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(B) The intersection points of $y^2 = x$ and $x^2 + y^2 = 2$ are found by substituting $y^2 = x$ into the circle equation: $x^2 + x - 2 = 0$,which giv

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$9\pi - 2 : 3\pi + 2$

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(B) The intersection points of $y^2 = x$ and $x^2 + y^2 = 2$ are found by substituting $y^2 = x$ into the circle equation: $x^2 + x - 2 = 0$,which gives $(x+2)(x-1) = 0$. Since $x \ge 0$,we have $x = 1$. Thus,the intersection points are $(1, 1)$ and $(1, -1)$.The area of the circle is $A_{total} = \pi r^2 = 2\pi$.The area of the region bounded by the parabola and the circle to the right of the $y$-axis is given by:$A_1 = 2 \int_{0}^{1} \sqrt{x} \, dx + 2 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx$Calculating the first integral:$2 \int_{0}^{1} x^{1/2} \, dx = 2 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{4}{3}$.Calculating the second integral:$2 \int_{1}^{\sqrt{2}} \sqrt{2 - x^2} \, dx = 2 \left[ \frac{x}{2} \sqrt{2 - x^2} + \frac{2}{2} \sin^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_{1}^{\sqrt{2}}$$= 2 \left[ (0 + \sin^{-1}(1)) - (\frac{1}{2} \sqrt{1} + \sin^{-1} \left( \frac{1}{\sqrt{2}} \right)) \right]$$= 2 \left[ \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} \right] = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \right] = \frac{\pi}{2} - 1$.So,$A_1 = \frac{4}{3} + \frac{\pi}{2} - 1 = \frac{\pi}{2} + \frac{1}{3} = \frac{3\pi + 2}{6}$.The other area is $A_2 = A_{total} - A_1 = 2\pi - \frac{3\pi + 2}{6} = \frac{12\pi - 3\pi - 2}{6} = \frac{9\pi - 2}{6}$.The ratio of the areas is $A_2 : A_1 = \frac{9\pi - 2}{6} : \frac{3\pi + 2}{6} = 9\pi - 2 : 3\pi + 2$.

The parabola $y^2 = x$ divides the circle $x^2 + y^2 = 2$ into two parts whose areas are in the ratio

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