Application of Derivatives 2 Question 15
15. The function $f(x)=\frac{\log (\pi+x)}{\log (e+x)}$ is
(1995, 1M)
(a) increasing on $(0, \infty)$
(b) decreasing on $(0, \infty)$ (c) increasing on $(0, \pi / e)$, decreasing on $(\pi / e, \infty)$
(d) decreasing on $(0, \pi / e)$, increasing on $(\pi / e, \infty)$
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Answer:
Correct Answer: 15. (b)
Solution:
- Given, $f(x)=\frac{\log (\pi+x)}{\log (e+x)}$
$$ f^{\prime}(x)=\frac{\log (e+x) \cdot \frac{1}{\pi+x}-\log (\pi+x) \cdot \frac{1}{e+x}}{[\log (e+x)]^{2}} \ldots(i) $$
On multiplying Eqs. (ii) and (iii), we get
$$ \frac{\log (\pi+x)}{e+x}>\frac{\log (e+x)}{\pi+x} $$
From Eqs. (i) and (iv), $f^{\prime}(x)<0$
$\therefore f(x)$ is decreasing for $x \in(0, \infty)$.
