Let f(x) = frac{tan^n x}{sum_{r=0}^{2n} tan^r | vedclass

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(A) Let $	an x = t$. Since $x \in [0, \frac{\pi}{2})$,$t \in [0, \infty)$.$f(x) = \frac{t^n}{1 + t + t^2 + \dots + t^{2n}}$.For $t=0$,$f(x) = 0$.For

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$f(x)$ is bounded and it takes both of its bounds and the range of $f(x)$ contains exactly one integral point.

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(A) Let $	an x = t$. Since $x \in [0, \frac{\pi}{2})$,$t \in [0, \infty)$.$f(x) = \frac{t^n}{1 + t + t^2 + \dots + t^{2n}}$.For $t=0$,$f(x) = 0$.For $t > 0$,divide numerator and denominator by $t^n$:$f(x) = \frac{1}{t^n + t^{n-1} + \dots + 1 + \dots + \frac{1}{t^{n-1}} + \frac{1}{t^n}}$.Using the $AM$-$GM$ inequality,$t^k + \frac{1}{t^k} \ge 2$ for $t > 0$.Thus,the denominator is $\ge 2 + 2 + \dots + 2 + 1 = 2n + 1$.Therefore,$f(x) \le \frac{1}{2n+1}$.The equality holds when $t=1$,i.e.,$x = \frac{\pi}{4}$.Since $f(0) = 0$ and the maximum value is $\frac{1}{2n+1}$,the range is $[0, \frac{1}{2n+1}]$.Since $n \in N$,$2n+1 \ge 3$,so $\frac{1}{2n+1} < 1$. The only integer in the range $[0, \frac{1}{2n+1}]$ is $0$. Thus,it contains exactly one integral point.

Let $f(x) = \frac{\tan^n x}{\sum_{r=0}^{2n} \tan^r x}$,$n \in N$,where $x \in [0, \frac{\pi}{2})$.

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