Question
The function $$f\left( x \right) = \frac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}}$$ is
A.
increasing on $$\left( {0,\infty } \right)$$
B.
decreasing on $$\left( {0,\infty } \right)$$
C.
increasing on $$\left( {0,\frac{\pi }{e}} \right),$$ decreasing on $$\left( {\frac{\pi }{e},\infty } \right)$$
D.
decreasing on $$\left( {0,\frac{\pi }{e}} \right),$$ increasing on $$\left( {\frac{\pi }{e},\infty } \right)$$
Answer :
decreasing on $$\left( {0,\infty } \right)$$
Solution :
$$\eqalign{
& {\text{We}}\,{\text{have}}\,\,f\left( x \right) = \frac{{\ln \left( {\pi + x} \right)}}{{\ln \left( {e + x} \right)}} \cr
& \therefore \quad {f^\prime }(x) = \frac{{\left( {\frac{1}{{\pi + x}}} \right)\ln (e + x) - \frac{1}{{(e + x)}}\ln (\pi + x)}}{{{{[\ln (e + x)]}^2}}} \cr
& = \frac{{(e + x)\ln (e + x) - (\pi + x)\ln (\pi + x)}}{{(e + x)(\pi + x){{(\ln (e + x))}^2}}} \cr
& < 0{\text{ on }}(0,\infty ){\text{ since }}1 < e < \pi \cr
& \therefore f\left( x \right)\,{\text{decraeses}}\,{\text{on}}\,(0,\infty ). \cr} $$