AMC12 2014 A

AMC12 2014 A · Q21

AMC12 2014 A · Q21. It mainly tests Exponents & radicals, Inequalities with floors/ceilings (basic).

For every real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$, and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?

对于每个实数 $x$，令 $\lfloor x\rfloor$ 表示不超过 $x$ 的最大整数，并令 \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] 所有满足 $1\leq x<2014$ 且 $f(x)\leq 1$ 的数的集合是若干不相交区间的并集。这些区间的长度之和是多少？

(A) 1 1

(B) \dfrac{\log 2015}{\log 2014} \dfrac{\log 2015}{\log 2014}

(D) \dfrac{2014}{2013} \dfrac{2014}{2013}

(E) 2014^{\frac1{2014}}\qquad 2014^{\frac1{2014}}\qquad

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$. Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$. In order for this to be less than or equal to $1$, we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$. Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$, and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$. We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]

设 $\lfloor x\rfloor=k$，其中整数 $1\leq k\leq 2013$。则可以将 $f(x)$ 重写为 $k(2014^{x-k}-1)$。为了使之小于等于 $1$，需 $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$。结合 $\lfloor x\rfloor =k$ 的条件，得 $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$，故区间长度为 $\log_{2014}\left(\dfrac{k+1}k\right)$。我们要求所有满足不等式的可能区间的长度之和；由于这些区间互不相交，故可从 $k=1$ 到 $k=2013$ 求和，得所需和为 \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]

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