OABC is a unit square where O is the origin a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 $a_1+2a_2+3a_3=$

Similar Questions

The area bounded by the curve $y=\cos x$,the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi / 4, \cos (\pi / 4))$ and $(0,2)$ is

Find the area lying between the curves $y^{2}=2x$ and $y=x$.

Let $S = \{(x, y) \in R \times R : x \geq 0, y \geq 0, y^2 \leq 4x, y^2 \leq 12 - 2x \text{ and } 3y + \sqrt{8}x \leq 5\sqrt{8}\}$. If the area of the region $S$ is $\alpha \sqrt{2}$,then $\alpha$ is equal to

Let $f: R \to R$ be a function such that $f(x) + 3f(\frac{\pi}{2} - x) = \sin x, x \in R$. Let the maximum value of $f$ on $R$ be $\alpha$. If the area of the region bounded by the curves $g(x) = x^2$ and $h(x) = \beta x^3, \beta > 0$, is $\alpha^2$, then $30\beta^3$ is equal to ————

The area enclosed by the parabola ${y^2} = 4ax$ and the straight line $y = 2ax$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module