Let f : 2 , 4 → ℝ be a differentiable function such that x log e x f ' x + log e x f x + f x…

Given,

Domain and range of function,

$f:\left[2,4\right]\to \mathrm{ℝ}$

$\left(x{\log }_{e}x\right)f^{\text{'}}\left(x\right)+\left({\log }_{e}x\right)f\left(x\right)+f\left(x\right)\ge 1, x\in \left[2,4\right]$

$\Rightarrow \left(x{\log }_{e}x\right)\frac{d}{dx}f\left(x\right)+\left({\log }_{e}x\right)f\left(x\right)\frac{d}{dx}\left(x\right)+xf\left(x\right)\frac{d}{dx}\left({\log }_{e}x\right)\ge 1$

$\Rightarrow \frac{d}{dx}\left[x\ln xf\left(x\right)\right]\ge 1$

$\Rightarrow \frac{d}{dx}\left[x\ln xf\left(x\right)\right]\ge \frac{d}{dx}\left(x\right)$

$\Rightarrow \frac{d}{dx}\left[x\ln xf\left(x\right)-x\right]\ge 0$

Hence, $h\left(x\right)=x\left(\ln x\right)f\left(x\right)-x$ is a increasing function,

$\therefore h\left(x\right)\ge h\left(2\right),x\in [2,4]$

$\Rightarrow x\ln x\times f\left(x\right)-x\ge 2\ln 2\times f\left(2\right)-2$

$\Rightarrow x\ln xf\left(x\right)-x\ge \ln 2-2$

Similarly, $h\left(x\right)\le h\left(4\right)$

$\Rightarrow x\ln xf\left(x\right)-x\le \ln 4-4$

So, $\frac{\ln 2-2}{x\ln x}+\frac{1}{\ln x}\le f\left(x\right)\le \frac{\ln 4-4}{x\ln x}+\frac{1}{\ln x}$

Now for $x\in \left[2,4\right]$

$\frac{\ln 4-4}{x\ln x}+\frac{1}{\ln x}\le \frac{\ln 4-4}{2\ln 2}+\frac{1}{\ln 2}=1-\frac{1}{\ln 2}<1$

$\Rightarrow f\left(x\right)\le 1 \text{for} x\in \left[2,4\right]$

Now for $x\in \left[2,4\right]$

$\frac{\ln 2-2}{x\ln x}+\frac{1}{\ln x}\ge \frac{\ln 2-2}{4\ln 4}+\frac{1}{\ln 4}=\frac{\ln 2-2}{4\ln 4}+\frac{1}{\ln 4}=\frac{1}{8}+\frac{1}{4\ln 2}>\frac{1}{8}$

Hence, $f\left(x\right)\ge \frac{1}{8}$

Hence both $A$ & $B$ are correct.
Note this question was bonus in Jee Main 2023 April session, as LMVT on $f\left(x\right)·x\ln x$ can't be satisfied.

Hence, no such $f\left(x\right)$ exist.