)
A.0<a<b<1   
B.b<a<0
C.1<a<b
D.a=b
考试时可没有画图软件的,也没学过导数.
QQ截图20220121103236张政治.png

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original poster realnumber posted 2022-1-21 10:45
$3\times3^{x-1}+4x>3\times4^{x-1}+4x>4\times4^{x-1}+3x,0<x<1$,排序不等式以及$4^x\ge4x$都不妥.

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战巡 posted 2022-1-21 11:15
回复 1# realnumber

它只问“可能”,又没让你证明这个东西,你能举出例子就说明可能了

比如A,令$b=0.5$,右边是$3.5$,左边如果弄$a=0.5$,会得到$\sqrt{3}+2>3.5$,那只能更小,说明$0<a<0.5$,根据函数连续这个值肯定存在啊

其他都是同理的

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realnumber + 1 对哦 ,才意识到

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original poster realnumber posted 2022-1-21 13:56
回复 3# 战巡


    选项C ,不可能总该证明了吧

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kuing posted 2022-1-21 15:50
回复 4# realnumber

那就是要证明 `4^x + 3 x > 3^x+ 4 x` 当 `x>1` 时恒成立。
禁导数确实不好弄,不然用勃撸力 `(4/3)^x=(1+1/3)^x>1+x/3` 得 `4^x>3^x+3^{x-1}x>3^x+x`。

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realnumber + 1 不努力不等式也不可啊,才高一,估计得凭指数 ...

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kuing posted 2022-1-21 18:17
长评论请用回帖写,评分那里显示不完,不知道你后面说凭指数什么?

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kuing posted 2022-1-21 18:19
勃撸力你可以先简单介绍一下,用 n=1,2,3 做实例,然后就说实际上对 >=1 都成立,证明暂且按下,下学期你就会了

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isee posted 2022-1-22 21:42
Last edited by isee 2022-1-22 22:22我现在是在“上帝”视角,这题出题人的目的应该是感知指数的函数的变化是增加很快,远远的将一次函数甩在后面,大于 1 时.

这是有多巧,知乎刷到个类题

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original poster realnumber posted 2022-1-23 18:02
意思是根据指数增长速度猜

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