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mathdeptv2/工具/文本文件/metadata.txt

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ans
14511
$(-\infty,2)$
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$(-3,5)$
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$80$
14514
$-2$
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$\sqrt{10}$
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$10.8$
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$\dfrac{7}{25}$
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$\dfrac{\sqrt{3}}3\pi$
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$\dfrac{2\sqrt{6}}5$
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$\dfrac{3}{10}$
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$10$
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$\dfrac{3\pi}{8}$
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C
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D
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B
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A
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(1) $2n-1$; (2) $4n^2+\dfrac{9^n}{4}-\dfrac 14$
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(1) 证明略; (2) $\dfrac{2\sqrt{3}}3$
14529
(1) 如设$AE=x$百米, 则$y=4x+8-2\sqrt{x^2-9}$($3<x<5$); 如设$\angle MAE=x$, 则$y=\dfrac{12}{\cos x}+8-6\tan x$($0<x<\arctan\dfrac 43$)等; (2) 约$18.39$百米
14530
(1) $\dfrac{x^2}8+\dfrac{y^2}4=1$; (2) 证明略; (3) 斜率的最小值为$\dfrac{\sqrt{6}}2$, 此时直线$l_1$的方程为$y=\dfrac{\sqrt{6}}6x+\dfrac{2\sqrt{7}}7$
14531
(1) $f(x)=2^x$是关于$(0,+\infty)$的同变函数, 不是关于$(0,1)$的同变函数; (2) $f(x)=\sqrt{2(x-2k)}+2k$, 当$x=2k+\dfrac 12$($k\in\mathbf{Z}$)时, $f(x)=x+\dfrac 12$, 当$x\ne 2k+\dfrac 12$($k\in\mathbf{Z}$)时, $f(x)<x+\dfrac 12$; (3) 证明略