AMC12 2020 B

AMC12 2020 B · Q22

AMC12 2020 B · Q22. It mainly tests Quadratic equations, Exponents & radicals.

What is the maximum value of \[\frac{t(2^t - 3t)}{4^t}\] for real values of $t$?

实数 $t$ 的 \[\frac{t(2^t - 3t)}{4^t}\] 的最大值是多少？

(A) \dfrac{1}{16} \dfrac{1}{16}

(B) \dfrac{1}{15} \dfrac{1}{15}

(D) \dfrac{1}{10} \dfrac{1}{10}

(E) \dfrac{1}{9} \dfrac{1}{9}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $x=\dfrac{3t}{2^t}$. Then the given expression is $y=\dfrac{1}{3}x-\dfrac{1}{3}x^2$, whose graph is a downward-opening parabola with vertex $\left(\dfrac{1}{2},\dfrac{1}{12}\right)$. Because $x=0$ when $t=0$ and $x=\dfrac{3}{2}$ when $t=1$, there must be some value of $t$ between 0 and 1 for which $x=\dfrac{1}{2}$, and therefore the maximum $y=\dfrac{1}{12}$ is attained. (There is also a value of $t>1$ for which $x=\dfrac{1}{2}$.)

答案（C）：令 $x=\dfrac{3t}{2^t}$。则所给表达式为 $y=\dfrac{1}{3}x-\dfrac{1}{3}x^2$，其图像是一条开口向下的抛物线，顶点为 $\left(\dfrac{1}{2},\dfrac{1}{12}\right)$。因为当 $t=0$ 时 $x=0$，当 $t=1$ 时 $x=\dfrac{3}{2}$，所以在 0 与 1 之间必存在某个 $t$ 使得 $x=\dfrac{1}{2}$，因此最大值 $y=\dfrac{1}{12}$ 能取得。（此外也存在某个 $t>1$ 使得 $x=\dfrac{1}{2}$。）

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