Question

Let $f\left(x\right)\underset{n\to \infty }{lim}{\left(\frac{n^n(x+n)(x+\frac{n}{2})...(x+\frac{n}{n})}{n!(x^2+n^2)(x^2+\frac{n^2}{4})...(x^2+\frac{n^2}{n^2})}\right)}^{\frac{x}{n}}$, for all x>0. Then

• A. $f\left(\frac{1}{2}\right)\ge f\left(1\right)$
• B. $f\left(\frac{1}{3}\right)\le f\left(\frac{2}{3}\right)$
• C. $f^{\prime }\left(2\right)\le 0$
• D. $\frac{f^{\prime }\left(3\right)}{f\left(3\right)}\ge \frac{f^{\prime }\left(2\right)}{f\left(2\right)}$

Answer 1

Answer: (B), (C)

$f\left(\frac{1}{3}\right)\le f\left(\frac{2}{3}\right)$
$f^{\prime }\left(2\right)\le 0$

$f\left(x\right)=\underset{n\to \infty }{lim}{\left\{\frac{n^{2n}(\frac{x}{n}+1))(\frac{x}{n}+2))...(\frac{x}{n}+\frac{1}{n})}{n!n^{2n}(\frac{x^2}{n^1}+1)(\frac{x^2}{n^2}+\frac{1}{2^2})...(\frac{x^2}{n^2}+\frac{1}{n^2})}\right\}}^{x/n}$

$\ell nf\left(x\right)=\underset{n\to \infty }{lim}\frac{x}{n}\{\sum _{r=1}^n\ell n(\frac{x}{n}+\frac{1}{r})-\sum _{r=1}^n\ell n(\frac{r{x}^2}{n^2}+\frac{1}{r})\}=\underset{n\to \infty }{lim}\frac{x}{n}\{\sum _{r=1}^n\ell n(1+\frac{rx}{n})-\sum _{r=1}^n\ell n(1+\frac{r^2{x}^2}{n^2})\}$

$\ell n\left(f\right(c\left)\right)=x{\int }_0^1\ell n(1+xy)dy-x{\int }_0^1\ell n(1+x^2{y}^2)dy$
Let $xy=t$

$\ell b\left(f\right(x\left)\right)={\int }_0^x\ell n(1+t)dt-{\int }_0^x\ell n(1+t^2)dt$

$\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\ell n\left(\frac{1+x}{1+x^2}\right)$

$\frac{f^{\prime }\left(2\right)}{f\left(2\right)}=\ell n\left(\frac{3}{5}\right)<0\Rightarrow f^{\prime }\left(2\right)<0$

$\frac{f^{\prime }\left(3\right)}{f\left(3\right)}=\ell n\left(\frac{4}{10}\right)=\ell n\left(\frac{2}{5}\right)\Rightarrow \frac{f^{\prime }\left(2\right)}{f\left(2\right)}\ge \frac{f^{\prime }\left(3\right)}{f\left(3\right)}$

Now $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}>0$ in (0, 1) and $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}<0$ in $(1,\infty )$

f(x) is increasing in (0, 1) & decreasing in $[1,\infty )$ (as f(x) is positive)
Hence $f\left(1\right)\ge f\left(\frac{1}{2}\right)$ and $f(\frac{1}{3}\le f(\frac{2}{3})$