SATHEE: Area Question 28

Area Question 28

Question 28

If the line $x = α$ divides the area of region $Missing or unrecognized delimiter for \left$ into two equal parts, then

(2017 Adv.) (a) $2 α^{4} - 4 α^{2} + 1 = 0$ (b) $α^{4} + 4 α^{2} - 1 = 0$ (c) $\frac{1}{2} < α < 1$ (d) $0 < α \leq \frac{1}{2}$

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Solution:

$\int_{0}^{1} (x - x^{3}) d x = 2 \int_{0}^{α} (x - x^{3}) d x$

$\begin{aligned} \frac{1}{4} & = 2 \frac{α^{2}}{2} - \frac{α^{4}}{4} \Rightarrow 2 α^{4} - 4 α^{2} + 1 & = 0 α^{2} & = \frac{4 - \sqrt{16 - 8}}{4} α^{2} & = 1 - \frac{1}{\sqrt{2}} \end{aligned}$

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Area Question 28

Question 28

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