JEE Main 20202 Sep, Morning shiftMathematicsMediumIntegral Calculus369× since 2002Area (in sq. units) of the region outside ∣x∣2+∣y∣3=1{{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 12∣x∣+3∣y∣=1 and inside the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 14x2+9y2=1 is :AOption A. 6(4−π)6\left( {4 - \pi } \right)6(4−π)BOption B. 3(4−π)3\left( {4 - \pi } \right)3(4−π)COption C. 6(π−2)6\left( {\pi - 2} \right)6(π−2)DOption D. 3(π−2)3\left( {\pi - 2} \right)3(π−2)CheckSkip · show solution
