Let x^k+y^k=a^k,a,k>0 and dy/dx+yx^13=0, then k is

Correct answer: (C) 2/3. Differentiating the given equation with respect to x k.x^(k-1)+k.y^(k-1)dy/dx=0 ⇒dy/dx=-xy^(k-1) ⇒dy/dx+xy^(k-1)=0 ⇒k-1=-1/3 ⇒k=1-1/3=2/3

Q31 / 75 · JEE Main 2020 (07 Jan Shift 1) · Mathematics

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
$\frac{4}{3}$
Correct answer
C
$\frac{2}{3}$
D
$\frac{1}{3}$

Solution

Differentiating the given equation with respect to

x

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

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

$\Rightarrow \frac{d y}{d x} + {(\frac{x}{y})}^{k - 1} = 0$

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

$\Rightarrow k = 1 - \frac{1}{3} = \frac{2}{3}$

Hence, the correct answer is option: 'C'.

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Let xk+yk=ak,a,k>0 and dydx+yx13=0, then k is

Solution

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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