JEE Main 20207 Jan, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let y = y(x) be a function of x satisfying y1−x2=k−x1−y2y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}y1−x2=k−x1−y2 where k is a constant and y(12)=−14y\left( {{1 \over 2}} \right) = - {1 \over 4}y(21)=−41. Then dydx{{dy} \over {dx}}dxdy at x = 12{1 \over 2}21, is equal to :AOption A. 25{2 \over {\sqrt 5 }}52BOption B. −52- {{\sqrt 5 } \over 2}−25COption C. 52{{\sqrt 5 } \over 2}25DOption D. −54- {{\sqrt 5 } \over 4}−45CheckSkip · show solution
