Let f(x)= _ n ( n^n(x+n) (x+n 2 ) (x+n n ) n! (x^2+n^2 ) (x^2+n^2 4 ) (x^2+n^2 n^2 ) )^ x n , for all x>0. Then (A) f (1 2 ) f(1) (B) f (1 3 ) f (2 3 ) (C) f^ (2) 0 (D) f^ (3) f(3) f^ (2) f(2)

Let f(x)= _ n ( n^n(x+n) (x+n 2 ) (x+n n ) n! (x^2+n^2 ) (x^2+n^2…

MCQ

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

($A$) $f\left(\frac{1}{2}\right) \geq f(1)$

($B$) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$

($C$) $f^{\prime}(2) \leq 0$

($D$) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$

A
$B,A$
✓
$B,C$
C
$B,D$
D
$B,C,D$

✓

Answer

Correct option: B.

$B,C$

b
We have,

$f(x)=\lim _{n \rightarrow \infty}\left\{\frac{n^n(x+n)\left(x+\frac{n}{2}\right) \ldots\left(x+\frac{n}{2}\right)}{n!\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right) \ldots\left(x^2+\frac{n^2}{n^2}\right)}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \log \left\{\prod_{r=1}^n \frac{\left(x+\frac{n}{r}\right)}{\left(x^2+\frac{n^2}{r^2}\right)} \cdot \frac{n}{r}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{2} \sum_{r=1}^n \log \left\{\left(\frac{x+\frac{n}{r}}{x^2+\frac{n^2}{r^2}}\right) \cdot \frac{1}{\frac{r}{n}}\right\}$

$\Rightarrow \log f(x)=\lim _{n \rightarrow \infty} \frac{x}{n} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\}$

$\Rightarrow \log f(x)=x \lim _{n \rightarrow \infty} \sum_{r=1}^n \log \left\{\frac{1+\frac{r}{n} x}{1+\left(\frac{r}{n} x\right)^2}\right\} \frac{1}{n}$

$\Rightarrow \log f(x)=x \int_0^1 \log \left\{\frac{1+t x}{1+(t x)^2}\right\} d t$

$\Rightarrow \log f(x)=\int_0^x \log \left(\frac{1+u}{1+u^2}\right) d u \text {, where } u=t tx$

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

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.