माना mathrm{f}:[2,4] rightarrow mathbb{R} एक | vedclass

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Question

$x \operatorname{lnxf} f^{\prime}(x)+\ln x f(x)+f(x) \geq I, x \in[2,4]$

And $f (2)=\frac{1}{2}, f (4)=\frac{1}{4}$

Now $x \ln x \frac{d y}{d x}

Vedclass · Accepted Answer

($A$) तथा ($B$) दोनों कथन सत्य हैं

Vedclass · Answer

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

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