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JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Differentiation

Let

x^{k} + y^{k} = a^{k}, (a, k > 0)

and

\frac{d y}{d x} + (\frac{y}{x})^{\frac{1}{3}} = 0

, then

k

is:

(A) $\frac{3}{2}$
(B) $\frac{4}{3}$
(C) $\frac{1}{3}$
(D) $\frac{2}{3}$ ✓ Correct

Answer: (D)

Solution

Differentiating

x^{k} + y^{k} = a^{k}

implicitly gives

k x^{k - 1} + k y^{k - 1} \frac{d y}{d x} = 0

, so

\frac{d y}{d x} = - (\frac{x}{y})^{k - 1}

. Comparing with the given

\frac{d y}{d x} = - (\frac{y}{x})^{1/3} = - (\frac{x}{y})^{- 1/3}

, the exponents must match:

k - 1 = - \frac{1}{3}

, so

k = \frac{2}{3}

.

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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