The polynomial f(x) satisfies the condition f | vedclass

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(A) Given $f(x + 1) = x^2 + 4x$. To find $f(x - 1)$,substitute $x$ with $x - 2$ in the given equation: $f((x - 2) + 1) = (x - 2)^2 + 4(x - 2)$ $f(x -

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$\frac{16\sqrt{2}}{3}$

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(A) Given $f(x + 1) = x^2 + 4x$. To find $f(x - 1)$,substitute $x$ with $x - 2$ in the given equation: $f((x - 2) + 1) = (x - 2)^2 + 4(x - 2)$ $f(x - 1) = x^2 - 4x + 4 + 4x - 8 = x^2 - 4$. Now,we find the intersection points of $y = x^2 - 4$ and $y = -x^2$: $x^2 - 4 = -x^2$ $2x^2 = 4 \implies x^2 = 2 \implies x = \pm\sqrt{2}$. The area $A$ enclosed by the two curves is given by: $A = \int_{-\sqrt{2}}^{\sqrt{2}} [(-x^2) - (x^2 - 4)] \, dx$ $A = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx = 2 \int_{0}^{\sqrt{2}} (4 - 2x^2) \, dx$ $A = 2 [4x - \frac{2x^3}{3}]_{0}^{\sqrt{2}} = 2 [4\sqrt{2} - \frac{2(2\sqrt{2})}{3}]$ $A = 2 [4\sqrt{2} - \frac{4\sqrt{2}}{3}] = 2 [\frac{12\sqrt{2} - 4\sqrt{2}}{3}] = 2 [\frac{8\sqrt{2}}{3}] = \frac{16\sqrt{2}}{3}$.

The polynomial $f(x)$ satisfies the condition $f(x + 1) = x^2 + 4x$. The area enclosed by $y = f(x - 1)$ and the curve $x^2 + y = 0$ is

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