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