question_answer

The domain of the function\[f(x)=\frac{1}{\sqrt{9-{{x}^{2}}}}\]is

A)  \[-3\le x\le 3\]                 

B)  \[-3<x<3\]

C) \[-9\le x\le 9\]  

D)         \[-9<x<9\]

E)  \[-\infty <x<\infty \]

Correct Answer: B

Solution :

\[f(x)=\frac{1}{\sqrt{9-{{x}^{2}}}}\] For existence of \[f(x),\]                 \[9-{{x}^{2}}>0\] \[\Rightarrow \]               \[{{x}^{2}}-9<0\] \[\Rightarrow \]               \[(x-3)(x+3)<0\]                 \[\Rightarrow \]               \[x\in (-3,3)\]or \[-3<x<3\] Which is the required domain of\[f(x)\].