If the line x=alpha divides the area of regio | vedclass

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(A) The total area of the region $R$ is given by:$A = \int_0^1 (x - x^3) dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(A) The total area of the region $R$ is given by:$A = \int_0^1 (x - x^3) dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} ight]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.Since the line $x = \alpha$ divides the area into two equal parts,the area from $0$ to $\alpha$ must be half of the total area:$\int_0^{\alpha} (x - x^3) dx = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.Evaluating the integral:$\left[ \frac{x^2}{2} - \frac{x^4}{4} ight]_0^{\alpha} = \frac{1}{8}$$\frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8}$Multiply by $4$:$2 \alpha^2 - \alpha^4 = \frac{1}{2}$$4 \alpha^2 - 2 \alpha^4 = 1$$2 \alpha^4 - 4 \alpha^2 + 1 = 0$. This confirms option $[C]$.Let $f(\alpha) = 2 \alpha^4 - 4 \alpha^2 + 1$. We have $f(0) = 1$ and $f(1) = 2 - 4 + 1 = -1$.Since $f(\alpha)$ is continuous and changes sign between $0$ and $1$,there exists a root $\alpha \in (0, 1)$.Also,$f(1/\sqrt{2}) = 2(1/4) - 4(1/2) + 1 = 0.5 - 2 + 1 = -0.5 Since $f(0) = 1 > 0$ and $f(1/\sqrt{2}) Since $1/\sqrt{2} \approx 0.707$,and $1/2 = 0.5$,we check $f(1/2) = 2(1/16) - 4(1/4) + 1 = 1/8 - 1 + 1 = 1/8 > 0$.Since $f(1/2) > 0$ and $f(1/\sqrt{2}) < 0$,the root $\alpha$ lies in $(1/2, 1/\sqrt{2})$,which is a subset of $(1/2, 1)$. Thus,option $[B]$ is also true.

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