Let f be a function such that f(x) = sum_{r=1 | vedclass

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(B) $f(x) = \sum_{r=1}^n r + \sum_{r=1}^n [\cos(\frac{x}{r})]$.Since $r$ is an integer,$[r + \cos(\frac{x}{r})] = r + [\cos(\frac{x}{r})]$.$f(x) = \fr

Vedclass · Accepted Answer

$\{\frac{n^2+n-2}{2}, \frac{n^2+n}{2}, \frac{n^2+3n}{2}\}$

Vedclass · Answer

(B) $f(x) = \sum_{r=1}^n r + \sum_{r=1}^n [\cos(\frac{x}{r})]$.Since $r$ is an integer,$[r + \cos(\frac{x}{r})] = r + [\cos(\frac{x}{r})]$.$f(x) = \frac{n(n+1)}{2} + \sum_{r=1}^n [\cos(\frac{x}{r})]$.For $x \in [0, \pi]$,$0 \le \frac{x}{r} \le \pi$. Thus,$\cos(\frac{x}{r}) \in [-1, 1]$.If $x=0$,then $\cos(\frac{x}{r}) = 1$ for all $r$,so $[\cos(\frac{x}{r})] = 1$. Thus,$f(0) = \frac{n(n+1)}{2} + n = \frac{n^2+3n}{2}$.If $x \in (0, \pi]$,then $\cos(\frac{x}{r}) If $x = \frac{\pi}{2}$,$\cos(\frac{\pi}{2r}) \ge 0$ for all $r \ge 1$,so $[\cos(\frac{\pi}{2r})] = 0$. Thus,$f(\frac{\pi}{2}) = \frac{n(n+1)}{2}$.If $x = \pi$,$\cos(\frac{\pi}{r}) = -1$ for $r=1$ and $\cos(\frac{\pi}{r}) \in (-1, 1)$ for $r > 1$. Thus,$[\cos(\pi)] = -1$ and $[\cos(\frac{\pi}{r})] = 0$ for $r > 1$. Thus,$f(\pi) = \frac{n(n+1)}{2} - 1 = \frac{n^2+n-2}{2}$.Therefore,the range is $\{\frac{n^2+n-2}{2}, \frac{n^2+n}{2}, \frac{n^2+3n}{2}\}$.

Let $f$ be a function such that $f(x) = \sum_{r=1}^n [r + \cos(\frac{x}{r})]$,where $[.]$ denotes the greatest integer function and $x \in [0, \pi]$. Then the range of $f(x)$ is:

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