The domain of the function f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}

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Answer: B. [2, 3) f\left( x ight) = {{{{\sin }^{ - 1}}\left( {x - 3} ight)} \over {\sqrt {9 - {x^2}} }} is defined if (i) \,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4 and (ii) 9 - {x^2} > 0 \Rightarrow - 3 < x < 3 Taking common solution of \left( i ight) and \left( {ii} ight), we get 2 \le x < 3 	herefore Domain = \left[ {2,\left. 3 ight)} ight.

The domain of the function $$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$

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