Mathematics · 2022
JEE Main · 27 July 2022, Shift 2 · Q6
Let f(x)=2+|x|-|x-1|+|x+1|, x ∈ R. Consider (S1): f^′(-3/2)+f^′(-1/2)+f^′(1/2)+f^′(3/2)=2 (S2): ∫_-2^2 f(x) d x=12 Then,
Let $\displaystyle f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$.
Consider
(S1) : $\displaystyle f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$
(S2) : $\displaystyle \int_{-2}^2 f(x) \mathrm{d} x=12$
Then,
Official answer
(4)
only (S2) is correct
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.