JEE MAIN - Mathematics (2003 - No. 13)
The trigonometric equation $${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$ has a solution for :
$$\left| a \right| \ge {1 \over {\sqrt 2 }}$$
$${1 \over 2} < \left| a \right| < {1 \over {\sqrt 2 }}$$
all real values of $$a$$
$$\left| a \right| \le {1 \over {\sqrt 2 }}$$
Explanation
Given that,
$${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$
We know,
$$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$
$$\therefore$$ $$\,\,\,$$ $$ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$$
$$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$$
$$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$$
$$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$$
$$\therefore$$ $$\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$$
$${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$
We know,
$$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$
$$\therefore$$ $$\,\,\,$$ $$ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$$
$$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$$
$$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$$
$$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$$
$$\therefore$$ $$\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$$
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