Let f: [2, 4] → ℝ be a differentiable function such that (x _ e x)...

Given,

Domain and range of function,

$f: \left[2, 4\right] \to \mathbb{R}$

\left(x{\log}_{e}x\right){f}^{'}\left(x\right) + \left(\log_{e}x\right)f\left(x\right) + f\left(x\right) \ge 1, x \in \left[2, 4\right]$</p><p>$⇒ \left(x{\log}_{e}x\right)\dfrac{d}{dx}f\left(x\right) + \left(\log_{e}x\right)f\left(x\right)\dfrac{d}{dx}\left(x\right) + xf\left(x\right)\dfrac{d}{dx}\left(\log_{e}x\right) \ge 1$</p><p>$⇒ \dfrac{d}{dx}\left[xlnxf\left(x\right)\right] \ge 1$</p><p>$⇒ \dfrac{d}{dx}\left[xlnxf\left(x\right)\right] \ge \dfrac{d}{dx}\left(x\right)$</p><p>$⇒ \dfrac{d}{dx}\left[xlnxf\left(x\right) - x\right] \ge 0$</p><p>Hence,$h ( x ) = x\left(\ln x\right)f\left(x\right) - x$is a increasing function,</p><p>$∴ h ( x ) \ge h ( 2 ) , x \in [ 2, 4 ]$</p><p>$⇒ xlnx \times f\left(x\right) - x \ge 2\ln 2 \times f\left(2\right) - 2$</p><p>$⇒ xlnxf ( x ) - x \ge \ln 2 - 2$</p><p>Similarly,$h\left(x\right) \le h\left(4\right)$</p><p>$⇒ xlnxf\left(x\right) - x \le \ln 4 - 4$</p><p>So,$\dfrac{\ln 2 - 2}{xlnx} + \dfrac{1}{\ln x} \le f ( x ) \le \dfrac{\ln 4 - 4}{xlnx} + \dfrac{1}{\ln x}$</p><p>Now for$x \in \left[2, 4\right]$</p><p>$\dfrac{\ln 4 - 4}{xlnx} + \dfrac{1}{\ln x} \le \dfrac{\ln 4 - 4}{2\ln 2} + \dfrac{1}{\ln 2} = 1 - \dfrac{1}{\ln 2} < 1$</p><p>$⇒ f \left(x\right)$\le 1$\text{for}x \in \left[2, 4\right]

Now$for x \in \left[2, 4\right]$

$\dfrac{\ln 2 - 2}{xlnx} + \dfrac{1}{\ln x}$ \ge $\dfrac{\ln 2 - 2}{4\ln 4} + \dfrac{1}{\ln 4} = \dfrac{\ln 2 - 2}{4\ln 4} + \dfrac{1}{\ln 4} = \dfrac{1}{8} + \dfrac{1}{4\ln 2}$ > $\dfrac{1}{8}</p><p>Hence,$f ( x ) \ge $\dfrac{1}{8}</p><p>Hence both A$ & $B$ are correct.
Note this question was bonus in Jee Main 2023 April session, as LMVT on $f ( x ) \cdot xlnx$ can't be satisfied.

Hence, no such $f ( x )$ exist.