The function $f(x)=\frac{\log (\pi+x)}{\log (e+x)}$ is
The function $f(x)=\frac{\log (\pi+x)}{\log (e+x)}$ is
- decreasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, increasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$
- increasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, decreasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$
- increasing on $(0, \infty)$
- decreasing on $(0, \infty)$
Solution
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\log (\pi+\mathrm{x})}{\log (\mathrm{e}+\mathrm{x})} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\frac{1}{\pi+\mathrm{x}} \log (\mathrm{e}+\mathrm{x})-\log (\pi+\mathrm{x}) \times \frac{1}{\mathrm{e}+\mathrm{x}}}{\{\log (\mathrm{e}+\mathrm{x})\}^2} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})}{(\pi+\mathrm{x})(\mathrm{e}+\mathrm{x})\{\log (\mathrm{e}+\mathrm{x})\}^2}\end{aligned}$
Let $g(x)=(e+x) \log (e+x)-(\pi+x) \log (\pi+x)$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\log (\mathrm{e}+\mathrm{x})-\log (\pi+\mathrm{x})<0$
So, $\mathrm{g}(\mathrm{x})$ is decreasing
$\begin{aligned}
& \Rightarrow \mathrm{g}(\mathrm{x})<\mathrm{g}(0) \forall \mathrm{x} \in(0, \infty) \\
& \Rightarrow(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})<\mathrm{e} \log \mathrm{e}-\pi \log \pi<0 \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})<0 \forall \mathrm{x} \in(0, \infty) \\
& \Rightarrow \mathrm{f}(\mathrm{x}) \text { is decreasing on }(0, \infty)
\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 1)
Practice more Applications of Derivatives questions on Aicharya