Mathematics · 2020
JEE Main · 2 September 2020, Shift 1 · Q62
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 (in sq. units) of the region outside $\displaystyle \frac{|x|}{2}+\frac{|y|}{3}=1$ and inside the ellipse $\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$ is :
Official answer
(3)
$\displaystyle 6(\pi-2)$
More from Area Under the Curve
- The area (in square units) of the region enclosed by the ellipse x^2+3 y^2=18 in the first quadrant below the line y=x is2024
- Let f:(1, ∞) → R be a function defined as f(x)=(x-1)/(x+1). Let f^i+1(x)=f(f^i(x)), i=1,2, …, 25, where f^1(x)=f(x). If g(x)+f^26(x)=0, x ∈(1, ∞),…2026
- The area of the region given by A={(x, y): x^2 ≤ y ≤ min {x+2,4-3 x}} is:2022
- The area (in square units) of the region bounded by the parabola y^2=4(x-2) and the line y=2 x-8, is:2024
- The area of the region A={(x, y): 4 x^2+y^2 ≤ 8. and.y^2 ≤ 4 x} is:2026
- If the area of the region {(x, y): a/x^2 ≤ y ≤ 1/x, 1 ≤ x ≤ 2,0<a<1} is ( log_e 2)-1/7 then the value of 7 a-3 is equal to:2024
- Given: f(x)= {x, 0 ≤ x<1/2; 1/2, x=1/2; 1-x, 1/2<x ≤ 1} and g(x)=(x-1/2)^2, x ∈ R. Then the area (in sq. units) of the region bounded by the curves,…2020
- Let A_1 be the area of the region bounded by the curves y= sin x, y= cos x and y -axis in the first quadrant. Also, let A_2 be the area of the region…2021
JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.