• # 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) $\frac{3}{2}$

$-\frac{\pi }{2}\le {{\sin }^{-1}}x\le \frac{\pi }{2}-\frac{\pi }{2}\le {{\sin }^{-1}}y\le \frac{\pi }{2}$ $-\pi \le {{\sin }^{-1}}x+{{\sin }^{-1}}y\le \pi$ 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.