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It is given $f_n: \mathbb{(0,\infty)} \to \mathbb{R}$ for any positive integer $n,$

If $f_n(x) =\tan^{-1}\frac{n}{1+x(x+n)} $

I have one query $\mathbb{R}$ is $(-\infty,\infty)$, but the range of $f_n(x)$ will be $(0,\frac{\pi}{2})$ for given domain $(0,\infty)$.

Because this question is from a reputed examination, i would like to know whether the given range is correct.

Michael Hardy
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Samar Imam Zaidi
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    The $\Bbb R$ that appears in the declaration of $f_n$, i.e. $f_n:(0,\infty)\to \Bbb R$, is not the range, but rather is the codomain. The range is a subset of the codomain. – Dave Jun 03 '18 at 04:15
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    If we want to be pedantic, the range of $f_n(x)$ for a specified value of $n$ would be $(0,\arctan(n))$, not $(0,\frac{\pi}{2})$ noting that $\arctan(n)<\frac{\pi}{2}$ for any finite value of $n$ and that for every value of $n$ the function is monotonic decreasing over the interval $(0,\infty)$. The union of all of the ranges of $f_n$ would be $(0,\frac{\pi}{2})$. – JMoravitz Jun 03 '18 at 04:26

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