A) \[y+{{\sin }^{-1}}x=c\]
B) \[{{y}^{2}}+2{{\sin }^{-1}}x+c=0\]
C) \[x+{{\sin }^{-1}}y=0\]
D) \[{{x}^{2}}+2{{\sin }^{-1}}y=1\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}=-\frac{1}{\sqrt{1-{{x}^{2}}}}\]Þ \[dy=-\frac{1}{\sqrt{1-{{x}^{2}}}}dx\] On integrating, we get \[y={{\cos }^{-1}}x+c\] Þ \[y=\frac{\pi }{2}-{{\sin }^{-1}}x+c\] Þ \[y+{{\sin }^{-1}}x=c\].
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