JEE MAIN - Mathematics (2004 - No. 1)
The domain of the function
$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$
$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$
[1, 2]
[2, 3)
[1, 2)
[2, 3]
Explanation
$$f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$$ 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 $$\left( i \right)$$ and $$\left( {ii} \right),$$
we get $$2 \le x < 3$$
$$\therefore$$ Domain $$ = \left[ {2,\left. 3 \right)} \right.$$
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 $$\left( i \right)$$ and $$\left( {ii} \right),$$
we get $$2 \le x < 3$$
$$\therefore$$ Domain $$ = \left[ {2,\left. 3 \right)} \right.$$
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