JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be a solution of the differential equation, \(\sqrt{1-\mathrm{x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{1-\mathrm{y}^{2}}=0,|\mathrm{x}|<1\) If \(\mathrm{y}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2},\) then \(\mathrm{y}\left(\frac{-1}{\sqrt{2}}\right)\) is equal to
- A \(-\frac{\sqrt{3}}{2}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\frac{\sqrt{3}}{2}\)
- D \(-\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}\) so, \(\frac{d y}{\sqrt{1-y^{2}}}+\frac{d x}{\sqrt{1-x^{2}}}=0\) Integrating, \(\sin ^{-1} x+\sin ^{-1} y=c\) so, \(\frac{\pi}{6}+\frac{\pi}{3}=c\) Hence, \(\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}\) Put…
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