The area bounded by the curves y = |x| - 1 and `y = -|x| +1` is equal to

Options

A. (a) 1 sq. unit
B. (b) 2 sq. units
C. (c) `2sqrt2 ` sq. unit
D. (d) 4 sq. unit

Answer

To find the area bounded by the curves \( y = |x| - 1 \) and \( y = -|x| + 1 \), we can follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ y = |x| - 1 \]
This can be expressed as:
- For \( x \geq 0 \): \( y = x - 1 \)
- For \( x < 0 \): \( y = -x - 1 \)

The second curve is:
\[ y = -|x| + 1 \]
This can be expressed as:
- For \( x \geq 0 \): \( y = -x + 1 \)
- For \( x < 0 \): \( y = x + 1 \)

### Step 2: Find points of intersection
To find the points where the curves intersect, we set the equations equal to each other.

For \( x \geq 0 \):
\[ x - 1 = -x + 1 \]
Solving this gives:
\[ 2x = 2 \implies x = 1 \]
Substituting \( x = 1 \) back into either equation gives \( y = 0 \). So one point of intersection is \( (1, 0) \).

For \( x < 0 \):
\[ -x - 1 = x + 1 \]
Solving this gives:
\[ -2x = 2 \implies x = -1 \]
Substituting \( x = -1 \) back into either equation gives \( y = 0 \). So another point of intersection is \( (-1, 0) \).

### Step 3: Sketch the curves
Plotting the curves \( y = |x| - 1 \) and \( y = -|x| + 1 \) will show that they intersect at the points \( (1, 0) \) and \( (-1, 0) \). The region bounded by these curves forms a shape that can be analyzed further.

### Step 4: Calculate the area
The area can be calculated by integrating the difference between the upper curve and the lower curve from \( x = -1 \) to \( x = 1 \).

The upper curve in this region is \( y = -|x| + 1 \) and the lower curve is \( y = |x| - 1 \).

Thus, the area \( A \) is given by:
\[ A = \int_{-1}^{1} \left((-|x| + 1) - (|x| - 1)\right) \, dx \]
This simplifies to:
\[ A = \int_{-1}^{1} (2 - 2|x|) \, dx \]

### Step 5: Evaluate the integral
We can break this integral into two parts:
\[ A = \int_{-1}^{0} (2 - 2(-x)) \, dx + \int_{0}^{1} (2 - 2x) \, dx \]
Calculating each part:
1. For \( x \in [-1, 0] \):
\[ A_1 = \int_{-1}^{0} (2 + 2x) \, dx = \left[2x + x^2\right]_{-1}^{0} = (0 + 0) - (-2 + 1) = 1 \]

2. For \( x \in [0, 1] \):
\[ A_2 = \int_{0}^{1} (2 - 2x) \, dx = \left[2x - x^2\right]_{0}^{1} = (2 - 1) - (0) = 1 \]

Thus, the total area is:
\[ A = A_1 + A_2 = 1 + 1 = 2 \]

### Conclusion
The area bounded by the curves \( y = |x| - 1 \) and \( y = -|x| + 1 \) is \( 2 \) square units.