JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\mathrm{x}^{\mathrm{k}}+\mathrm{y}^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}},(\mathrm{a}, \mathrm{K}>0)\) and \(\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{1}{3}}=0\) then \(\mathrm{k}\) is
- A \(\frac{3}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{2}{3}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{x}^{\mathrm{k}}+\mathrm{y}^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}}(\mathrm{a}, \mathrm{k}>0)\) \(\mathrm{kx}^{\mathrm{k}-1}+\mathrm{ky}^{\mathrm{k}-1} \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=0\)…
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