question_answer

The area nearer to origin bounded by \[x={{x}_{1}},y={{y}_{1}}\] and \[y=-{{(x+1)}^{2}}\] where \[{{x}_{1}},{{y}_{1}}\] are the values of x, y satisfying the equation \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=-\pi \] will be

A) \[\frac{1}{3}\]                                   

B) \[\frac{2}{3}\]

C)  1                                            

D) \[\frac{3}{2}\]

Correct Answer: B

Solution :

\[-\frac{\pi }{2}\le {{\sin }^{-1}}x\le \frac{\pi }{2}-\frac{\pi }{2}\le {{\sin }^{-1}}y\le \frac{\pi }{2}\] \[-\pi \le {{\sin }^{-1}}x+{{\sin }^{-1}}y\le \pi \] So for\[{{\sin }^{-1}}x+{{\sin }^{-1}}y=-\pi \] \[\Rightarrow \]\[{{\sin }^{-1}}x=-\frac{\pi }{2}\And si{{n}^{-1}}y=-\frac{\pi }{2}\]x = - and y = - 1 area required \[=1-\left| \int\limits_{-1}^{0}{-{{(x+1)}^{2}}dx} \right|\] \[=1-\frac{1}{3}=\frac{2}{3}\]sq. units.