A) \[(2,\,\,3)\]
B) \[[2,\,\,3)\]
C) \[(2,\,\,3]\]
D) None of these
Correct Answer: B
Solution :
\[\sqrt{9-{{x}^{2}}}\]is defined for \[9-{{x}^{2}}\ge 0\Rightarrow \,\,(3-x)(3+x)\ge 0\] \[\Rightarrow \] \[(x-3)(x+3)\le 0\] ... (i) \[\Rightarrow \] \[-3\le x\le 3\] \[{{\sin }^{-1}}(3-x)\] defined for \[-1\le 3-x<1\] \[\Rightarrow \] \[-4\le -x\le -2\] \[\Rightarrow \] \[2\le x\le 4\] ... (ii) Also,\[{{\sin }^{-1}}(3-x)\ne 0\] \[3-x\ne 0\]or\[x\ne 3\] ... (iii) From Eqs. (i), (ii) and (iii), we get The domain of \[f=([-3,\,\,3]\cap [2,\,\,4)-\{3\}=[2,\,\,3)\]
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