JEE Main 20207 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Let x^k + y^k = a^k, (a, k > 0 ) and dydx+(yx)13=0{{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0dxdy+(xy)31=0, then k is:AOption A. 13{1 \over 3}31BOption B. 23{2 \over 3}32COption C. 43{4 \over 3}34DOption D. 32{3 \over 2}23CheckSkip · show solution