JEE MAIN - Mathematics (2002 - No. 4)
The domain of $${\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right]$$ is
[1, 9]
[-1, 9]
[9, 1]
[-9, -1]
Explanation
$$f\left( x \right) = {\sin ^{ - 1}}\left( {{{\log }_3}\left( {{x \over 3}} \right)} \right)$$ exists
if $$\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1$$
$$ \Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}$$
$$ \Leftrightarrow 1 \le x \le 9$$
or $$\,\,\,\,x \in \left[ {1,9} \right]$$
if $$\,\,\,\, - 1 \le {\log _3}\left( {{x \over 3}} \right) \le 1$$
$$ \Leftrightarrow {3^{ - 1}} \le {x \over 3} \le {3^1}$$
$$ \Leftrightarrow 1 \le x \le 9$$
or $$\,\,\,\,x \in \left[ {1,9} \right]$$
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