Question

$f\left(x\right)=2x-{\tan }^{-1}x-\log (x+\sqrt{1+x^2})(x>0)$ is increasing in

• A. (1, 2)
• B. $(0,1)U(2,\infty )$
• C. $(0,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is C $(0,\infty )$

$f\left(x\right)=2x-ta{n}^{-1}x-\log (x+\sqrt{1+x^2})$
$f^{\prime }\left(x\right)=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(1+2x^2)-\sqrt{1+x^2}}{1+x^2}$
For f(x) to be increasing, f'(x)>0
$\Rightarrow \frac{(1+2x^2)-\sqrt{1+x^2}}{1+x^2}>0$
$\Rightarrow (1+2x^2)-\sqrt{1+x^2}>0$
$\Rightarrow (2x^2+1{)}^2>1+x^2$
$\Rightarrow x^2(4x^2+3)>0$
$\Rightarrow x>0$
$\Rightarrow x\in (0,\infty )$