In the interval $\left(\frac{1}{\mathrm{e}}, \mathrm{e}\right)$, a decreasing function among the following…
In the interval $\left(\frac{1}{\mathrm{e}}, \mathrm{e}\right)$, a decreasing function among the following functions is
$f(x)=\frac{\log x}{x}$
$f(x)=x^2 \log x$
$f(x)=x \log x$
$f(x)=x^{-x}$
Solution
The best way to approach this question is, find $1^{\text {st }}$ derivative of all the function at $x \in\left(\frac{1}{e}, e\right)$ you can choose $x=\frac{2}{e}, \frac{3}{e} \in\left(\frac{1}{e}, e\right) \cdot$ If $f^{\prime}(x)$ is -ve at any point in the given interval then that $f(x)$ is decreasing
For option (a): $f(x)=\frac{\log (x)}{x} \Rightarrow f^{\prime}(x)=\frac{1-\log x}{x^2}$. when $x=\frac{2}{e} \in\left(\frac{1}{e}, e\right) \Rightarrow f^{\prime}\left(\frac{2}{e}\right)>0$
Hence $f(x)$ is not decreasing.
For option (b) : $f(x)=x^2 \log (x)$
$f^{\prime}(x)>0$ when $x \in\left(\frac{1}{e}, e\right)$
For option (c): $f(x)=x \log x$
$f^{\prime}(x)>0$ when $x \in\left(\frac{1}{e}, e\right)$
For option (d) : $f(x)=x^{-} x$
$\Rightarrow \quad \log f(x)=-x \log (x)$ ...(i)
Hence $f^{\prime}(x)=-\mathrm{x}-x[1+\log x]$ ...(ii)
Let $x=\frac{2}{e} \in\left(\frac{1}{e}, e\right)$
$
f^{\prime}\left(\frac{2}{e}\right)=-\left(\frac{2}{e}\right)^{-2 / e}\left[1+\log \frac{2}{e}\right]=\frac{-1}{\left(\frac{2}{e}\right)^{2 / e}}[\log 2]
$
Since $\log 2=0.3010$
and $\left(\frac{2}{e}\right)^{2 / e}$ are positive terms.
Hence $f^{\prime}\left(\frac{2}{e}\right)=-\left[\frac{1}{\left(\frac{2}{e}\right)^{\left(\frac{2}{e}\right)}} \cdot \log 2\right] < 0$
Hence $f(x)$ is decreasing