Correct option: C.Both the statement $(A)$ and $(B)$ are true
c
$x \operatorname{lnxf} f^{\prime}(x)+\ln x f(x)+f(x) \geq I, x \in[2,4]$
And $f (2)=\frac{1}{2}, f (4)=\frac{1}{4}$
Now $x \ln x \frac{d y}{d x}+(\ln +1) y \geq 1$
$\frac{ d }{ dx }( y \cdot x \ln x ) \geq 1$
$\frac{ d }{ dx }( f ( x ) x \ln x ) \geq 1$
$\Rightarrow \frac{ d }{ dx }( x \ln xf ( x )- x ) \geq 0, x \in[2,4]$
$\Rightarrow \text { The function } g(x)=x \ln x f(x)-x \text { is increasing in }$
$\Rightarrow$ The function $g(x)=x \ln x f(x)-x$ is increasing in $[2.4]$
And $g(2)=2 \ln 2 f(2)-2=\ln 2-2$
$g(4)=4 \ln 4 f(4)-4=\ln 4-4$
$=2(\ln 2-2)$
Now $\quad g (2) \leq g ( x ) \leq g (4)$
$\ln 2-2 \leq x \ln x f(x)-x \leq 2(\ln 2-2)$
$\frac{\ln 2-2}{x \ln x}+\frac{1}{\ln x} \leq f(x) \leq \frac{2(\ln 2-2)}{x \ln x}+\frac{1}{\ln x}$
Now for $x \in[2,4]$
$\frac{2(\ln 2-2)}{ x \ln x}+\frac{1}{\ln x}<\frac{2(\ln 2-2)}{2 \ln 2}+\frac{1}{\ln 2}=1-\frac{1}{\ln 2} < 1$
$\Rightarrow f ( x ) \leq 1 \text { for } x \in[2,4]$
Also for $x \in[2,4]$ :
$\frac{\ln 2-2}{x \ln x}+\frac{1}{\ln x} \geq \frac{\ln 2-2}{4 \ln 4}+\frac{1}{\ln 4}$$=\frac{1}{8}+\frac{1}{2 \ln 2}>\frac{1}{8}$
$\Rightarrow f(x) \geq \frac{1}{8} \text { for } x \in[2,4]$
Hence both $A$ and $B$ are true.
LMVT on $(yx (lnx))$ not satisfied.
Hence no such function exists.
Therefore it should be bonus.