A) \[\left[ -\frac{1}{4},\,\,\frac{1}{2} \right]\]
B) \[\left[ -\frac{1}{2},\,\,\frac{1}{2} \right]\]
C) \[\left[ -\frac{1}{2},\,\,\frac{1}{9} \right]\]
D) \[[0,\,\,\pi ]\]
Correct Answer: A
Solution :
Here,\[f(x)=\sqrt{{{\sin }^{-1}}(2x)+\frac{\pi }{6}}\], to find domain, we must have, \[{{\sin }^{-1}}(2x)+\frac{\pi }{6}\ge 0\] \[\left( but\,\,-\frac{\pi }{2}\le {{\sin }^{-1}}\theta \le \frac{\pi }{2} \right)\] \[\therefore \] \[-\frac{\pi }{6}\le {{\sin }^{-1}}(2x)\le \sin \frac{\pi }{2}\] \[\Rightarrow \] \[\sin \left( -\frac{\pi }{6} \right)\le 2x\le \sin \left( \frac{\pi }{2} \right)\] \[\Rightarrow \] \[-\frac{1}{2}\le 2x\le 1\] \[\Rightarrow \] \[-\frac{1}{4}\le x\le \frac{1}{2}\] \[\Rightarrow \] \[x\in \left[ -\frac{1}{4},\,\,\frac{1}{2} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
