JEE Mains · Maths · STD 12 - 5. continuity and differentiation
For the curve \(C :\) \(\left(x^{2}+y^{2}-3\right)+\left(x^{2}-y^{2}-1\right)^{5}=0\), the value of \(3 y^{\prime}-y^{3} y^{\prime \prime}\), at the point \((\alpha, \alpha), \alpha>0\), on \(C\), is equal to.
- A \(18\)
- B \(15\)
- C \(16\)
- D \(14\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\((\alpha, \alpha)\) lies on \(C: x^{2}+y^{2}-3+x^{2}-y^{2}-1^{5}=0\) \(\operatorname{Put}(\alpha, \alpha), 2 \alpha^{2}-3+-1^{5}=0\) \(\alpha=\sqrt{2}\) Now, differentiate \(C\) \(2 x+2 y \cdot y^{\prime}+5\left(x^{2}-y^{2}-1\right)^{4}\left(2 x-2 y y^{\prime}\right)=0\) At…
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