Question:
Mark the correct alternative in the following:
$f(x)=2 x-\tan ^{-1} x-\log \left\{x+\sqrt{x^{2}+1}\right\}$ is monotonically increasing when
A. $x>0$
B. $x<0$
C. $x \in R$
D. $X \in R-\{0\}$
Solution:
Formula:- (i) The necessary and sufficient condition for differentiable function defined on (a,b) to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)>0$ for all $x \in(a, b)$
Given:-
$f(x)=2 x-\tan ^{-1} x-\log \left\{x+\sqrt{x^{2}+1}\right\}$
$\frac{d f(x)}{d x}=2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}}=f^{\prime}(x)$
For increasing function $f^{\prime}(x)>0$
$\Rightarrow 2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{x^{2}+1}}>0$
$x \in R$