Let x k + y k = a k , a , k > 0 and d y d x + y x 1 3 = 0 , then k is

Let xk+yk=ak,a,k>0 and dydx+yx13=0, then k is
  1. 32
  2. 43
  3. 23
  4. 13

Solution

Differentiating the given equation with respect to x

k.xk-1+k.yk-1dydx=0

dydx=-xyk-1

dydx+xyk-1=0

k-1=-13

k=1-13=23

Asked in: JEE Main 2020 (07 Jan Shift 1)

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