JEE Main 20264 Apr, Morning shiftMathematicsEasyDifferential Equations163× since 2002Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx=(1+x+x2)(1−y+y2)\frac{dy}{dx}=(1+x+x^2)(1-y+y^2)dxdy=(1+x+x2)(1−y+y2), y(0)=12y(0)=\frac12y(0)=21. Then (2y(1)−1)(2y(1)-1)(2y(1)−1) is equal toAOption A. 3tan(1136)\sqrt3\tan\left(\frac{11\sqrt3}{6}\right)3tan(6113)BOption B. 32tan(11312)\frac{\sqrt3}{2}\tan\left(\frac{11\sqrt3}{12}\right)23tan(12113)COption C. 3tan(11312)\sqrt3\tan\left(\frac{11\sqrt3}{12}\right)3tan(12113)DOption D. 32tan(1136)\frac{\sqrt3}{2}\tan\left(\frac{11\sqrt3}{6}\right)23tan(6113)CheckStep-by-step hintSkip · show solution