A) \[{{\sin }^{-1}}\left( \frac{x-2}{2} \right)-\frac{\pi }{6}\]
B) \[{{\sin }^{-1}}\left( \frac{x-2}{2} \right)+\frac{\pi }{6}\]
C) \[\frac{2\pi }{3}+{{\cos }^{-1}}\left( \frac{x-2}{2} \right)\]
D) none of these
Correct Answer: B
Solution :
[b] \[y=f(x)=\sqrt{3}\sin x-\cos x+2\] \[=2\sin \left( x-\frac{\pi }{6} \right)+2\] ...(1) Since \[f(x)\]is one-one and onto, f is invertible. From (1), \[\sin \left( x-\frac{\pi }{6} \right)=\frac{y-2}{2}\] Or \[x={{\sin }^{-1}}\frac{y-2}{2}+\frac{\pi }{6}\] Or \[{{f}^{-1}}(x)=si{{n}^{-1}}\left( \frac{x-2}{2} \right)+\frac{\pi }{6}\]
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