Q94 / 150 · MHT CET 2022 (06 Aug Shift 1) · Mathematics

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

  • A
    decreasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, increasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$
  • B
    increasing on $\left(0, \frac{\pi}{\mathrm{e}}\right)$, decreasing on $\left(\frac{\pi}{\mathrm{e}}, \infty\right)$
  • C
    increasing on $(0, \infty)$
  • Correct answer
    D
    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 $$ \begin{aligned} & \Rightarrow \mathrm{g}(\mathrm{x})<\mathrm{g}(0) \forall \mathrm{x} \in(0, \infty) \\ & \Rightarrow(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})<\mathrm{e} \log \mathrm{e}-\pi \log \pi<0 \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})<0 \forall \mathrm{x} \in(0, \infty) \\ & \Rightarrow \mathrm{f}(\mathrm{x}) \text { is decreasing on }(0, \infty) \end{aligned} $$

Hence, the correct answer is option: 'D'.

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