Mathematics · 2025
JEE Main · 23 January 2025, Shift 2 · Q16
If the area of the region {(x, y):-1 ≤ x ≤ 1,0 ≤ y ≤ a + e^|x|- e^-x, a >0} is (e^2+8 e +1)/e, then the value of a is:
If the area of the region $\displaystyle \left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}>0\right\}$ is $\displaystyle \frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of a is :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 5$
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
- 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
- Let f: R → R be a twice differentiable function such that f(x+y)=f(x) f(y) for all x, y ∈ R. If f^′(0)=4 a and f satisfies f^′ ′(x)-3 a…2025
JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.