Discussion Forum

Let f(x)$=\lim_{n \rightarrow \infty}\left[\frac{n^{n}(x+n)(x+\frac{n}{2})....(x+\frac{n}{n})}{n!(x^{2}+n^{2})(x^{2}+\frac{n^{2}}{4}).....(x^{2}+\frac{n^{2}}{n^{2}})}\right]^{\frac{x}{n}}$

for all x=0, then

Answer:

Explanation:

Here, f(x)

  $=\lim_{n \rightarrow \infty}\left[\frac{n^{n}(x+n)(x+\frac{n}{2})....(x+\frac{n}{n})}{n!(x^{2}+n^{2})(x^{2}+\frac{n^{2}}{4}).....(x^{2}+\frac{n^{2}}{n^{2}})}\right]^{\frac{x}{n}}$

     x>0

Taking  log on both sides, we get  $\log_{e}[f(x)]$

             $=\lim_{n \rightarrow \infty}\log\left[\frac{n^{n}(x+n)(x+\frac{n}{2})....(x+\frac{n}{n})}{n!(x^{2}+n^{2})(x^{2}+\frac{n^{2}}{4}).....(x^{2}+\frac{n^{2}}{n^{2}})}\right]^{\frac{x}{n}}$

$=\lim_{n \rightarrow \infty}\frac{x}{n}.\log \left[\frac{\prod_{r=1}^{n}(x+\frac{1}{r/n})}{\prod_{r=1}^n(x^{2}+\frac{1}{(r/n)^{2}})\prod_{r=1}^{n}(r/n)}\right]$

$=x\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r=1}^{n}\log\left[\frac{x+\frac{n}{r}}{(x^{2}+\frac{n^{2}}{r^{2}})(r/n)}\right]$

$=x\lim_{n \rightarrow \infty}\frac{1}{n}\sum_1^n \log \left[\frac{\frac{r}{n}.x+1}{\frac{r^{2}}{n^{2}}.x^{2}+1}\right]$

Converting summation into definite integration, we get  $\log_{e}[f(x)]$

$=x\int_{0}^{1} \log (\frac{xt+1}{x^{2}t^{2}+1})dt$

put $tx=z\Rightarrow xdt=dz$

$\therefore  \log_{e}[f(x)]=x\int_{0}^{x} \log(\frac{1+z}{1+z^{2}})\frac{dz}{x}$

$\Rightarrow  \log_{e}[f(x)]=\int_{0}^{x} \log(\frac{1+z}{1+z^{2}})dz$

Using Newton-Leibnitz formula, we get

$\frac{1}{f(x)}.f'(x)=\log(\frac{1+x}{1+x^{2}})......(i)$

Here x=1,

   $\frac{f'(1)}{f(1)}=\log(1)=0$

$\therefore$          F'(1)=0

Now,sign scheme of f'(x) is shown below

               $\therefore$  At x=1, function attains maximum

 Since, f(x) increases on (0,1)

                         f(1)>f(1/2)

$\therefore$   Option (a) is incorrect

                               f(1/3)<f(2/3)

$\therefore$    Option (b) is correct

 Also ,f'(x)<0, when x>1

          $\Rightarrow$   f'(2)<0

$\therefore$ Option(c) is correct

         Also,  $\frac{f'(x)}{f(x)}=\log( \frac{1+x}{1+x^{2}})$

$\therefore$   $\frac{f'(3)}{f(3)}-\frac{f'(2)}{f(2)}=\log(\frac{4}{10})-\log(\frac{3}{5})$

                              =  $\log(2/3)<0$

$\Rightarrow  \frac{f'(3)}{f(3)}<\frac{f'(2)}{f(2)}$

    $\therefore$  Option (d) is incorrect