Mathematics · 2024
JEE Main · 9 April 2024, Shift 2 · Q13
Let ∫_0^x √(1-(y^′(t))^2) d t=∫_0^x y(t) d t, 0 ≤ x ≤ 3, y ≥ 0, y(0)=0. Then at x=2, y^′ ′+y+1 is equal to
Let $\displaystyle \int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$. Then at $\displaystyle x=2, y^{\prime \prime}+y+1$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 1$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.