$$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$$

$$\therefore$$ $$f(x) = \left\{ {\matrix{ { - x} & , & {x < - 1} \cr {x + 2} & , & { - 1 \le x < 0} \cr {3x + 2} & , & {0 \le x < 1} \cr {x + 4} & , & {x \ge 1} \cr } } \right.$$

$$\therefore$$ $$f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$$

and $$\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $$

$$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$$

$$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$$

$$\therefore$$ Only (S2) is correct