MHT CET · Maths · Application of Derivatives
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 \((0, \infty)\)
- D decreasing on \((0, \infty)\)
Answer & Solution
Correct Answer
(D) decreasing on \((0, \infty)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(\mathrm{x})=\frac{\log (\pi+\mathrm{x})}{\log (\mathrm{e}+\mathrm{x})} \)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\frac{1}{\pi+\mathrm{x}} \log (\mathrm{e}+\mathrm{x})-\log (\pi+\mathrm{x}) \times \frac{1}{\mathrm{e}+\mathrm{x}}}{\{\log (\mathrm{e}+\mathrm{x})\}^2} \)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})}{(\pi+\mathrm{x})(\mathrm{e}+\mathrm{x})\{\log (\mathrm{e}+\mathrm{x})\}^2}\)
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
\(\Rightarrow \mathrm{g}(\mathrm{x})<\mathrm{g}(0) \forall \mathrm{x} \in(0, \infty) \)
\( \Rightarrow(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})<\mathrm{e}\) \(\log \mathrm{e}\) \(-\pi \log \pi<0 \)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})<0 \forall \mathrm{x} \in(0, \infty) \)
\( \Rightarrow \mathrm{f}(\mathrm{x}) \text { is decreasing on }(0, \infty)\)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\frac{1}{\pi+\mathrm{x}} \log (\mathrm{e}+\mathrm{x})-\log (\pi+\mathrm{x}) \times \frac{1}{\mathrm{e}+\mathrm{x}}}{\{\log (\mathrm{e}+\mathrm{x})\}^2} \)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})}{(\pi+\mathrm{x})(\mathrm{e}+\mathrm{x})\{\log (\mathrm{e}+\mathrm{x})\}^2}\)
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
\(\Rightarrow \mathrm{g}(\mathrm{x})<\mathrm{g}(0) \forall \mathrm{x} \in(0, \infty) \)
\( \Rightarrow(\mathrm{e}+\mathrm{x}) \log (\mathrm{e}+\mathrm{x})-(\pi+\mathrm{x}) \log (\pi+\mathrm{x})<\mathrm{e}\) \(\log \mathrm{e}\) \(-\pi \log \pi<0 \)
\( \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})<0 \forall \mathrm{x} \in(0, \infty) \)
\( \Rightarrow \mathrm{f}(\mathrm{x}) \text { is decreasing on }(0, \infty)\)
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