Mathematics · 2024
JEE Main · 30 January 2024, Shift 1 · Q11
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:
The area (in square units) of the region bounded by the parabola $\displaystyle y^2=4(x-2)$ and the line $\displaystyle y=2 x-8$, is:
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 9$
More from Area Under the Curve
- 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 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
- The area of the region R={(x, y): x y ≤ 27,1 ≤ y ≤ x^2} is equal to:2026
- The area of the region bounded by the curve y= max {|x|, x|x-2|}, the x -axis and the lines x=-2 and x=4 is equal to ____2025
- A line passing through the point A(-2,0), touches the parabola P: y^2=x-2 at the point B in the first quadrant. The area, of the region bounded by…2025
- Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y^2=x and y=|α x-5|-|1-α x|+α x^2. Then (f(0)+f(1)) is equal to2026
- The area of the region enclosed between the circles x^2+y^2=4 and x^2+(y-2)^2=4 is:2026
JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.