Question

$OABC$ 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 $OABO,OBO$ and $OBCO$. If $a_1,a_2,a_3$ are the areas (in sq units) of these parts respectively, then $a_1+2a_2+3a_3=$

• A. 1
• B. 2
• C. 6
• D. 64

Answer 1

Answer: (B)

$\because a_1+a_2+a_3=1$ ...(i)
and due to symmetry $a_1=a_3$ ...(ii)
Now, $a_2=\underset{0}{\overset{1}{\int }}\left(\sqrt{x}-x^2\right)dx$
$={\left[\frac{2}{3}{x}^{3/2}-\frac{x^3}{3}\right]}_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

So, $a_1+a_3+\frac{1}{3}=1$
$\Rightarrow a_1+a_3=\frac{2}{3}$ ...(iii)
From Eqs. (ii) and (iii),
$a_1=a_3=\frac{1}{3}=a_2$
So, $a_1+2a_2+3a_3=\frac{1}{3}+\frac{2}{3}+\frac{3}{3}=2$