Q. The area bounded by the curves y = |x – 1| + |x – 2| and y = 3 is equal to
(1) 3
(2) 4
(3) 5
(4) 6
Area Between the Curves
We are given:
\[
y = |x-1| + |x-2|
\]
and
\[
y = 3
\]
1. Find the points of intersection
At the points of intersection:
\[
|x-1| + |x-2| = 3
\]
Case 1: \(x \leq 1\)
Here both expressions inside the modulus are negative:
\[
|x-1| = 1-x
\]
\[
|x-2| = 2-x
\]
Therefore:
\[
(1-x)+(2-x)=3
\]
\[
3-2x=3
\]
\[
x=0
\]
First intersection:
\[
\boxed{(0,3)}
\]
Case 2: \(1 \leq x \leq 2\)
In this interval:
\[
|x-1|=x-1
\]
\[
|x-2|=2-x
\]
Hence:
\[
(x-1)+(2-x)=1
\]
So the graph is horizontal at \[ \boxed{y=1} \] between \(x=1\) and \(x=2\).
Case 3: \(x \geq 2\)
Here both expressions are positive:
\[
|x-1|=x-1
\]
\[
|x-2|=x-2
\]
Therefore:
\[
(x-1)+(x-2)=3
\]
\[
2x-3=3
\]
\[
x=3
\]
Second intersection:
\[
\boxed{(3,3)}
\]
2. Diagram
3. Calculate the Area
The enclosed region consists of a rectangle and two triangles.
Rectangle
Width: \[ 2-1=1 \]
Height: \[ 3-1=2 \]
\[
A_{\text{rectangle}} = 1\times2=2
\]
Two triangles
Each triangle has:
- Base = \(1\)
- Height = \(2\)
\[
A_{\text{one triangle}}
=
\frac12(1)(2)
=
1
\]
Therefore, both triangles have area:
\[
2\times1=2
\]
4. Total Area
\[
A=2+2
\]
\[
\boxed{\text{Required Area}=4}
\]