Why is the graph flat between x = 1 and x = 2?

For any point between 1 and 2, the sum of distances |x - 1| + |x - 2| is constant and equal to the distance between 1 and 2, which is 2 - 1 = 1.

Integral Calculus — JEE Main 6 Apr 2023 Shift 2 Q9

A · 3

Believing that the area bounded is a simple triangle of base $3$ and height $2$ , or omitting the rectangular core between $x = 1$ and $x = 2$ . Plot the curve by intervals: between $x = 1$ and $x = 2$ , the function is constant at $y = 1$ , so the bounded region is a trapezoid rather than a triangle.

B · 4

Correct option. Identifying the trapezoidal shape with parallel horizontal boundaries at $y = 1$ and $y = 3$ . The top boundary has length $3 - (- 1) = 4$ , the bottom flat segment has length $2 - 1 = 1$ , and the height is $3 - 1 = 2$ . The area is $\frac{1}{2} (4 + 1) (2) = 5$ minus the offset, or more directly $\frac{1}{2} (a + b) h = \frac{1}{2} (4 + 1) (2)$ wait: check lengths: $y = ∣ x - 1 ∣ + ∣ x - 2 ∣ = 3$ . For $x \geq 2$ , $2 x - 3 = 3 ⟹ x = 3$ . For $x \leq 1$ , $3 - 2 x = 3 ⟹ x = 0$ . Base at $y = 3$ extends from $x = 0$ to $x = 3$ , length is $3 - 0 = 3$ , not $4$ . Wait! Let's recompute: for $x \leq 1$ , $∣ x - 1 ∣ + ∣ x - 2 ∣ = 1 - x + 2 - x = 3 - 2 x$ . Set $3 - 2 x = 3 ⟹ x = 0$ . For $x \geq 2$ , $x - 1 + x - 2 = 2 x - 3 = 3 ⟹ x = 3$ . Length of top base is $3 - 0 = 3$ . Length of bottom base is $2 - 1 = 1$ . Height is $3 - 1 = 2$ . Area of trapezoid is $\frac{1}{2} (3 + 1) (2) = 4$ .

C · 5

Incorrectly solving for the intersection points by taking the outer domain interval as $[- 1, 3]$ (length $4$ ) instead of $[0, 3]$ (length $3$ ). Solve $∣ x - 1 ∣ + ∣ x - 2 ∣ = 3$ carefully: for $x \leq 1$ , $- (x - 1) - (x - 2) = 3 - 2 x = 3 ⟹ x = 0$ , not $x = - 1$ .

D · 6

Multiplying the base length $3$ by height $2$ without applying the trapezoid factor, treating the region as a full bounding rectangle. Remember that the side boundaries slope inward from $y = 3$ down to $y = 1$ , forming a trapezoid of area $\frac{1}{2} (b_{1} + b_{2}) h$ , not a rectangle of area $b_{1} h$ .

Area Bounded by a Piecewise Linear Modulus Function and a Horizontal Line

What feels right?

The region bounded by $y = ∣ x - 1 ∣ + ∣ x - 2 ∣$ and $y = 3$ is a trapezoid with parallel sides of lengths $4$ and $1$ and height $2$ , yielding an area of $4$ .

Why each option works or fails

Solution

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