JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 12)
The number of distinct solutions of the equation
$${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$ in the interval [0, 2$$\pi $$], is ____.
$${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$ in the interval [0, 2$$\pi $$], is ____.
Answer
8
Explanation
$${\log _{{1 \over 2}}}\left| {\sin x} \right| = 2 - {\log _{{1 \over 2}}}\left| {\cos x} \right|$$
$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left| {\sin x} \right|$$ + $${\log _{{1 \over 2}}}\left| {\cos x} \right|$$ = 2
$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left( {\left| {\sin x\cos x} \right|} \right)$$ = 2
$$ \Rightarrow $$ $${\left| {\sin x\cos x} \right| = {1 \over 4}}$$
$$ \Rightarrow $$ sin 2x = $$ \pm $$ $${1 \over 2}$$_9th_January_Morning_Slot_en_12_1.png)
$$ \therefore $$ Number of distinct solution = 8
$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left| {\sin x} \right|$$ + $${\log _{{1 \over 2}}}\left| {\cos x} \right|$$ = 2
$$ \Rightarrow $$ $${\log _{{1 \over 2}}}\left( {\left| {\sin x\cos x} \right|} \right)$$ = 2
$$ \Rightarrow $$ $${\left| {\sin x\cos x} \right| = {1 \over 4}}$$
$$ \Rightarrow $$ sin 2x = $$ \pm $$ $${1 \over 2}$$
$$ \therefore $$ Number of distinct solution = 8
Comments (0)
