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Let $S$ be the set of all points $(x,y)$ in the $x-y$ plane such that $|x|+|y|\leq 2$ and $|x|\geq 1$. Then, the area, in square units, of the region represented by $S$ equals _____

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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