Let f:[2,4] R be a differentiable function such that (x _ex)f^'(x)+( _ex)f(x)+f(x) 1,x [2,…

Question

Let $f:[2,4]	o R$ be a differentiable function such that $(x{\log }_{e}x){f}^{'}(x)+({\log }_{e}x)f(x)+f(x)\ge 1,x\in [2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements :

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

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

Then,

Accepted Answer

Given, $$\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$$

$$

\left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1$$

$$ \Rightarrow $$ $$
\frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1$$

$$
\begin{aligned}
&\Rightarrow  \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\
& \forall x \in[2,4]
\end{aligned}
$$

Let $g(x)=x \log _e x f(x)-x$

As $g(x) \geq 0, \forall x \in[2,4],
g(x)$ is an increasing function in $[2,4]$

$$
\begin{aligned}
g(2) & =2 \log _e 2 f(2)-2 \\
& =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right]
\end{aligned}
$$

$$
\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}
$$

As, $g(x)$ is an increasing function,

$$
\begin{aligned}
& g(2) \leq g(x) \leq g(4) \\
& \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\
& \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right)
\end{aligned}
$$

$$
\frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x}
$$

$$
\begin{aligned}
& \text {Now for } x \in[2,4] \\
& \begin{aligned}
\frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}\frac{1}{8}
\end{aligned}
\end{aligned}$$

$$
\therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4]
$$

Hence, both statements $A$ and $B$ are true.

Note : LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied.

Hence no such function exists.

Therefore it should be bonus.

Solutions · 1