If ${cos}^{- 1} (2 x - 1 - x^{2}) + {sin}^{- 1} (y) = 0$ then the value of $x^{2} + y^{2} + x y$ is equal to

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Solution

The correct option is A $0$
${cos}^{- 1} (x - 1)^{2} + \frac{π}{2} - {cos}^{- 1} (y) = 0$
${cos}^{- 1} ((x - 1)^{2} (y) + \sqrt{1 - (x - 1)^{2}} \sqrt{1 - y^{2}} = π / 2$
${cos}^{- 1} (x - 1)^{2} y + \sqrt{1 - x^{2} + 2 x + 1} \sqrt{1 - y^{2}} = π / 2$
$(x - 1)^{2} y = \sqrt{x (2 - x)} \sqrt{1 - y^{2}}$
$(x - 1)^{2} y^{2} = x (2 - x) .1 - y^{2}$
$(x - 1)^{2} (x^{2} y^{2} - 2 x y^{2} + y^{2}) = 2 x - 2 x y^{2} - x^{2} + x^{2} y^{2}$
by solving
$x^{2} + y^{2} + x y = 0$

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