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If $x^k+y^k=a^k(a, k>0)$ and $\frac{d y}{d x}+\left(\frac{y}{x}\right)^{\frac{1}{b}}-0$, then $\mathrm{k}$ has the value
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Verified Answer
The correct answer is:
$\frac{2}{3}$
$x^{\mathrm{k}}+y^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \mathrm{k} x^{k-1}+\mathrm{k} y^{\mathrm{k}-1} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{k} x^{\mathrm{k}-1}}{\mathrm{k} y^{\mathrm{k}-1}} \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{x}{y}\right)^{\mathrm{k}-1} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{1-\mathrm{k}}=0 \\
& \text { But } \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0 ...[Given]
\end{aligned}$
Comparing above equations, we get
$\begin{aligned}
1-k & =\frac{1}{3} \\
& 1-\frac{1}{3}=k \\
\therefore \quad k & =\frac{2}{3}
\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
& \mathrm{k} x^{k-1}+\mathrm{k} y^{\mathrm{k}-1} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{\mathrm{k} x^{\mathrm{k}-1}}{\mathrm{k} y^{\mathrm{k}-1}} \\
\therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=-\left(\frac{x}{y}\right)^{\mathrm{k}-1} \\
& \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{1-\mathrm{k}}=0 \\
& \text { But } \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0 ...[Given]
\end{aligned}$
Comparing above equations, we get
$\begin{aligned}
1-k & =\frac{1}{3} \\
& 1-\frac{1}{3}=k \\
\therefore \quad k & =\frac{2}{3}
\end{aligned}$
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