A) \[\frac{2}{1+{{x}^{2}}}\], when \[-1<x<1\]
B) \[\frac{2}{1+{{x}^{2}}}\] when \[x<-1\,or\,x>1\]
C) \[\frac{-2}{1+{{x}^{2}}}\]when\[-1<x<1\]
D) None of these
Correct Answer: A
Solution :
[a] \[\because y=si{{n}^{-}}^{1}\left( \frac{2x}{1+{{x}^{2}}} \right)\] Putting \[x=tan\theta \And \theta ={{\tan }^{-1}}x\,\therefore y={{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)\]\[=si{{n}^{-1}}\left( sin2\theta \right)=2\theta =2tan_{x}^{-1}\frac{dy}{dx}\] \[=2.\frac{1}{1+{{x}^{2}}}=\frac{2}{1+{{x}^{2}}}\frac{dy}{dx}=\frac{-2}{1+{{x}^{2}}}\] when \[x<-1\text{ }or\text{ }x>1\]Hence/ option [a] is correct.
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