If the line x=αdivides the area of region R={(x,y)R2:x3≤x,0≤x≤1}into equal parts, then: 2α4−4α2+1=0 α4+4α2−1=00<α≤21 (d) 21<α<1

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If the line {x}= the area of region {R}={{({x},{y})}{R}^{{2}}:

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If the line $x = α$ divides the area of region $R = {(x, y) R^{2} : x^{3} \leq x, 0 \leq x \leq 1}$ into equal parts, then: $2 α^{4} - 4 α^{2} + 1 = 0$ $α^{4} + 4 α^{2} - 1 = 0$ $0 < α \leq \frac{1}{2}$ (d) $\frac{1}{2} < α < 1$

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William Discussed

If the line

x = α

divides the area of region

R = {(x, y) R^{2} : x^{3} \leq x, 0 \leq x \leq 1}

into equal parts, then:

2 α^{4} - 4 α^{2} + 1 = 0

α^{4} + 4 α^{2} - 1 = 0

0 < α \leq \frac{1}{2}

(d)

\frac{1}{2} < α < 1

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Question Text	If the line $x = α$ divides the area of region $R = {(x, y) R^{2} : x^{3} \leq x, 0 \leq x \leq 1}$ into equal parts, then: $2 α^{4} - 4 α^{2} + 1 = 0$ $α^{4} + 4 α^{2} - 1 = 0$ $0 < α \leq \frac{1}{2}$ (d) $\frac{1}{2} < α < 1$
Answer Type	Video solution: 1
Upvotes	150