Let f(x) = 2 + |x| - |x - 1| + |x + 1|,x in R | vedclass

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(A) Given $f(x) = 2 + |x| - |x - 1| + |x + 1|$.We define $f(x)$ in different intervals:For $x < -1$: $f(x) = 2 - x - (1 - x) - (x + 1) = 2 - x - 1 + x

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both $(S1)$ and $(S2)$ are correct

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(A) Given $f(x) = 2 + |x| - |x - 1| + |x + 1|$.We define $f(x)$ in different intervals:For $x For $-1 \le x For $0 \le x For $x \ge 1$: $f(x) = 2 + x - (x - 1) + (x + 1) = 2 + x - x + 1 + x + 1 = x + 4$.Checking $(S1)$:$f^{\prime}(x) = -1$ for $x  1$.$f^{\prime}(-3/2) = -1$,$f^{\prime}(-1/2) = 1$,$f^{\prime}(1/2) = 3$,$f^{\prime}(3/2) = 1$.Sum $= -1 + 1 + 3 + 1 = 4$. Thus,$(S1)$ is correct.Checking $(S2)$:$\int_{-2}^{2} f(x) dx = \int_{-2}^{-1} (-x) dx + \int_{-1}^{0} (x + 2) dx + \int_{0}^{1} (3x + 2) dx + \int_{1}^{2} (x + 4) dx$$= [-\frac{x^2}{2}]_{-2}^{-1} + [\frac{x^2}{2} + 2x]_{-1}^{0} + [\frac{3x^2}{2} + 2x]_{0}^{1} + [\frac{x^2}{2} + 4x]_{1}^{2}$$= (-1/2 + 2) + (0 - (1/2 - 2)) + (3/2 + 2 - 0) + (2 + 8 - (1/2 + 4))$$= 1.5 + 1.5 + 3.5 + 5.5 = 12$. Thus,$(S2)$ is correct.Both are correct.

Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. Consider:
$(S1): f^{\prime}\left(-\frac{3}{2}\right) + f^{\prime}\left(-\frac{1}{2}\right) + f^{\prime}\left(\frac{1}{2}\right) + f^{\prime}\left(\frac{3}{2}\right) = 4$
$(S2): \int_{-2}^{2} f(x) dx = 12$
Then,

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Let $f(x) = 2 + |x| - |x - 1| + |x + 1|$,$x \in R$. Consider:$(S1): 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) = 4$$(S2): \int_{-2}^{2} f(x) dx = 12$Then,