A) 1
B) 0
C) 2
D) none of these
Correct Answer: D
Solution :
[d] we know that \[Co{{s}^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=\left\{ \begin{matrix} 2{{\tan }^{-1}}x,x\ge 0 \\ -2{{\tan }^{-1}}x,x\le 0 \\ \end{matrix} \right.\] Or \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{x}{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2{{\tan }^{-1}}x}{x}=2\] And \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{1}{x}{{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ -\frac{2{{\tan }^{-1}}x}{x} \right]=-2\]
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