A) [-1, 1]
B) R-{-.1,1}
C) R - (-1, 1)
D) None of these
Correct Answer: B
Solution :
Since, \[f(x)={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[\therefore \]\[f'(x)=\frac{1}{\sqrt{1-\left( \frac{2x}{1+{{x}^{2}}} \right)}}.\frac{2(1-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}\] \[=\frac{(1+{{x}^{2}})}{\sqrt{{{(1-{{x}^{2}})}^{2}}}}.\frac{2(1-{{x}^{2}})}{{{(1+{{x}^{2}})}^{2}}}\] \[=\frac{2(1-{{x}^{2}})}{(1+{{x}^{2}})|1-{{x}^{2}}|}\] \[=\left\{ \begin{matrix} \frac{2}{1+{{x}^{2}}}, & \text{if}\,|x|<1 \\ -\frac{2}{1+{{x}^{2}}}, & \text{if}\,|x|>1 \\ \end{matrix} \right.\] \[\therefore \]f? (x) does not exist for |x| = 1 ie, x = ± 1 Hence, /(x) is differentiable on R - {-1,1}.
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