Q. If $f(x) = \frac{1}{x + 1} - \log{(1 + x)},\ x > 0$, then $f$ is
A) an increasing function
B) a decreasing function
C) both increasing and decreasing function
D) None of the above
Explanation
(b)
Given curve is $f(x) = \frac{1}{x + 1} - \log(1 + x)$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = - \frac{1}{(x + 1)^{2}} - \frac{1}{1 + x}$
$\Rightarrow \ \ f^{'}(x) = - \left\lbrack \frac{1}{x + 1} + \frac{1}{(x + 1)^{2}} \right\rbrack$
$\Rightarrow \ \ \ f^{'}(x) = -$ve, when $x > 0$
$\therefore\ \ f(x)$ is a decreasing function
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.