($A$) $f\left(\frac{1}{2}\right) \geq f(1)$
($B$) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$
($C$) $f^{\prime}(2) \leq 0$
($D$) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$
$f(x)=\lim _{n \rightarrow \infty}\left\{\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \ldots\left(x+\frac{n}{2}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) \ldots\left(x^2+\frac{n^2}{n^2}\right)}\right\}$
$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \log \left\{\prod_{r=1}^n \frac{\left(x+\frac{n}{r}\right)}{\left(x^2+\frac{n^2}{r^2}\right)} \cdot \frac{n}{r}\right\}$
$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{2} \sum_{r=1}^n \log \left\{\left(\frac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}}\right) \cdot \frac{1}{\frac{r}{n}}\right\}$
$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\}$
$\Rightarrow \log f(x)=x \lim _{n \rightarrow \infty} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\} \frac{1}{n}$
$\Rightarrow \log f(x)=x \int_0^1 \log \left\{\frac{1+t x}{1+(t x)^2}\right\} d t$
$\Rightarrow \log f(x)=\int_0^x \log \left(\frac{1+u}{1+u^2}\right) d u \text {, where } u=t tx$
$\Rightarrow \frac{f^{\prime}(x)}{f(x)}=\log \left(\frac{1+x}{1+x^2}\right)$
We observe that $f(x)>0$ for all $a>0$.
Also,
$\log \left(\frac{1+x}{1+x^2}\right) > 0 \Leftrightarrow \frac{1+x}{1+x^2} > 1 \Leftrightarrow x>x^2 \Leftrightarrow 0$
$ < x < 1$
$\therefore \frac{f^{\prime}(x)}{f(x)}>0 \text { for } 0 $ < 0 \text { for } x > 1$ $\Rightarrow f(x)$ is increasing on $(0,1)$ and decreasing on $(1, \infty)$. $\Rightarrow f\left(\frac{1}{2}\right) \leq f(1) \text { and } f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$ So option$(b)$ is correct and option $(a)$ is incorrect. From$(ii)$, we obtain $f^{\prime}(2) < 0$. So,option $(c)$ is correct. From$(i)$, we obtain $\frac{f^{\prime}(3)}{f(3)}-\frac{f^{\prime}(2)}{f(2)}=\frac{\log (4)}{10}-\frac{\log (3)}{5}=\log \left(\frac{2}{3}\right) < 0$ $\Rightarrow \frac{f^{\prime}(3)}{f(3)} < \frac{f^{\prime}(2)}{f(2)} \text {. So, option (d) is not correct. }$
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