JEE Mathematics 2023 Question with Solution

Given: $y = |x - 1| + |x - 2|$ and $y = 3$.

Find: The area bounded by these curves.

Break the absolute value expression into intervals:

Now find the points of intersection with $y=3$.

For $x<1$:

For $1\le x\le 2$, the curve is $y=1$, so it does not intersect $y=3$ inside this interval.

For $x>2$:

So the bounded region lies from $x=0$ to $x=3$.

Compute the area between the line and the curve:

Simplifying each part:

Evaluate:

Therefore,

So, the area bounded by the curves is $4$ square units. The correct option is B.

Given: $y = |x - 1| + |x - 2|$ and $y = 3$.

Find: The bounded area.

The graph has a horizontal segment $y=1$ from $x=1$ to $x=2$, and straight-line portions on both sides. The intersections with $y=3$ occur at $x=0$ and $x=3$.

So the shaded region can be seen as a rectangle from $x=1$ to $x=2$ of height $3-1=2$, together with two congruent right triangles on the left and right.

Rectangle area:

Each triangle has base $1$ and height $2$, so each area is:

Hence total area is:

Therefore, the bounded area is $4$ square units, so the correct option is B.