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Given that,

  • $ \left | x \right | + \left | y \right | \leq 2 \quad \longrightarrow (1) $
  • $ \left | x \right | \geq 1 \quad \longrightarrow (2) $

We know that, $ \left | x \right | = \left\{\begin{matrix} x \; ; &  x \geq 0 & \\ -x\; ; & x < 0 & \end{matrix}\right.$

Now, from the equation $(1),$

  • $ x+y = 2 \quad \longrightarrow (3) $
  • $ x-y = 2 \quad \longrightarrow (4) $
  • $ -x+y = 2 \quad \longrightarrow (5) $
  • $ -x-y = 2 \quad \longrightarrow (6) $

On solving equation $(3)$, and $(4),$ we get

$\begin{array}{c} x+y = 2 \\ x-y = 2 \\\hline 2x = 4 \end{array}$

$ \Rightarrow x=2,$ and $y=0$ 

$ \Rightarrow \boxed{(x,y) = (2,0)}$

On solving equation $(3)$, and $(5),$ we get

$\begin{array}{c} x+y = 2 \\ -x+y = 2 \\\hline 2y = 4 \end{array}$

$ \Rightarrow y=2,$ and $x=0$ 

$ \Rightarrow \boxed{(x,y) = (0,2)}$

On solving equation $(4)$, and $(6)$, we get

$\begin{array}{c} x-y = 2 \\ -x-y = 2 \\\hline -2y = 4 \end{array}$

$ \Rightarrow y=-2,$ and $x=0$ 

$ \Rightarrow \boxed{(x,y) = (0,-2)}$

On solving equation $(5)$, and $(6)$, we get

$\begin{array}{c} -x+y = 2 \\ -x-y = 2 \\\hline -2x = 4 \end{array}$

$ \Rightarrow x=-2,$ and $y=0$ 

$ \Rightarrow \boxed{(x,y) = (-2,0)}$

Now, from the equation $(2),$

  • $ x=1 \quad \longrightarrow (7)$
  • $ -x=1 \Rightarrow x=-1 \quad \longrightarrow (8)$

Put $\boxed {x=1}$ in equation $(3)$, and $(4)$, we get 

  • $ x+y = 2 \Rightarrow 1+y = 2 \Rightarrow \boxed{y=1}$

$ \qquad \qquad \qquad \Rightarrow \boxed{(x,y) = (1,1)}$

  • $ x-y = 2 \Rightarrow 1-y = 2 \Rightarrow \boxed{y=-1}$

$ \qquad \qquad \qquad \Rightarrow \boxed{(x,y) = (1,-1)}$

Put $ \boxed {x=-1}$ in equation $(5)$, and $(6)$, we get 

  • $ -x+y = 2 \Rightarrow -(-1)+y = 2 \Rightarrow \boxed{y=1}$

$ \qquad \qquad \qquad \Rightarrow \boxed{(x,y) = (-1,1)}$

  • $ -x-y = 2 \Rightarrow -(-1)-y = 2 \Rightarrow \boxed{y=-1}$

$ \qquad \qquad \qquad \Rightarrow \boxed{(x,y) = (-1,-1)}$

Now, we can draw the $x\text{-}y$ plane.

We know that, area of triangle $ = \frac{1}{2} \times \text{Base} \times \text{Height}$

Here, area of shaded triangle $ 1 = 2 \left( \frac{1}{2} \times 1 \times 1  \right)  = 1 \; \text {square unit} $

And, area of shaded triangle $ 2= 2 \left( \frac{1}{2} \times 1 \times 1  \right) = 1 \; \text {square unit} $

The total area of the region represented by $S =$ area of  shaded triangle $1 +$ area of shaded triangle $2$

$\qquad = 1+1 = 2 \; \text {square units} $

$\therefore$ The total area of the region represented by $S$ is $2 \; \text {square units.} $

Correct Answer $:2$

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