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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

`वक्र \(C :\left( x ^2+ y ^2-3\right)+\left( x ^2- y ^2-1\right)^5=0\), के बिंन्दु \((\alpha, \alpha), \alpha>0\), पर \(3 y^{\prime}-y^3 y^{\prime \prime}\), का मान बराबर है \(...........\)

  1. A \(18\)
  2. B \(15\)
  3. C \(16\)
  4. D \(14\)
Verified Solution

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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