Medium MCQ +4 / -1 PYQ · JEE Mains 2020

Let x^k + y^k = a^k, (a, k > 0 ) and ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$, then k is:

A ${1 \over 3}$
B ${2 \over 3}$ Correct answer
C ${4 \over 3}$
D ${3 \over 2}$

Solution

xk + yk = ak $\Rightarrow$ kxk - 1 + kyk - 1${{{dy} \over {dx}}}$ = 0 $\Rightarrow$ ${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$ = 0 ...(1) Given ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$ ...(2) Comparing (1) and (2), we get k - 1 = $- {1 \over 3}$ $\Rightarrow$ k = ${2 \over 3}$

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Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions

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

Solution

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Let x^k + y^k = a^k, (a, k > 0 ) and ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$, then k is: