JEE Main 20208 Jan, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be a solution of the differential equation, 1−x2dydx+1−y2=0\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 01−x2dxdy+1−y2=0, |x| < 1. If y(12)=32y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}y(21)=23, then y(−12)y\left( { - {1 \over {\sqrt 2 }}} \right)y(−21) is equal to :AOption A. −32- {{\sqrt 3 } \over 2}−23BOption B. None of thoseCOption C. 12{{1 \over {\sqrt 2 }}}21DOption D. −12-{{1 \over {\sqrt 2 }}}−21CheckSkip · show solution
