JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q6

MathematicsSingle correct+4 / -1Question 6 of 90

The area enclosed by $y^2=8x$ and $y=\sqrt2x$ that lies outside the triangle formed by $y=\sqrt2x$ , $x=1$ , $y=2\sqrt2$ , is equal to :

\dfrac{16\sqrt2}6

\dfrac{11\sqrt2}6

\dfrac{13\sqrt2}6

\dfrac{5\sqrt2}6

Answer and worked solution

Official answer: C

Whole region between $y=\sqrt2x$ and $y^2=8x$ (meeting at $x=4$ ): $\int_0^4\left(2\sqrt2\sqrt x-\sqrt2x\right)dx=\frac{8\sqrt2}3$ .

Triangle with vertices $(1,\sqrt2)$ , $(1,2\sqrt2)$ , $(2,2\sqrt2)$ lies inside it, area $\frac{\sqrt2}2$ .

Outside. $\frac{8\sqrt2}3-\frac{\sqrt2}2=\frac{13\sqrt2}6$ .

(C)

Topics:Area between curves

JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 6