Mathematics · 2026
JEE Main · 6 April 2026, Shift 2 · Q19
The area of the region {(x, y): x^2-8 x ≤ y ≤-x} is:
The area of the region $\displaystyle \left\{(x, y): x^2-8 x \leq y \leq-x\right\}$ is:
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle \frac{343}{6}$
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…2026
- The area of the region R={(x, y): x y ≤ 27,1 ≤ y ≤ x^2} is equal to:2026
- If the area of the region bounded by 16 x^2-9 y^2=144 and 8 x-3 y=24 is A, then 3( A+6 log_e(3)) is equal to…2026
- The area of the region {(x, y): y ≤ π-|x|, y ≤|x sin x|, y ≥ 0} is:2026
- The area of the region bounded by the curves x+3 y^2=0 and x+4 y^2=1 is equal to:2026
- Let e be the base of natural logarithm and let f:{1,2,3,4} →{1, e, e^2, e^3} and g:{1, e, e^2, e^3} →{1, 1/2,…2026
- The area of the region {(x, y): 0 ≤ y ≤ 6-x, y^2 ≥ 4 x-3, x ≥ 0} is:2026
- Let f: R → R be a function such that f(x)+3 f((π)/2-x)= sin x, x ∈ R. Let the maximum value of f on R be α.…2026
JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.