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JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2020 · 2 September · Morning Shift · Mathematics · Area Under Curves

Area (in sq. units) of the region outside (|x|)/(2) + (|y|)/(3) = 1...

Area (in sq. units) of the region outside

\dfrac{\left|x\right|}{2} + \dfrac{\left|y\right|}{3} = 1

and inside the ellipse

\dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{9} = 1

is

Area (in sq. units) of the region outside (|x|)/(2) + (|y|)/(3) = 1 and inside the ellipse x ^ 2 4 + y ^ 2 9 = 1 is

Options

A $6\left(\pi - 2\right)$
B $3\left(\pi - 2\right)$
C $3\left(4 - \pi\right)$
D $6\left(4 - \pi\right)$

Correct answer

A. $6\left(\pi - 2\right)$

Step-by-step solution

Area (in sq. units) of the region outside (|x|)/(2) + (|y|)/(3) = 1 and inside the ellipse x ^ 2 4 + y ^ 2 9 = 1 is

Area of ellipse $= \piab = 6\pi$
Area enclosed by $\dfrac{| x |}{2} + \dfrac{| y |}{3} = 1$ is
$= 4 \times \dfrac{1}{2}\left(2\right)\left(3\right) = 12$
So, required area $= 6\pi - 12$ sq. units

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