JEE Main 202228 Jul, Evening shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let x(t)=22costsin2tx(t)=2 \sqrt{2} \cos t \sqrt{\sin 2 t}x(t)=22costsin2t and y(t)=22sintsin2t,t∈(0,π2)y(t)=2 \sqrt{2} \sin t \sqrt{\sin 2 t}, t \in\left(0, \frac{\pi}{2}\right)y(t)=22sintsin2t,t∈(0,2π). Then 1+(dydx)2d2ydx2\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}dx2d2y1+(dxdy)2 at t=π4t=\frac{\pi}{4}t=4π is equal to :AOption A. −223\frac{-2 \sqrt{2}}{3}3−22BOption B. 23\frac{2}{3}32COption C. 13\frac{1}{3}31DOption D. −23\frac{-2}{3}3−2CheckSkip · show solution
