If the area bounded by the corve `y=x^(2)+1, y=x` and the pair of lines `x^2+y^2+2xy-4x-4y+3=0` is K units, then the area of the region bounded by the curve `y=x^2+1,y=sqrt(x-1)` and the pair of lines `(x+y-1)(x+y-3)=0` is

Options

A. (a)K
B. (b)2K
C. (c)`K/2`
D. (d)None of these

Answer

To solve the problem, we need to find the area of the region bounded by the curves and lines specified in the question. Let's break it down step by step.

### Step 1: Identify the curves and lines

We have two curves:
1. \( y = x^2 + 1 \)
2. \( y = \sqrt{x - 1} \)

And two lines:
1. \( x + y - 1 = 0 \) (which can be rewritten as \( y = 1 - x \))
2. \( x + y - 3 = 0 \) (which can be rewritten as \( y = 3 - x \))

### Step 2: Find the intersection points of the curves and lines

**For the first curve \( y = x^2 + 1 \) and the line \( y = 1 - x \):**
Set them equal to find the intersection:
\[ x^2 + 1 = 1 - x \]
\[ x^2 + x = 0 \]
Factoring gives:
\[ x(x + 1) = 0 \]
Thus, \( x = 0 \) or \( x = -1 \).

**For the second curve \( y = x^2 + 1 \) and the line \( y = 3 - x \):**
Set them equal:
\[ x^2 + 1 = 3 - x \]
\[ x^2 + x - 2 = 0 \]
Factoring gives:
\[ (x - 1)(x + 2) = 0 \]
Thus, \( x = 1 \) or \( x = -2 \).

**For the curve \( y = \sqrt{x - 1} \) and the line \( y = 1 - x \):**
Set them equal:
\[ \sqrt{x - 1} = 1 - x \]
Squaring both sides:
\[ x - 1 = (1 - x)^2 \]
Expanding gives:
\[ x - 1 = 1 - 2x + x^2 \]
Rearranging gives:
\[ x^2 - 3x + 2 = 0 \]
Factoring gives:
\[ (x - 1)(x - 2) = 0 \]
Thus, \( x = 1 \) or \( x = 2 \).

**For the curve \( y = \sqrt{x - 1} \) and the line \( y = 3 - x \):**
Set them equal:
\[ \sqrt{x - 1} = 3 - x \]
Squaring both sides:
\[ x - 1 = (3 - x)^2 \]
Expanding gives:
\[ x - 1 = 9 - 6x + x^2 \]
Rearranging gives:
\[ x^2 - 7x + 10 = 0 \]
Factoring gives:
\[ (x - 2)(x - 5) = 0 \]
Thus, \( x = 2 \) or \( x = 5 \).

### Step 3: Determine the area bounded by the curves and lines

The area can be calculated as follows:
1. The area between \( y = x^2 + 1 \) and \( y = 1 - x \) from \( x = -1 \) to \( x = 0 \).
2. The area between \( y = x^2 + 1 \) and \( y = 3 - x \) from \( x = 1 \) to \( x = 2 \).
3. The area between \( y = \sqrt{x - 1} \) and the lines from \( x = 1 \) to \( x = 2 \) and from \( x = 2 \) to \( x = 5 \).

### Step 4: Calculate the areas

1. **Area between \( y = x^2 + 1 \) and \( y = 1 - x \)** from \( x = -1 \) to \( x = 0 \):
\[ \text{Area}_1 = \int_{-1}^{0} ((1 - x) - (x^2 + 1)) \, dx = \int_{-1}^{0} (-x^2 - x) \, dx \]

2. **Area between \( y = x^2 + 1 \) and \( y = 3 - x \)** from \( x = 1 \) to \( x = 2 \):
\[ \text{Area}_2 = \int_{1}^{2} ((3 - x) - (x^2 + 1)) \, dx = \int_{1}^{2} (2 - x - x^2) \, dx \]

3. **Area between \( y = \sqrt{x - 1} \) and the lines** from \( x = 1 \) to \( x = 2 \) and from \( x = 2 \) to \( x = 5 \):
\[ \text{Area}_3 = \int_{1}^{2} ((3 - x) - \sqrt{x - 1}) \, dx + \int_{2}^{5} ((3 - x) - \sqrt{x - 1}) \, dx \]

### Step 5: Combine the areas

Finally, sum the areas calculated in the previous steps to find the total area.

### Final Result

The area of the region bounded by the curves and lines is \( 2K \).