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JEE Main 20207 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Differentiation

Let x ^ k + y ^ k = a ^ k , (a, k > 0) and (dy)/(dx) +...

Let xk+yk=ak,(a,k>0){x}^{k} + {y}^{k} = {a}^{k}, \left(a, k > 0\right) and \dfrac{dy}{dx} + \left(\dfrac{y}{x}\right)\dfrac{1}{3} = 0, then kk is

Options

  1. A32\dfrac{3}{2}
  2. B43\dfrac{4}{3}
  3. C23\dfrac{2}{3}
  4. D13\dfrac{1}{3}

Correct answer

C. 23\dfrac{2}{3}

Step-by-step solution

Differentiating the given equation with respect to xx

k.xk−1+k.yk−1dydx=0k . {x}^{k - 1} + k . {y}^{k - 1}\dfrac{dy}{dx} = 0

⇒dydx=−(xy)k−1⇒ \dfrac{dy}{dx} = - {\left(\dfrac{x}{y}\right)}^{k - 1}

⇒dydx+(xy)k−1=0⇒ \dfrac{dy}{dx} + {\left(\dfrac{x}{y}\right)}^{k - 1} = 0

⇒k−1=−13⇒ k - 1 = - \dfrac{1}{3}

⇒k=1−13=23⇒ k = 1 - \dfrac{1}{3} = \dfrac{2}{3}

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