Let $f:[2,4] \rightarrow \mathbb{R}$ be a differentiable function such that $$\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements :

(A) : $f(x) \leq 1$, for all $x \in[2,4]$

(B) : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,

Given, $$\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$$ <br/><br/>$\left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1$ <br/><br/>$\Rightarrow$ $\frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1$ <br/><br/>$$ \begin{aligned} &\Rightarrow \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\\\ & \forall x \in[2,4] \end{aligned} $$ <br/><br/>Let $g(x)=x \log _e x f(x)-x$ <br/><br/>As $g(x) \geq 0, \forall x \in[2,4], g(x)$ is an increasing function in $[2,4]$ <br/><br/>$$ \begin{aligned} g(2) & =2 \log _e 2 f(2)-2 \\\\ & =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right] \end{aligned} $$ <br/><br/>$$ \begin{aligned} & g(4)=4 \log _e 4 f(4)-4=\log _e 4-4 \\\\ &=2\left(\log _e 2-2\right)\\\\ & {\left[\therefore f(4)=\frac{1}{4}\right] } \end{aligned} $$ <br/><br/>As, $g(x)$ is an increasing function, <br/><br/>$$ \begin{aligned} & g(2) \leq g(x) \leq g(4) \\\\ & \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\\\ & \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right) \end{aligned} $$ <br/><br/>$$ \frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x} $$ <br/><br/>$$ \begin{aligned} & \text {Now for } x \in[2,4] \\\\ & \begin{aligned} \frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}<1 \end{aligned} \end{aligned} $$ <br/><br/>$\Rightarrow \mathrm{f}(\mathrm{x}) \leq 1 \text { for } \mathrm{x} \in[2,4]$ <br/><br/>$\text { Also for } \mathrm{x} \in[2,4] \text { : }$ <br/><br/>Now, <br/><br/>$$ \begin{aligned} \frac{2\left(\log _e 2-2\right)+2}{2 \log _e 2} & \geq \frac{\log _e 2-2+4}{4 \log _e 4} =\frac{1}{8}+\frac{1}{2 \log _e 2}>\frac{1}{8} \end{aligned} $$ <br/><br/>$\therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4]$ <br/><br/>Hence, both statements $A$ and $B$ are true. <br/><br/><b>Note :</b> LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied. <br/><br/>Hence no such function exists. <br/><br/>Therefore it should be bonus.