Let f(x) = 2 + |x| - |x - 1| + |x + 1|, x ∈ R . Consider (S1): f ^...

Given f\left(x\right) = 2 + \left|x\right| - \left|x - 1\right| + \left|x + 1\right|, x \in R ⇒ f\left(x\right) = \left\{\begin{cases}- x & , & x < - 1 \\ x + 2 & , & - 1 \le x < 0 \\ 3x + 2 & , & 0 \le x < 1 \\ x + 4 & , & x \ge 1\end{cases}\right)</p><p>⇒ {f}^{'}\left(x\right) = \left\{$\begin{cases}- 1 & , & x 1\end{cases}$\right)

$∴ {f}^{'}\left(- \dfrac{3}{2}\right) + {f}^{'}\left(- \dfrac{1}{2}\right) + {f}^{'}\left(\dfrac{1}{2}\right) + {f}^{'}\left(\dfrac{3}{2}\right)$

$= - 1 + 1 + 3 + 1 = 4$

i.e. \left(S1\right)$is incorrect.</p></blockquote><blockquote><p>Now${∫}_{- 2}^{2}f\left(x\right)dx = {∫}_{- 2}^{- 1}f\left(x\right)dx + {∫}_{- 1}^{0}f\left(x\right)dx + {∫}_{0}^{1}f\left(x\right)dx + {∫}_{1}^{2}f\left(x\right)dx$</p><p>$= {\left[- \dfrac{{x}^{2}}{2}\right]}_{- 2}^{- 1} + {\left[\dfrac{{\left(x + 2\right)}^{2}}{2}\right]}_{- 1}^{0} + {\left[\dfrac{{\left(3x + 2\right)}^{2}}{6}\right]}_{0}^{1} + {\left[\dfrac{{\left(x + 4\right)}^{2}}{2}\right]}_{1}^{2}$</p><p>\left(- \dfrac{1}{2} + 2\right) +$\left(2 - \dfrac{1}{2}\right) + \left(\dfrac{25}{6} - \dfrac{4}{6}\right) + \left(18 - \dfrac{25}{2}\right)$</p><p>$= \dfrac{3}{2} + \dfrac{3}{2} + \dfrac{7}{2} + \dfrac{11}{2} = 12$</p><p>i.e. \left(S2\right)$ is correct.