A) \[(-1,1)\]
B) \[\left[ \frac{1}{3},\infty \right)\]
C) \[\left( \frac{-1}{3},\frac{1}{3} \right)\]
D) None of these
Correct Answer: B
Solution :
[b] \[{{\cos }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)=\frac{-\pi }{2}+2{{\tan }^{-1}}\left( 3x \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{\pi }{2}-{{\sin }^{-1}}\left( \frac{6x}{1+9{{x}^{2}}} \right)=-\frac{\pi }{2}+2{{\tan }^{-1}}3x\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{\sin }^{-1}}\left( \frac{2(3x)}{1+{{(3x)}^{2}}} \right)=\pi -2{{\tan }^{-1}}(3x)\] Let \[3x=\tan \theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)=\pi -2\theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{\sin }^{-1}}(\sin 2\theta )=\pi -2\theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{\pi }{2}\le 2\theta \le \frac{3\pi }{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{\pi }{4}\le \theta \le \frac{3\pi }{4}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{\pi }{4}\le {{\tan }^{-1}}3x\le \frac{3\pi }{4}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{\pi }{4}\le {{\tan }^{-1}}3x<\frac{\pi }{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,1\le 3x<\infty \] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}\le x<\infty \]
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