JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 90

The area of the region enclosed between the parabolas $y^2=2x-1$ and $y^2=4x-3$ is

\dfrac13

\dfrac16

\dfrac23

\dfrac34

Answer and worked solution

Official answer: A

In terms of $y$ . $x=\frac{y^2+1}2$ and $x=\frac{y^2+3}4$ meet at $y=\pm1$ ; the second lies to the right between them.

Area. $\displaystyle\int_{-1}^1\left(\frac{y^2+3}4-\frac{y^2+1}2\right)dy=\int_{-1}^1\frac{1-y^2}4dy=\frac13$ .

(A)

Topics:Area between curves Parabola

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 6