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(B) The unit square $OABC$ has an area of $1 	ext{ sq unit}$. The curves $y^2=x$ and $x^2=y$ intersect at $(0,0)$ and $(1,1)$.Let $a_1, a_2, a_3$ be

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$2$

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(B) The unit square $OABC$ has an area of $1 	ext{ sq unit}$. The curves $y^2=x$ and $x^2=y$ intersect at $(0,0)$ and $(1,1)$.Let $a_1, a_2, a_3$ be the areas of the three parts. By symmetry,$a_1 = a_3$.The total area is $a_1 + a_2 + a_3 = 1 \quad \dots(i)$.Due to symmetry,$a_1 = a_3 \quad \dots(ii)$.The area $a_2$ is the area between the curves $y = \sqrt{x}$ and $y = x^2$ from $x=0$ to $x=1$:$a_2 = \int_0^1 (\sqrt{x} - x^2) dx = \left[ \frac{2}{3}x^{3/2} - \frac{x^3}{3} ight]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$.Substituting $a_2 = \frac{1}{3}$ into equation $(i)$:$a_1 + \frac{1}{3} + a_3 = 1 \implies a_1 + a_3 = \frac{2}{3}$.Since $a_1 = a_3$,we have $2a_1 = \frac{2}{3} \implies a_1 = \frac{1}{3}$ and $a_3 = \frac{1}{3}$.Thus,$a_1 = a_2 = a_3 = \frac{1}{3}$.We need to find $a_1 + 2a_2 + 3a_3$:$a_1 + 2a_2 + 3a_3 = \frac{1}{3} + 2\left(\frac{1}{3}ight) + 3\left(\frac{1}{3}ight) = \frac{1+2+3}{3} = \frac{6}{3} = 2$.

$OABC$ is a unit square where $O$ is the origin and $B=(1,1)$. The curves $y^2=x$ and $x^2=y$ divide the area of the square into three parts $a_1, a_2, a_3$. If $a_1, a_2, a_3$ are the areas (in sq units) of these parts respectively,then $a_1+2a_2+3a_3=$

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