Mathematics · 2024
JEE Main · 4 April 2024, Shift 1 · Q10
Let f(x)= {-2, -2 ≤ x ≤ 0; x-2, 0<x ≤ 2} and h (x)=f(|x|)+|f(x)|. Then ∫_-2^2 h (x) d x is equal to:
Let $\displaystyle f(x)=\left\{\begin{array}{ll}-2, & -2 \leq x \leq 0 \\ x-2, & 0<x \leq 2\end{array}\right.$ and $\displaystyle \mathrm{h}(x)=f(|x|)+|f(x)|$. Then $\displaystyle \int_{-2}^2 \mathrm{~h}(x) \mathrm{d} x$ is equal to:
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 2$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.