Question
$$f\left( x \right) = \frac{{\log \left( {\pi + x} \right)}}{{\log \left( {e + x} \right)}}{\text{ is :}}$$
A.
increasing in $$\left[ {0,\,\infty } \right)$$
B.
decreasing in $$\left[ {0,\,\infty } \right)$$
C.
decreasing in $$\left[ {0,\,\frac{\pi }{e}} \right]$$ & increasing in $$\left[ {\frac{\pi }{e},\,\infty } \right)$$
D.
increasing in $$\left[ {0,\,\frac{\pi }{e}} \right]$$ & decreasing in $$\left[ {\frac{\pi }{e},\,\infty } \right)$$
Answer :
decreasing in $$\left[ {0,\,\infty } \right)$$
Solution :
We have $$e < \pi $$ and
$$\eqalign{
& f'\left( x \right) = \frac{{\frac{1}{{\pi + x}}\log \left( {e + x} \right) - \frac{1}{{e + x}}\log \left( {\pi + x} \right)}}{{{{\left\{ {\log \left( {e + x} \right)} \right\}}^2}}} \cr
& = \frac{{\left( {e + x} \right)\log \left( {e + x} \right) - \left( {\pi + x} \right)\log \left( {\pi + x} \right)}}{{\left( {\pi + x} \right)\left( {e + x} \right){{\left\{ {\log \left( {e + x} \right)} \right\}}^2}}} \cr} $$
$${\text{In}}\left[ {0,\,\infty } \right),$$ denominator $$ > 0$$ and numerator $$ < 0,$$
Since, $$e + x = \pi + x.$$
Hence, $$f\left( x \right)$$ is decreasing in $$\left[ {0,\,\infty } \right).$$