Area bounded by the lines y = |x| – 2 and y = 1 – | x – 1 | is equal to

The correct option is 4 sq. units.

To find the area bounded by the two curves, we first analyze their equations:
1. $y=\left|x\right|-2$
2. $y=1-|x-1|$

Let us analyze the shape and vertices of both functions:
- The first curve, $y=\left|x\right|-2$, is a V-shaped curve with its vertex at $(0,-2)$ that opens upwards.
- The second curve, $y=1-|x-1|$, is an inverted V-shaped curve with its vertex at $(1,1)$ that opens downwards.

Now, we find the points of intersection of the two curves by analyzing them in different intervals:
Case 1: $x<0$
Here, $\left|x\right|=-x$ and $|x-1|=1-x$.
The equations become:
$y=-x-2$
$y=1-(1-x)=x$
Equating them: $-x-2=x\Rightarrow 2x=-2\Rightarrow x=-1$.
Thus, one intersection point is $(-1,-1)$.

Case 2: $x>1$
Here, $\left|x\right|=x$ and $|x-1|=x-1$.
The equations become:
$y=x-2$
$y=1-(x-1)=2-x$
Equating them: $x-2=2-x\Rightarrow 2x=4\Rightarrow x=2$.
Thus, the other intersection point is $(2,0)$.

The bounded region is a quadrilateral with vertices at the intersection points and the vertices of the two V-graphs:
- Vertex $A(-1,-1)$
- Vertex $B(0,-2)$
- Vertex $C(2,0)$
- Vertex $D(1,1)$

We can calculate the area of this quadrilateral $ABCD$ by dividing it into two triangles along the diagonal $AC$ or by using the coordinates formula for the area of a polygon:
$\text{Area}=\frac{1}{2}\left|x_A\right(y_B-y_D)+x_B(y_C-y_A)+x_C(y_D-y_B)+x_D(y_A-y_C\left)\right|$
Let's compute the area using the formula:
$\text{Area}=\frac{1}{2}\left|\right(-1\left)\right(-2-1)+0(0-(-1))+2(1-(-2))+1(-1-0\left)\right|$
$\text{Area}=\frac{1}{2}\left|\right(-1\left)\right(-3)+0+2(3)+1(-1\left)\right|$
$\text{Area}=\frac{1}{2}|3+0+6-1|$
$\text{Area}=\frac{1}{2}\left|8\right|=4\text{sq. units}$

Therefore, the area bounded by the lines is indeed 4 sq. units.