Area Question 28
Question 28
- If the line $x=\alpha$ divides the area of region $R=\left{(x, y) \in R^{2}: x^{3} \leq y \leq x, 0 \leq x \leq 1\right}$ into two equal parts, then
(2017 Adv.) (a) $2 \alpha^{4}-4 \alpha^{2}+1=0$ (b) $\alpha^{4}+4 \alpha^{2}-1=0$ (c) $\frac{1}{2}<\alpha<1$ (d) $0<\alpha \leq \frac{1}{2}$
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Solution:
- $\int_{0}^{1}\left(x-x^{3}\right) d x=2 \int_{0}^{\alpha}\left(x-x^{3}\right) d x$
$$ \begin{aligned} \frac{1}{4} & =2 \frac{\alpha^{2}}{2}-\frac{\alpha^{4}}{4} \ \Rightarrow \quad 2 \alpha^{4}-4 \alpha^{2}+1 & =0 \ \alpha^{2} & =\frac{4-\sqrt{16-8}}{4} \ \alpha^{2} & =1-\frac{1}{\sqrt{2}} \end{aligned} $$
