If f(x) = 2x + cot^{-1}x + log(sqrt{1 + x^2} | vedclass

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(A) Given $f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)$.First,we differentiate $f(x)$ with respect to $x$:$f'(x) = 2 - \frac{1}{1 + x^2} + \frac

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Increases in $[0, \infty)$

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(A) Given $f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)$.First,we differentiate $f(x)$ with respect to $x$:$f'(x) = 2 - \frac{1}{1 + x^2} + \frac{1}{\sqrt{1 + x^2} - x} \cdot \left( \frac{x}{\sqrt{1 + x^2}} - 1 ight)$Simplify the third term:$\frac{1}{\sqrt{1 + x^2} - x} \cdot \left( \frac{x - \sqrt{1 + x^2}}{\sqrt{1 + x^2}} ight) = -\frac{1}{\sqrt{1 + x^2}}$Thus,$f'(x) = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}}$$f'(x) = \frac{2(1 + x^2) - 1 - \sqrt{1 + x^2}}{1 + x^2} = \frac{2x^2 + 1 - \sqrt{1 + x^2}}{1 + x^2}$Since $\sqrt{1 + x^2} > 1$ for $x 
eq 0$ and $2x^2 + 1 > \sqrt{1 + x^2}$ for all $x$,$f'(x) > 0$.Therefore,$f(x)$ is an increasing function on $[0, \infty)$.

If $f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)$,then $f(x)$

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