Question
फलन $f(x) = \frac{{{\rm{ln}}(\pi + x)}}{{{\rm{ln}}(e + x)}}$ है
$\therefore f'(x) = \frac{{\ln (e + x) \times \frac{1}{{\pi + x}} - \ln (\pi + x)\frac{1}{{e + x}}}}{{{{\ln }^2}(e + x)}}$
$ = \frac{{(e + x)\ln (e + x) - (\pi + x)\ln (\pi + x)}}{{{{\ln }^2}(e + x) \times (e + x)(\pi + x)}}$
$ \Rightarrow f'(x) < 0$, सभी $x \ge 0$ के लिए
अत: $f(x)$, अंतराल $[0,\infty )$ में हृासमान है।
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