is decreasing
A)
B)
C)
D)
Correct Answer: B
Solution :
\[\operatorname{y}=co{{s}^{-1}}\,\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\,\,=\,\,\left\{ \begin{matrix} 2\,{{\tan }^{-1}}x,\,\,if\,\,0\le x<\infty \\ -2\,{{\tan }^{-1}}x,\,\,if\,\,-\infty <x<0 \\ \end{matrix} \right.\] \[f'(x)=\frac{d}{dx}{{\cos }^{-1}}\,\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=\frac{2x}{\left| \,x\, \right|(1+{{x}^{2}})}\] \[\operatorname{f}(x)<0\,\,if\,\,x<0\] \[\operatorname{x}\in \left( -\,\infty ,\,\,0 \right)\]
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