Q. The function $f(x) = \log{(1 + x)} - \frac{2x}{2 + x}$ is increasing on
A) $(0,\ \infty)$
B) $( - \infty,\ 0)$
C) $( - \infty,\ \infty)$
D) None of these
Explanation
(a)
Given, $g(x)1 + x) - \frac{2x}{2 + x}$
$\therefore f^{'}(x) = \frac{1}{1 + x} - \frac{(2 + x)2 - 2x}{(2 + x)^{2}} = \frac{x^{2}}{(1 + x)(x + 2)^{2}}$
Clearly $f^{'}(x) > 0$ for all $x > 0.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.