JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be a function of \(\mathrm{x}\) satisfying \(y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}}\) where \(k\) is a constant and \(y\left(\frac{1}{2}\right)=-\frac{1}{4} .\) Then \(\frac{d y}{d x}\) at \(x=\frac{1}{2},\) is equal to:
- A \(\frac{\sqrt{5}}{2}\)
- B \(-\frac{\sqrt{5}}{2}\)
- C \(\frac{2}{\sqrt{5}}\)
- D \(-\frac{\sqrt{5}}{4}\)
Answer & Solution
Correct Answer
(B) \(-\frac{\sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
Put \(x=\sin \theta, y=\sin \alpha\) \(y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}}\) \(\Rightarrow \sin \alpha \cdot \cos \theta+\cos \alpha \cdot \sin \theta=\mathrm{k}\) \(\Rightarrow \sin (\alpha+\theta)=k\) \(\Rightarrow \alpha+\theta=\sin ^{-1} k\)…
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