The function $f(x)=\frac{\log _e(\pi+x)}{\log _e(e+x)}$ is

The function $f(x)=\frac{\log _e(\pi+x)}{\log _e(e+x)}$ is
  1. increasing on $(0, \infty)$.
  2. increasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, decreasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$.
  3. decreasing on $(0, \infty)$.
  4. decreasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, increasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$

Solution

$\begin{aligned} & \quad \text { Let } \mathrm{f}(x)=\frac{\ln (\pi+x)}{\ln (\mathrm{e}+x)} \\ & \begin{aligned} \therefore \quad \mathrm{f}^{\prime}(x) & =\frac{\ln (\mathrm{e}+x) \times \frac{1}{\pi+x}-\ln (\pi+x) \times \frac{1}{\mathrm{e}+x}}{[\ln (\mathrm{e}+x)]^2} \\ & =\frac{(\mathrm{e}+x) \ln (\mathrm{e}+x)-(\pi+x) \ln (\pi+x)}{[\ln (\mathrm{e}+x)]^2 \times(\mathrm{e}+x)(\pi+x)} \\ \Rightarrow \quad \Rightarrow f^{\prime}(x) & \lt 0 \text { for all } x\gt0 \\ \therefore \quad \mathrm{f}(x) & \text { is decreasing on }(0, \infty) .\end{aligned}\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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