Let y=p(x) be the parabola passing through th | vedclass

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(D) The parabola $y=p(x)$ passes through $(-1,0), (0,1), (1,0)$. Let $p(x) = ax^2+bx+c$. Substituting the points: $c=1$,$a-b+1=0$,$a+b+1=0$. Solving g

Vedclass · Accepted Answer

$16$

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(D) The parabola $y=p(x)$ passes through $(-1,0), (0,1), (1,0)$. Let $p(x) = ax^2+bx+c$. Substituting the points: $c=1$,$a-b+1=0$,$a+b+1=0$. Solving gives $a=-1, b=0, c=1$. Thus,$p(x) = 1-x^2$. The region is defined by $(x+1)^2+(y-1)^2 \leq 1$ (a circle with center $(-1, 1)$ and radius $1$) and $y \leq 1-x^2$. Let $X = x+1$,then $x = X-1$. The parabola becomes $y = 1-(X-1)^2 = 1-(X^2-2X+1) = 2X-X^2$. The circle is $X^2+(y-1)^2 = 1$,so $y = 1 \pm \sqrt{1-X^2}$. The region is $X^2+(y-1)^2 \leq 1$ and $y \leq 2X-X^2$. Intersection: $X^2+(2X-X^2-1)^2 = 1 \implies X^2+(X^2-2X+1)^2 = 1 \implies X^2+(X-1)^4 = 1$. Solving $X^2+(X-1)^4=1$ gives $X=0$ and $X=1$. The area $A$ is the area of the circular segment cut by the parabola. Calculating the integral or using geometric properties,the area $A$ is found to be $\frac{\pi}{4} - \frac{2}{3}$. Thus,$12(\pi - 4A) = 12(\pi - 4(\frac{\pi}{4} - \frac{2}{3})) = 12(\pi - \pi + \frac{8}{3}) = 12 	imes \frac{8}{3} = 32$. Wait,re-evaluating the intersection and area: The area $A$ is $\frac{\pi}{4} - \frac{1}{3}$. Then $12(\pi - 4(\frac{\pi}{4} - \frac{1}{3})) = 12(\pi - \pi + \frac{4}{3}) = 16$.

Let $y=p(x)$ be the parabola passing through the points $(-1,0), (0,1)$ and $(1,0)$. If the area of the region $\{(x, y) : (x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\}$ is $A$,then $12(\pi-4A)$ is equal to $.........$.

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