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[不等式] 估值与放缩相关的一个选择题

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facebooker posted 2022-4-20 17:56 |Read mode
12. 下列不等式正确的有:$
A. \frac{101^{90}}{100^{91}}>\frac{3}{125}

B. \left(\frac{5}{4}\right)^{\sqrt{2}}>\left(\frac{6}{5}\right)^{\sqrt{3}}

C. e^{2-\sqrt{e}}>\frac{3}{2}

D. \frac{\pi}{2}>\tan1>\frac{3}{2}
$

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战巡 posted 2022-4-21 16:06
回复 1# facebooker


A、
这个等价于证明
\[(1+\frac{1}{100})^{90}>\frac{3\cdot 100}{125}=\frac{12}{5}\]
\[90\ln(1+\frac{1}{100})>\ln(\frac{12}{5})\]
这里面有
\[x-\frac{x^2}{2}<\ln(1+x)<x\]
故此
\[90\ln(1+\frac{1}{100})>90(\frac{1}{100}-\frac{1}{100^2\cdot 2})=0.8955\]
另一边则有
\[\ln(\frac{12}{5})=2\ln(2)+\ln(3)-\ln(5)\approx 2\cdot 0.693+1.099-1.609=0.876\]
故此成立

B
两边取对数,就变成证明
\[\sqrt{2}\ln(\frac{5}{4})>\sqrt{3}\ln(\frac{6}{5})\]
注意
\[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}<\ln(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}\]
这就有
\[\frac{1}{4}-\frac{1}{4^2\cdot 2}+\frac{1}{4^3\cdot 3}-\frac{1}{4^4\cdot 4}<\ln(\frac{5}{4})<\frac{1}{4}-\frac{1}{4^2\cdot 2}+\frac{1}{4^3\cdot 3}-\frac{1}{4^4\cdot 4}+\frac{1}{4^5\cdot 5}\]
\[\frac{685}{3072}<\ln(\frac{5}{4})<\frac{857}{3840}\]
\[0.22298<\ln(\frac{5}{4})<0.22318\]
\[1.4142\cdot 0.22298<\sqrt{2}\ln(\frac{5}{4})<1.4143\cdot 0.22318\]
\[0.31534<\sqrt{2}\ln(\frac{5}{4})<0.31564\]
另一边有
\[\frac{1}{5}-\frac{1}{5^2\cdot 2}+\frac{1}{5^3\cdot 3}-\frac{1}{5^4\cdot 4}<\ln(\frac{6}{5})<\frac{1}{5}-\frac{1}{5^2\cdot 2}+\frac{1}{5^3\cdot 3}-\frac{1}{5^4\cdot 4}+\frac{1}{5^5\cdot 5}\]
\[\frac{1367}{7500}<\ln(\frac{6}{5})<\frac{34187}{187500}\]
\[0.182267<\ln(\frac{6}{5})<0.182331\]
\[1.732\cdot  0.182267<\sqrt{3}\ln(\frac{6}{5})<1.7321\cdot 0.182331\]
\[0.315686<\sqrt{3}\ln(\frac{6}{5})<0.315815\]
故此会有
\[\sqrt{2}\ln(\frac{5}{4})<\sqrt{3}\ln(\frac{6}{5})\]
这个是错的

C、
取对数得到
\[2-\sqrt{e}>\ln(\frac{3}{2})\]
其实就是
\[2-\ln(3)+\ln(2)>\sqrt{e}\]
\[2-\ln(3)+\ln(2)\approx 2-1.099+0.693=1.594\]
然后
\[1.594^2=2.54<2.718<e\]
故此是错的

D、
$\tan(x)$我们可以找一个接近$1$,而且正切值容易求的位置展开,比如说$x=\frac{\pi}{3}$就是个不错的点
于是有
\[\tan(x)=\sqrt{3}+4(x-\frac{\pi}{3})+4\sqrt{3}(x-\frac{\pi}{3})^2+o(x-\frac{\pi}{3})^2\]
当$x<\frac{\pi}{3}$时,会有
\[\sqrt{3}+4(x-\frac{\pi}{3})<\tan(x)<\sqrt{3}+4(x-\frac{\pi}{3})+4\sqrt{3}(x-\frac{\pi}{3})^2\]
套进去$x=1$会有
\[\sqrt{3}+4(1-\frac{\pi}{3})<\tan(1)<\sqrt{3}+4(1-\frac{\pi}{3})+4\sqrt{3}(1-\frac{\pi}{3})^2\]
\[1.543<\tan(1)<1.559\]
那当然是对的

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kuing posted 2022-4-21 16:10
回复 2# 战巡

竟然不说直接按计算器

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original poster facebooker posted 2022-4-21 19:01
回复 2# 战巡


    大开眼界!

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isee posted 2022-4-21 19:12
回复 3# kuing

+1

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