Let $f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$.

Consider

$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$

$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$

Then,

<p>$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$</p> <p>$\therefore$ $$f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.$$</p> <p>$\therefore$ $$f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$$</p> <p>and $$\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $$</p> <p>$$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$$</p> <p>$= {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$</p> <p>$\therefore$ Only (S2) is correct</p>