JEE MAIN - Mathematics (2019 - 12th April Morning Slot - No. 12)
The equation y = sinx sin (x + 2) – sin2
(x + 1) represents a straight line lying in :
first, second and fourth quadrants
first, third and fourth quadrants
second and third quadrants only
third and fourth quadrants only
Explanation
y = sinx.sin(x+2) - sin2(x+1)
$$ \Rightarrow {1 \over 2}\left\{ {2\sin \left( {x + 2} \right)\sin x - 2{{\sin }^2}(x + 1)} \right\}$$
$$ \Rightarrow {1 \over 2}\left\{ {\cos 2 - \cos (2x + 2) + cos(2x + 2) - 1} \right\} = - {\sin ^2}1 < 0$$
Hence the line passes through III and IV quadrant
$$ \Rightarrow {1 \over 2}\left\{ {2\sin \left( {x + 2} \right)\sin x - 2{{\sin }^2}(x + 1)} \right\}$$
$$ \Rightarrow {1 \over 2}\left\{ {\cos 2 - \cos (2x + 2) + cos(2x + 2) - 1} \right\} = - {\sin ^2}1 < 0$$
Hence the line passes through III and IV quadrant
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