JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 10)
If the line x = $$\alpha $$ divides the area of region R = {(x, y) $$ \in $$R2 : x3 $$ \le $$ y $$ \le $$ x, 0 $$ \le $$ x $$ \le $$ 1} into two equal parts, then
2$$\alpha $$4 $$-$$ 4$$\alpha $$2 + 1 =0
$$\alpha $$4 + 4$$\alpha $$2 $$-$$ 1 =0
$${1 \over 2} < \alpha < 1$$
0 < $$\alpha $$ $$ \le $$ $${1 \over 2}$$
Explanation
$$\int_0^1 {(x - {x^3})dx = 2\int_0^\alpha {x - {x^3})dx} } $$
$${1 \over 4} = 2\left( {{{{\alpha ^2}} \over 2} - {{{\alpha ^2}} \over 4}} \right)$$
$$2{\alpha ^2} - 4{\alpha ^2} + 1 = 0$$
$${\alpha ^2} = {{4 - \sqrt {16 - 8} } \over 4}$$ ($$ \because $$ $$\alpha $$$$ \in $$(0, 1))
$$ \because $$ $${\alpha ^2} = 1 - {1 \over {\sqrt 2 }}$$
$${1 \over 4} = 2\left( {{{{\alpha ^2}} \over 2} - {{{\alpha ^2}} \over 4}} \right)$$
$$2{\alpha ^2} - 4{\alpha ^2} + 1 = 0$$
$${\alpha ^2} = {{4 - \sqrt {16 - 8} } \over 4}$$ ($$ \because $$ $$\alpha $$$$ \in $$(0, 1))
$$ \because $$ $${\alpha ^2} = 1 - {1 \over {\sqrt 2 }}$$
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