If the line x=α divides the area of region R=(x,y)∈R2:x3≤y≤x,0≤x≤1 into two equal parts then

1 2 < α < 1 , 2 α 4 − 4 α 2 + 1 = 0

If the line x=α divides the area of region R=(x,y)∈R2:x3≤y≤x,0≤x≤1 into two equal parts then

2 ∫ 0 α x − x 3 d x = ∫ 0 1 x − x 3 d x 2 α 2 2 − α 4 4 = 1 2 − 1 4 2 α 4 − 4 α 2 + 1 = 0 α 2 = 4 ± 2 2 4 ⇒ α 2 = 1 - 1 2 ⇒ 1 2 < α < 1 <img 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">

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$If the line x = α divides the area of region R = \{(x, y) \in R^{2} : x^{3} \leq y \leq x, 0 \leq x \leq 1\} into two equal parts$ then

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$α^{4} + 4 α^{2} - 1 = 0$

$\frac{1}{2} < α < 1$

$0 < α \leq \frac{1}{2}$

$2 α^{4} - 4 α^{2} + 1 = 0$

answer is B, D.

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Detailed Solution

$\begin{matrix} 2 \int_{0}^{α} (x - x^{3}) d x = \int_{0}^{1} (x - x^{3}) d x \\ 2 [\frac{α^{2}}{2} - \frac{α^{4}}{4}] = \frac{1}{2} - \frac{1}{4} \\ 2 α^{4} - 4 α^{2} + 1 = 0 \\ α^{2} = \frac{4 \pm 2 \sqrt{2}}{4} \\ \Rightarrow α^{2} = 1 - \frac{1}{\sqrt{2}} \Rightarrow \frac{1}{2} < α < 1 \end{matrix}$