First slide

Monotonicity

Question

$The function g (x) = e^{x^{2}} \frac{\log (π + x)}{\log (e + x)} (x \geq 0) is$

Moderate

A

$increasing on [0, \infty)$
B

$decreasing on [0, \infty)$
C

$increasing on [0, π / e) and decreasing on [π / e, \infty)$
D

$decreasing on [0, π / e,) and increasing on [π / e, \infty)$

Solution

$\begin{aligned} Since e^{x^{2}} increases on [0, \infty) so it is enough to consider \\ \begin{array}{l} f (x) = \frac{\log (π + x)}{\log (e + x)} \\ f^{'} (x) = \frac{\log (e + x) \times \frac{1}{π + x} - \log (π + x) \frac{1}{e + x}}{(\log (e + x))^{2}} \\ = f^{'} (x) = \frac{\log (e + x) \times (e + x) - (π + x) \log (π + x)}{(π + x) (e + x) (\log (e + x))^{2}} \end{array} \end{aligned}$
Since log function is an increasing function and
$\begin{array}{l} e < π, \log (e + x) < \log (π + x) \\ Thus (e + x), \log (e + x) \log (e + x) < (e + x) \log (π + x) < \\ (π + x) \log (π + x) for all x > 0 \\ Thus f^{'} (x) < 0 for \forall x > 0 \Rightarrow f (x) decreases on (0, \infty) \end{array}$

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