- AIncreasing in $(1,\,\,\infty )$
- BDecreasing in $(1,\,\,\infty )$
- ✓Increasing in $(1,\,e),$ decreasing in $(e,\infty )$
- DDecreasing in $(1,\,e),$ increasing in$(e,\infty )$
==>$\frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{{x^2}}} - \frac{{\log x}}{{{x^2}}} = \frac{{1 - \log x}}{{{x^2}}}$
==>$\frac{{dy}}{{dx}} = {x^{1/x}}\left( {\frac{{1 - \log x}}{{{x^2}}}} \right)$
Now, ${x^{1/x}} > 0$for all x and $\frac{{1 - \log x}}{{{x^2}}} > 0$ in $ (1, e) $ and
$\frac{{1 - \log x}}{{{x^2}}} < 0$ in $(e,\infty )$
$\therefore$ $f(x)$ is increasing in $ (1, e) $ and decreasing in $(e,\,\infty ).$
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where [ ] denotes the greatest integer function is :
$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$
$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$
$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$
$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$