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(D) Let $n = [a]$. Then $a = n + f$,where $0  [a]^k$,which means $[(n+f)^k] > n^k$. By the binomial expansion,$(n+f)^k

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$k \leq \frac{1}{a-[a]}+1$

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(D) Let $n = [a]$. Then $a = n + f$,where $0 We are given $[a^k] > [a]^k$,which means $[(n+f)^k] > n^k$. By the binomial expansion,$(n+f)^k = n^k + k n^{k-1} f + \binom{k}{2} n^{k-2} f^2 + \dots + f^k$. For the condition $[a^k] > n^k$ to hold,we must have $(n+f)^k \geq n^k + 1$. Using the first two terms of the expansion,$n^k + k n^{k-1} f > n^k$,which implies $k n^{k-1} f > 1$. Since $n \geq 1$,$n^{k-1} \geq 1$,so $k f > 1$ is a necessary condition for the inequality to hold for some $k$. Thus,$k > \frac{1}{f} = \frac{1}{a-[a]}$. The smallest such integer $k$ satisfies $k \leq \frac{1}{a-[a]} + 1$.

For a real number $r$,let $[r]$ denote the largest integer less than or equal to $r$. Let $a > 1$ be a real number which is not an integer,and let $k$ be the smallest positive integer such that $[a^k] > [a]^k$. Then,which of the following statements is always true?

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