Mathematics · 2017
JEE Main · 8 April 2017, Shift 1 · Q76
The area (in sq. units) of the smaller portion enclosed between the curves, x^2+y^2=4 and y^2=3 x, is:
The area (in sq. units) of the smaller portion enclosed between the curves, $\displaystyle x^2+y^2=4$ and $\displaystyle y^2=3 x$, is :
Official answer
(4)
$\displaystyle \frac{1}{\sqrt{3}}+\frac{4 \pi}{3}$
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
- If the area enclosed between the curves y=k x^2 and x=k y^2,(k>0), is 1 square unit. Then k is:2019
- 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
JEE Main 2017 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.