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:

  1. 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(\mathrm{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 $x \in(0, \infty)$.



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