SATHEE: Application of Derivatives 2 Question 15

Application of Derivatives 2 Question 15

####15. The function $f (x) = \frac{\log (π + x)}{\log (e + x)}$ is

(1995, 1M)

(a) increasing on $(0, \infty)$

(b) decreasing on $(0, \infty)$

(c) increasing on $(0, π / e)$ , decreasing on $(π / e, \infty)$

(d) decreasing on $(0, π / e)$ , increasing on $(π / e, \infty)$

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Answer:

Correct Answer: 15. (b)

Solution:

Given, $f (x) = \frac{\log (π + x)}{\log (e + x)}$

$f^{'} (x) = \frac{\log (e + x) \cdot \frac{1}{π + x} - \log (π + x) \cdot \frac{1}{e + x}}{[\log (e + x)]^{2}} \dots (i)$

On multiplying Eqs. (ii) and (iii), we get

$\frac{\log (π + x)}{e + x} > \frac{\log (e + x)}{π + x}$

From Eqs. (i) and (iv), $f^{'} (x) < 0$

$∴ f (x)$ is decreasing for $x \in (0, \infty)$ .

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Application of Derivatives 2 Question 15

Answer:

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