A) 0
B) \[\frac{1}{2}\]
C) 1
D) 2
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dy}{\sqrt{1-{{y}^{2}}}}\,=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}};}}\,\,\,{{\sin }^{-1}}y=\frac{-1}{x}+C\] \[y\left( \frac{2}{\pi } \right)=0\] gives \[C=\frac{\pi }{2}\] \[\therefore \,\,\frac{1}{x}\,=\frac{\pi }{2}\,-{{\sin }^{-1}}y={{\cos }^{-1}}(y)\] \[\Rightarrow \,y=\cos \left( \frac{1}{x} \right)\] \[\therefore \,y\left( \frac{3}{\pi } \right)=\cos \,\frac{\pi }{3}\,=\frac{1}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
