A) \[\frac{3}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: D
Solution :
[d] Given \[{{x}^{k}}+{{y}^{k}}={{a}^{k}}\] Differentiating, \[k\cdot {{x}^{k-1}}+k\cdot {{y}^{k-1}}\frac{dy}{dx}=0\] \[\therefore \frac{dy}{dx}=-{{\left( \frac{x}{y} \right)}^{k-1}}\] \[\left( \frac{dy}{dx} \right)+{{\left( \frac{y}{x} \right)}^{1-k}}=0\] \[\therefore 1-k=\frac{1}{3}\] \[\Rightarrow k=\frac{2}{3}\]
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