Let $$f(x) = 2 + |x| - |x-1| + |x+1|$$, $$x \in \mathbb{R}$$. Consider
$$(S_1): f'(-3/2) + f'(-1/2) + f'(1/2) + f'(3/2) = 2$$
$$(S_2): \int_{-2}^{2} f(x) dx = 12$$
Then,

Given,

$$f(x)=2+|x|-|x-1|+|x+1|$$

Break the function into intervals using critical points

$$x=-1,0,1$$

Case 1: $$x<-1$$

$$|x|=-x,\qquad |x-1|=1-x,\qquad |x+1|=-(x+1)$$

Hence,

$$f(x)=2-x-(1-x)-(x+1)=-x$$

Therefore,

$$f'(x)=-1$$

Case 2: $$-1\le x<0$$

$$|x|=-x,\qquad |x-1|=1-x,\qquad |x+1|=x+1$$

Hence,

$$f(x)=2-x-(1-x)+(x+1)=x+2$$

Therefore,

$$f'(x)=1$$

Case 3: $$0\le x<1$$

$$|x|=x,\qquad |x-1|=1-x,\qquad |x+1|=x+1$$

Hence,

$$f(x)=2+x-(1-x)+(x+1)=3x+2$$

Therefore,

$$f'(x)=3$$

Case 4: $$x\ge1$$

$$|x|=x,\qquad |x-1|=x-1,\qquad |x+1|=x+1$$

Hence,

$$f(x)=2+x-(x-1)+(x+1)=x+4$$

Therefore,

$$f'(x)=1$$

Now check $$(S_1):$$

$$f'(-3/2)+f'(-1/2)+f'(1/2)+f'(3/2)$$

$$=-1+1+3+1$$

$$=4$$

Since $$4\ne2,$$

$$(S_1)$$ is false.

Now check $$(S_2):$$

$$\int_{-2}^{2}f(x)\,dx$$

$$=\int_{-2}^{-1}(-x)\,dx+\int_{-1}^{0}(x+2)\,dx+\int_{0}^{1}(3x+2)\,dx+\int_{1}^{2}(x+4)\,dx$$

Now,

$$\int_{-2}^{-1}(-x)\,dx=\left[-\frac{x^2}{2}\right]_{-2}^{-1}=\frac32$$

$$\int_{-1}^{0}(x+2)\,dx=\left[\frac{x^2}{2}+2x\right]_{-1}^{0}=\frac32$$

$$\int_{0}^{1}(3x+2)\,dx=\left[\frac{3x^2}{2}+2x\right]_{0}^{1}=\frac72$$

$$\int_{1}^{2}(x+4)\,dx=\left[\frac{x^2}{2}+4x\right]_{1}^{2}=\frac{11}{2}$$

Therefore,

$$\int_{-2}^{2}f(x)\,dx=\frac32+\frac32+\frac72+\frac{11}{2}$$

$$=12$$

Hence, $$(S_2)$$ is true.

Therefore, only $$(S_2)$$ is correct.