A) \[\frac{\sqrt{3}}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[-\frac{\sqrt{3}}{2}\]
D) \[-\frac{1}{\sqrt{2}}\]
Correct Answer: A , B , C , D
Solution :
| (Bonus) |
| \[\frac{dy}{dx}=\frac{-\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] |
| \[\Rightarrow \int{\frac{dy}{\sqrt{1-{{y}^{2}}}}+\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}=0}}\] |
| \[\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=c\] |
| Given \[y\left( \frac{1}{2} \right)=\frac{\sqrt{3}}{2}\] \[\Rightarrow c=\frac{\pi }{2}\] |
| \[\Rightarrow {{\sin }^{-1}}y=\frac{\pi }{2}-{{\sin }^{-1}}x\] |
| \[\Rightarrow {{\sin }^{-1}}y={{\cos }^{-1}}x\] |
| Putting \[x=\frac{-1}{\sqrt{2}}\] we get \[{{\sin }^{-1}}y=\frac{3\pi }{4}\] |
Solution :
| (Bonus) |
| \[\frac{dy}{dx}=\frac{-\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] |
| \[\Rightarrow \int{\frac{dy}{\sqrt{1-{{y}^{2}}}}+\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}=0}}\] |
| \[\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=c\] |
| Given \[y\left( \frac{1}{2} \right)=\frac{\sqrt{3}}{2}\] \[\Rightarrow c=\frac{\pi }{2}\] |
| \[\Rightarrow {{\sin }^{-1}}y=\frac{\pi }{2}-{{\sin }^{-1}}x\] |
| \[\Rightarrow {{\sin }^{-1}}y={{\cos }^{-1}}x\] |
| Putting \[x=\frac{-1}{\sqrt{2}}\] we get \[{{\sin }^{-1}}y=\frac{3\pi }{4}\] |
Solution :
| (Bonus) |
| \[\frac{dy}{dx}=\frac{-\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] |
| \[\Rightarrow \int{\frac{dy}{\sqrt{1-{{y}^{2}}}}+\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}=0}}\] |
| \[\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=c\] |
| Given \[y\left( \frac{1}{2} \right)=\frac{\sqrt{3}}{2}\] \[\Rightarrow c=\frac{\pi }{2}\] |
| \[\Rightarrow {{\sin }^{-1}}y=\frac{\pi }{2}-{{\sin }^{-1}}x\] |
| \[\Rightarrow {{\sin }^{-1}}y={{\cos }^{-1}}x\] |
| Putting \[x=\frac{-1}{\sqrt{2}}\] we get \[{{\sin }^{-1}}y=\frac{3\pi }{4}\] |
Solution :
| (Bonus) |
| \[\frac{dy}{dx}=\frac{-\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] |
| \[\Rightarrow \int{\frac{dy}{\sqrt{1-{{y}^{2}}}}+\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}=0}}\] |
| \[\Rightarrow {{\sin }^{-1}}y+{{\sin }^{-1}}x=c\] |
| Given \[y\left( \frac{1}{2} \right)=\frac{\sqrt{3}}{2}\] \[\Rightarrow c=\frac{\pi }{2}\] |
| \[\Rightarrow {{\sin }^{-1}}y=\frac{\pi }{2}-{{\sin }^{-1}}x\] |
| \[\Rightarrow {{\sin }^{-1}}y={{\cos }^{-1}}x\] |
| Putting \[x=\frac{-1}{\sqrt{2}}\] we get \[{{\sin }^{-1}}y=\frac{3\pi }{4}\] |
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