Question

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

• A. increasing in $[0,\infty )$
• B. decreasing in $[0,\infty )$
• C. decreasing in $\left[0,\frac{\pi }{e}\right]\&$ increasing in $\left[\frac{\pi }{e},\infty \right)$
• D. increasing in $\left[0,\frac{\pi }{c}\right]$ \& decreasing in $\left[\frac{\pi }{e},\infty \right)$

Answer 1

Answer: (B)

We have $e<\pi$ and
$f^{\prime }(x)=\frac{\frac{1}{\pi +x}\log (e+x)-\frac{1}{e+x}\log (\pi +x)}{\{\log (e+x){\}}^2}$
$=\frac{(e+x)\log (e+x)-(\pi +x)\log (\pi +x)}{(\pi +x)(e+x)\{\log (e+x){\}}^2}$
In $[0,\infty )$, denominator $>0$ and numerator $<0$,
since, $e+x<\pi +x$.
Hence, $f(x)$ is decreasing in $[0,\infty )$.