JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q9

MathematicsSingle correct+4 / -1Question 9 of 90

The area of the region $S=\left\{(x,y):y^2\le8x,\ y\ge\sqrt2x,\ x\ge1\right\}$ is

\dfrac{13\sqrt2}6

\dfrac{11\sqrt2}6

\dfrac{5\sqrt2}6

\dfrac{19\sqrt2}6

Answer and worked solution

Official answer: B

Intersection. $y=\sqrt2x$ meets $y^2=8x$ at $x=4$ .

Area. $\displaystyle\int_1^4\left(2\sqrt2\sqrt x-\sqrt2x\right)dx=\frac{28\sqrt2}3-\frac{15\sqrt2}2=\frac{11\sqrt2}6$ .

(B)

Topics:Area between curves

JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 9