Question

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

• A. increasing in {0,$\pi$}
• B. $decreasingin(-\infty ,0)$
• C. $decreasingom[0,\frac{\pi }{e}]anddecreasingon[\frac{\pi }{e},\infty ]$
• D. $decreasingin[0,\frac{\pi }{e}]andincreasingon[\frac{\pi }{e},\infty ]$

Answer 1

The correct option is A increasing in {0,$\pi$}

$f\left(x\right)=\frac{\log (\pi +x)}{\log (e+x)}$

$\therefore f^{{}^{\prime }}\left(x\right)=\frac{log(e+x).\frac{1}{\pi +x}-\log (\pi +x).\frac{1}{(e+x)}}{\left(\log \right(e+x){)}^2}$

When, $x>0;\pi +x>e+x[\therefore \pi >e]\Rightarrow$