Question

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

• A. Increasing on $[0,\infty )$
• B. decreasing on $[0,\infty )$
• C. decreasing on $[0,\frac{\pi }{e})$ & Increasing on $[\frac{\pi }{e},\infty )$
• D. increasing on $[0,\frac{\pi }{e})$ & decreasing on $[\frac{\pi }{e},\infty )$

Answer 1

The correct option is B decreasing on $[0,\infty )$

$f\left(x\right)=\frac{\log (\pi +x)}{\log (e+x)}$
$f^{\prime }\left(x\right)=\frac{\log (e+x)\times \frac{1}{\pi +x}-\log (\pi +x)\frac{1}{e+x}}{{\left(\log (e+x)\right)}^2}$
$=\frac{\log (e+x)\times (e+x)-(\pi +x)\log (\pi +x)}{(\pi +x)(e+x){\left(\log (e+x)\right)}^2}$
Since log function is an increasing function and $e<\pi$, $\log (e+x)<\log (\pi +x)$.
Thus $(e+x)\log (e+x)<(e+x)\log (\pi +x)<(\pi +x)\log (\pi +x)$ for all $x>0$.
Thus, $f^{\prime }\left(x\right)<0$ for $\forall x>0$
$\Rightarrow f\left(x\right)$ decreases on $(0,\infty )$
Ans: B