The area (in sq. units) covered by `[x-y]=-3` with the coordinate axes is (where `[.]` is the greatest integer function)

Options

A. 2
B. 4
C. `(5)/(2)`
D. `(11)/(4)`

Answer

To find the area covered by the equation \([x - y] = -3\) with the coordinate axes, we will follow these steps:

### Step 1: Understand the Greatest Integer Function
The notation \([x - y] = -3\) means that the value of \(x - y\) lies in the interval \([-3, -2)\). This can be expressed as:
\[ -3 \leq x - y < -2 \]

### Step 2: Set Up the Inequalities
From the inequality \(x - y \geq -3\), we can rearrange it to find:
\[ x - y = -3 \implies y = x + 3 \]
And from the inequality \(x - y < -2\), we can rearrange it to find:
\[ x - y = -2 \implies y = x + 2 \]

### Step 3: Identify the Lines
We now have two lines:
1. \(y = x + 3\)
2. \(y = x + 2\)

### Step 4: Find Intersections with the Axes
Next, we will find where these lines intersect the coordinate axes.

**For \(y = x + 3\):**
- To find the x-intercept (set \(y = 0\)):
\[ 0 = x + 3 \implies x = -3 \quad \text{(Point: (-3, 0))} \]
- To find the y-intercept (set \(x = 0\)):
\[ y = 0 + 3 \implies y = 3 \quad \text{(Point: (0, 3))} \]

**For \(y = x + 2\):**
- To find the x-intercept (set \(y = 0\)):
\[ 0 = x + 2 \implies x = -2 \quad \text{(Point: (-2, 0))} \]
- To find the y-intercept (set \(x = 0\)):
\[ y = 0 + 2 \implies y = 2 \quad \text{(Point: (0, 2))} \]

### Step 5: Determine the Area of the Trapezium
The area we need to calculate is the area of the trapezium formed by the points \((-3, 0)\), \((-2, 0)\), \((0, 3)\), and \((0, 2)\).

The bases of the trapezium are the lengths of the segments on the y-axis:
- Length of the top base (from \(y = 3\) to \(y = 2\)): \(3 - 2 = 1\)
- Length of the bottom base (from \(y = 0\) to \(y = 0\)): \(0 - 0 = 0\)

The height of the trapezium is the distance between the two lines, which is the horizontal distance between \(x = -3\) and \(x = -2\):
- Height = \(-2 - (-3) = 1\)

### Step 6: Calculate the Area
Using the formula for the area of a trapezium:
\[ \text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height} \]
Substituting the values:
\[ \text{Area} = \frac{1}{2} \times (3 + 2) \times 1 = \frac{1}{2} \times 5 \times 1 = \frac{5}{2} \]

Thus, the area covered by \([x - y] = -3\) with the coordinate axes is:
\[ \boxed{\frac{5}{2}} \text{ square units} \]